Solution: The given system of equations may be written as follows: $\begin{array}{l} \frac{a x}{b}-\frac{b y}{a}-(a+b)=0\dots \dots(i) \\ a x-b y-2 a b=0\dots \dots(ii) \end{array}$ Here,...

### If 1 is added to both of the numerator and denominator of a fraction, it becomes . If however, 5 is subtracted from both numerator and denominator, the fraction becomes . Find the fraction.

Solution: Suppose the required fraction be $\frac{x}{y}$. Therefore, we have: $\begin{array}{l} \frac{x+1}{y+1}=\frac{4}{5} \\ \Rightarrow 5(x+1)=4(y+1) \\ \Rightarrow 5 x+5=4 y+4 \\ \Rightarrow 5...

### Solve for and

Solution: We have: $\frac{35}{x+y}+\frac{14}{x-y}=19$ and $\frac{14}{x+y}+\frac{35}{x-y}=37$ Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$. $\begin{array}{l} 35 u+14 v-19=0\dots (i) \\ 14 u+35...

### Find the angles of a cyclic quadrilateral in which , and

Solution: As it is given: In a cyclic quadrilateral $\mathrm{ABCD}$, we have: $\begin{array}{l} \angle \mathrm{A}=(4 \mathrm{x}+20)^{0} \\ \angle \mathrm{B}=(3 \mathrm{x}-5)^{0} \\ \angle...

### 5 pencils and 7 pens together cost Rs 195 while 7 pencils and 5 pens together cost Rs 153 . Find the cost of each one of the pencil and pen.

Solution: Suppose the cost of each pencil be Rs. $x$ and that of each pen be Rs. $y$. Therefore, we have: $\begin{array}{l} 5 x+7 y=195\dots (i) \\ 7 x+5 y=153\dots (ii) \end{array}$ Adding...

### In a , find the measure of each one of and .

Solution: Suppose $\angle \mathrm{A}=\mathrm{x}^{0}$ and $\angle \mathrm{B}=\mathrm{y}^{0}$ Therefore, $\angle \mathrm{C}=3 \angle \mathrm{B}=3 \mathrm{y}^{0}$ Now, we have: $\begin{array}{l} \angle...

### Find the value of for which the system of equations and has (i) a unique solution, (ii) no solution.

Solution: The given equations are: $\begin{array}{l} 3 x+y=1 \\ \Rightarrow 3 x+y-1=0\dots \dots (i) \\ k x+2 y=5 \\ \Rightarrow k x+2 y-5=0\dots \dots (ii) \end{array}$ The equations are of the...

### Solve: and

Solution: The given system of equations are as follows: $\begin{array}{l} 6 x+3 y=7 x y\dots (i) \\ 3 x+9 y=11 x y\dots (ii) \end{array}$ For eq.(i), we have: $\frac{6 x+3 y}{x y}=7$ $\Rightarrow...

### Solve: .

Solution: The given system of equations are as follows: $\begin{array}{l} 23 x+29 y=98\dots (i) \\ 29 x+23 y=110\dots (ii) \end{array}$ On adding equation(i) and equation(ii), we obtain:...

### The difference between two numbers is 26 and one number is three times the other. Find the numbers.

Solution: Suppose the larger number be $\mathrm{x}$ and the smaller number be $y$. Therefore, we have: $\begin{array}{l} x-y=26\dots (i) \\ x=3 y\dots (ii) \end{array}$ On substituting $\mathrm{x}=3...

### Show that the paths represented by the equations and are parallel.

Solution: The given system of equations is: $x-3 y-2=0$ and $-2 x+6 y-5=0$ Given equations are of the following form: $a_{1} x+b_{1 y}+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ Here, $a_{1}=1,...

### Solve the system of equations: .

Solution: The given system of equations are as follows: $\begin{array}{l} x-2 y=0\dots (i) \\ 3 x+4 y=20\dots (ii) \end{array}$ On multiplying equation(i) by 2 , we obtain: $2 x-4 y=0\dots (iii)$ On...

### Show that the equations have infinitely many solutions.

Solution: The given system of equations is: $9 x-10 y-21=0$ and $\frac{3 x}{2} x-\frac{5 y}{3}-\frac{7}{2}=0$ Given equations are of the following form: $a_{1} x+b_{1} x+c_{1}=0$ and $a_{2} x+b_{2}...

### For what values of is the system of equations inconsistent?

Solution: The given system of equations is: $\mathrm{kx}+3 \mathrm{y}-(\mathrm{k}-2)=0$ and $12 \mathrm{x}+\mathrm{ky}-\mathrm{k}=0$ Given equations are of the following form: $a_{1} x+b_{1}...

### Show that the system of equations and has a unique solution.

Solution: The given system of equations is as follows: $-x+2 y+2=0 \text { and } \frac{1}{2} x-\frac{1}{4} y-1=0$ Given equations are of the following form: $a_{1} x+b_{1} x+c_{1}=0$ and $a_{2}...

### If and , which of the following is wrong?

(a)

(b)

(c)

(d)

Answer: (d) $\frac{1}{x}-\frac{1}{y}=0$ Solution: It is given: $x=-y \text { and } y>0$ Now, we have: (i) $x^{2} y$ On substituting $x=-y$, we obtain: $(-y)^{2} y=y^{3}>0(\because y>0)$...

### The pair of equations has

(a) a unique solution

(b) two solutions

(c) no solution.

(d) infinitely many solutions

Answer: (a) a unique solution Solution: The given system can be written as: $2 x+y-5=0$ and $3 x+2 y-8=0$ Given equations are of the following form: $a_{1} x+b_{1 Y}+c_{1}=0$ and $a_{2} x+b_{2}...

### If and have an infinite number of solutions, then

(a)

(b)

(c)

(d)

Answer: $($ d) $a=-5, b=-1$ Solution: The given system can be written as: $2 x$ 3y $7-0$ and $(a+b) x \quad(a+b \quad 3) y \quad(4 a+b)=0$ Given equations are of the following form: $a_{1} x+b_{1}...

### Find the value of for which the system of equations and has a unique solution.

Solution: The given system of equations is $\begin{array}{ll} \mathrm{kx}-\mathrm{y}-2=0 \ldots \ldots(\mathrm{i}) \\ 6 \mathrm{x}-2 \mathrm{y}-3=0 \ldots \cdots (ii) \end{array}$ Here,...

### The sum of the digits of a two digit number is The number obtained by interchanging the digits exceeds the given number by The number is

(a) 96

(b) 69

(c) 87

(d) 78

Answer: (a) 96 Solution: Suppose the tens and the units digits of the required no. be $\mathrm{x}$ and $\mathrm{y}$, respectively. The required number $=(10 x+y)$ As per the question, we have:...

### The graphs of the equations and are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each other

Answer: (a) coincident Solution: The correct option is (a). The given system of equations can be written as follows: $5 x-15 y-8=0$ and $3 x-9 y-\frac{24}{5}=0$ Given equations are of the following...

### The graphs of the equations and are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each other

Answer: Solution: The given system of equations are as follows: $2 x+3 y-2=0$ and $x-2 y-8=0$ They are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ Here, $a_{1}=2,...

### The graphs of the equations 6x – 2y + 9 = 0 and 3x – y + 12 = 0 are two lines which are

(a) coincident

(b) parallel

(c) intersecting exactly at one point

(d) perpendicular to each other

Answer: (b) parallel Solution: The given system of equations are as follows: $6 x-2 y+9=0$ and $3 x-y+12=0$ They are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$...

The correct answer is: (a) / (b)/ (c)/ (d).

Answer: (c) Solution: The correct answer is option (C). It is clear that, Reason (R) is false. Upon solving $x+y=8$ and $x-y=2$, we obtain: $x=5$ and $y=3$ Therefore, the given system has a unique...

### 5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is

(a) 45 years

(b) 50 years

(c) 47 years

(d) 40 years

Answer: (d) 40 years Solution: Suppose the present age of the man be $\mathrm{x}$ years. And his son's present age be $y$ years. 5 years later: $\begin{array}{l} (x+5)=3(y+5) \\ \Rightarrow x+5=3...

### In a cyclic quadrilateral , it is being given that , and Then,

(a)

(b)

(c)

(d)

Answer: (b) $80^{\circ}$ Solution: Correct option is (b). In a cyclic quadrilateral $\mathrm{ABCD}$ : $\begin{array}{l} \angle A=(x+y+10)^{0} \\ \angle B=(y+20)^{0} \\ \angle C=(x+y-30)^{0} \\...

### In a , then

(a)

(b)

(c)

(d)

Answer: (b) $40^{0}$ Solution: Suppose $\angle \mathrm{A}=\mathrm{x}^{0}$ and $\angle \mathrm{B}=\mathrm{y}^{0}$ $\therefore \angle \mathrm{A}=3 \angle \mathrm{B}=(3 \mathrm{y})^{0}$ Now, $\angle...

### If a pair of linear equations is inconsistent, then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincident

Answer: (a) parallel Solution: If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their...

### If a pair of linear equations is consistent, then their graph lines will be

(a) parallel

(b) always coincident

(c) always intersecting

(d) intersecting or coincident

Answer: (d) intersecting or coincident Solution: If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincidence.

### The pair of equations and has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solution

Answer: (d) no solution Here, $a_{1}=3, b_{1}=2 k, c_{1}=-2, a_{2}=2, b_{2}=5$ and $c_{2}=1$ $\therefore \frac{a_{1}}{a_{4}}=\frac{3}{2}, \frac{b_{1}}{b_{L}}=\frac{2 k}{5}$ and...

### The pair of equations and has

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solution

Answer: (d) no solution Solution: We can write the given system of equations as: $x+2 y+5=0$ and $-3 x-6 y+1=0$ Given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2}...

### For what value of do the equations and represent two lines intersecting at a unique point?

(a)

(b)

(c)

(d) all real values except -6

Answer: (d) all real values except -6 Solution: We can write the given system of equations as follows: $\mathrm{kx}-2 y-3=0$ and $3 \mathrm{x}+\mathrm{y}-5=0$ Given equations are of the following...

### For the system of equations to have no solution, we must have:

(a)

(b)

(c)

(d)

Answer: (d) $\frac{15^{2}}{4}$ Solution: We can write the given system of equations as follows: $3 x+2 k y-2=0$ and $2 x+5 y+1=0$ Given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0$...

### The system and have no solution when?

(a)

(b)

(c)

(d)

Answer: (a) $\mathrm{k}=10$ We can write the given system of equations as follows: $x+2 y-3=0$ and $5 x+k y+7=0$ The given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2}...

### The system and have a unique solution only when ?

(a)

(b)

(c)

(d)

Answer: (b) $\mathrm{k} \neq-6$ Solution: The correct option is (b). We can write the given system of equations as follows: $x-2 y-3=0$ and $3 x+k y-1=0$ given equations are of the following form:...

### The system of and has a unique solution only when

(a)

(b)

(c)

(d)

Answer: (d) $k \neq 3$. Solution: The given system of equations are $\begin{array}{l} \mathrm{kx}-\mathrm{y}-2=0\dots \dots(i) \\ 6 \mathrm{x}-2 \mathrm{y}-3=0\dots \dots(ii) \end{array}$ Here,...

### If and then

(a)

(b)

(c)

(d)

Answer: (b) $x=\frac{2}{3}, y=1$ Solution: The given system of equations are $\begin{array}{l} \frac{2}{x}+\frac{3}{y}=6\dots (i) \\ \frac{1}{x}+\frac{1}{2 y}=2\dots (ii) \end{array}$ Multiplying...

### If then the value of is

(a)

(b)

(c) 0

(d) none of these

Answer: (e) 0 Solution: $\begin{array}{l} \because 2^{x+y}=2^{x-y}=\sqrt{8} \\ \therefore \mathrm{x}+\mathrm{y}=\mathrm{x}-\mathrm{y} \\ \Rightarrow \mathrm{y}=0 \end{array}$

### If and then

(a) x=1, y=2

(b) x=2, y=1

(c) x=3, y=2

(d) x=2, y=3

Answer: $(a) x=1, y=2$ Solution: The given system is $29 x+37 y=103\dots \dots(i)$ $37 x+29 y=95\dots \dots(ii)$ Adding equation(i) and equation(ii), we get $66 x+66 y=198$ $\Rightarrow x+y=3\dots...

### If and then

(a) x=2, y=3

(b) x=1, y=2

(c) x=3, y=4

(d) x=1, y=-1

Answer: (c) $x=3, y=4$ Solution: The given system of equations are $\begin{array}{l} 4 x+6 y=3 x y\dots (i) \\ 8 x+9 y=5 x y\dots (ii) \end{array}$ Dividing equation(i) and equation(ii) by $x y$, we...

### If and then

(a)

(b)

(c)

(d)

Answer: (b) $x=\frac{5}{2}, y=\frac{1}{2}$ Solution: The given system of equations are $\begin{array}{l} \frac{3}{x+y}+\frac{2}{x-y}=2\dots \dots(i) \\ \frac{9}{x+y}-\frac{4}{x-y}=1\dots \dots(ii)...

### If then

(a) x=1, y=1

(b) x=-1, y=-1

(c) x=1, y=2

(d) x=2, y=1

Answer: $($ a) $x=1, y=1$ Solution: Considering $\frac{2 x+y+2}{5}=\frac{3 x-y+1}{3}$ and $\frac{3 x-y+1}{3}=\frac{3 x+2 y+1}{3}$. Now, on simplifying these equations, we obtain $\begin{array}{l}...

### If and then

(a)

(b) x=-2, y=3

(c)

(d)

Answer: (d) $x=\frac{-1}{2}, y=\frac{1}{3}$ Solution: The given system is $\begin{array}{l} \frac{1}{x}+\frac{2}{y}=4\dots \dots(i) \\ \frac{3}{y}-\frac{1}{x}=11\dots \dots(ii) \end{array}$ Adding...

### If and then

(a) x=2, y=3

(b) x=-2, y=3

(c) x=2, y=-3

(d) x=-2, y=-3

Answer: $($ a) $x=2, y=3$ Solution: The given system is $\begin{array}{l} \frac{2 x}{3}-\frac{y}{2}=-\frac{1}{6}\dots \dots(i) \\ \frac{x}{2}+\frac{2 y}{3}=3\dots \dots(ii) \end{array}$ Multiplying...

### If and then

(a) x=4, y=2

(b) x=5, y=3

(c) x=6, y=4

(d) x=7, y=5

Answer: (c) $x=6, y=4$ Solution: The given system is $\begin{array}{l} x-y=2\dots (i) \\ x+y=10\dots (ii) \end{array}$ Adding equation(i) and equation(ii), we get $2 x=12 \Rightarrow x=6$ Now,...

### If and then

(a) x=2, y=3

(b) x=2, y=-3

(c) x=3, y=2

(d) x=3, y=-2

Answer: (c) $x=3, y=2$ Solution: The given system is $\begin{array}{l} 2 x+3 y=12\dots \dots(i) \\ 3 x-2 y=5\dots \dots(ii) \end{array}$ Multiplying equation(i) by 2 and equation(ii) by 3 and then...

### Solve for and

Solution: The given system is $\begin{array}{l} \frac{3}{x+y}+\frac{2}{x-y}=2\dots\dots(i) \\ \frac{9}{x+y}-\frac{4}{x-y}=1\dots \dots(ii) \end{array}$ Substituting $\frac{1}{x+y}=\mathrm{u}$ and...

### Find the value of for which the system of equations and is inconsistent.

Solution: The given system of equations is $\begin{array}{l} \mathrm{x}+2 \mathrm{y}-3=0\dots \dots(i) \\ 5 \mathrm{x}+\mathrm{ky}+7=0\dots \dots(ii) \end{array}$ Here, $a_{1}=1, b_{1}=2, c_{1}=-3,...

### Show that the system and has no solution.

Solution: The given system of equations is $\begin{array}{ll} 2 \mathrm{x}+3 \mathrm{y}-1=0 \ldots \ldots(\mathrm{i}) \\ 4 \mathrm{x}+6 \mathrm{y}-4=0 \ldots \ldots \text { (ii) } \end{array}$ Here,...

### Find the value of for which the system of equations and has infinite number of solutions.

Solution: The given system of equations is $\begin{array}{ll} 2 \mathrm{x}+3 \mathrm{y}-5=0\dots \dots(i) \\ 4 \mathrm{x}+\mathrm{ky}-10=0 \ldots \ldots(\mathrm{ii}) \\ \end{array}$ Here, $a_{1}=2,...

### Find the value of k for which the system of equations and has infinite nonzero solutions.

Solution: The given pair of equations is $\begin{array}{l} 3 x+5 y=0\dots \dots (i) \\ k x+10 y=0\dots \dots(ii) \end{array}$ This is a homogeneous system of linear differential equation, therefore...

### If 12x+17y=53 and 17x+12y=63 then find the value of

Solution: Given pair of equations is $\begin{array}{l} 12 x+17 y=53\dots \dots(i) \\ 17 x+12 y=63\dots \dots(ii) \end{array}$ Adding equation(i) and equation(ii), we obtain $29 x+29 y=116$...

### If and , then find the value of .

Solution: Given pair of equations is $\begin{array}{l} \frac{x}{4}+\frac{y}{3}=\frac{5}{12}\dots \dots(i) \\ \frac{x}{2}+y=1\dots \dots(ii) \end{array}$ Multiplying equation(i) by 12 and...

### If and , find the values of and .

Solution: Given pair of equation is $\begin{array}{l} \frac{2}{x}+\frac{3}{y}=\frac{9}{x y}\dots \dots(i) \\ \frac{4}{x}+\frac{9}{y}=\frac{21}{x y}\dots \dots(ii) \end{array}$ Multiplying...

### A man has some hens and cows. If the number of heads be 48 and number of feet by 140 . How many cows are there.

Solution: Suppose the no. of hens and cow be $x$ and $y$ respectively. According to the question $\begin{array}{l} x+y=48\dots \dots(i) \\ 2 x+4 y=140 \\ x+2 y=70\dots \dots(ii) \end{array}$...

### A number consists of two digits whose sum is 10. If 18 is subtracted form the number, its digits are reversed. Find the number.

Solution: Suppose the ones digit and tens digit be $x$ and $y$ respectively. Then according the question $\begin{array}{l} \mathrm{x}+\mathrm{y}=10\dots \dots(i) \\ (10 \mathrm{y}+\mathrm{x})-18=10...

### The sum of two numbers is 80 . The larger number exceeds four times the smaller one by 5 . Find the numbers.

Solution: Suppose the larger number be $\mathrm{x}$ and the smaller number be $\mathrm{y}$. Then according to the question $\begin{array}{l} x+y=80\dots \dots(i) \\ x=4 y+5 \\ x-4 y=5\dots \dots(ii)...

### The difference of two numbers is 5 and the difference between their squares is 65 . Find the numbers.

Solution: Suppose that the numbers are $x$ and $y$, where $x>y$. Then according to the question $\begin{array}{l} x-y=5\dots \dots(i) \\ x^{2}-y^{2}=65\dots \dots(ii) \end{array}$ Dividing...

### Find the values of k for which the system of equations , has a unique solution.

Solution: The given equations are $\begin{array}{l} 3 x+k y=0\dots \dots(i) \\ 2 x-y=0\dots \dots(ii) \end{array}$ Which is of the form $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$, where...

### Write the number of solutions of the following pair of linear equations:

,

Solution: The given equations are $\begin{array}{l} \mathrm{x}+3 \mathrm{y}-4=0 \text {......(i) } \\ 2 \mathrm{x}+6 \mathrm{y}-7=0 \ldots \ldots \text { (ii) } \end{array}$ Which is of the form...

### Find the value of k for which the system of linear equations has an infinite number of solutions.

Solution: The given equations are $\begin{array}{ll} 2 \mathrm{x}+3 \mathrm{y}-9=0\dots \dots(i) \\ 6 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}-(3 \mathrm{k}-2)=0\dots \dots(ii) \end{array}$ Which is of...

### Find the value of k for which the system of linear equations has an infinite number of solutions.

Solution: The given equations are $\begin{array}{l} 10 x+5 y-(k-5)=0\dots \dots(i) \\ 20 x+10 y-k=0 \ldots \ldots(i i) \end{array}$ Which is of the form $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2}...

### Find the value of k for which the system of linear equations has an infinite number of solutions.

Solution: Given equations are $\begin{array}{ll} 2 \mathrm{x}+3 \mathrm{y}-7=0\ldots \ldots(\mathrm{i}) \\ (\mathrm{k}-1) \mathrm{x}+(\mathrm{k}+2) \mathrm{y}-3 \mathrm{k}=0 & \ldots...

### Write the number of solutions of the following pair of linear equations:

Solution: The equations given are $\begin{array}{ll} \mathrm{x}+2 \mathrm{y}-8=0\dots \dots(i) \\ 2 \mathrm{x}+4 \mathrm{y}-16=0 \ldots \ldots \text { (ii) } \\ \end{array}$ Which is of the form...

### In a cyclic quadrilateral , it is given and . Find the four angles.

Solution: Opposite angles of cyclic quadrilateral are supplementary, therefore $\begin{array}{l} \angle \mathrm{A}+\angle \mathrm{C}=180^{\circ} \\ \Rightarrow(2 \mathrm{x}+4)^{0}+(2...

### In a and . Find the there angles.

Solution: $\begin{array}{l} \because \angle \mathrm{C}-\angle \mathrm{B}=9^{0} \\ \therefore \mathrm{y}^{0}-(3 \mathrm{x}-2)^{0}-9^{0} \\ \Rightarrow \mathrm{y}^{0}-3 \mathrm{x}^{0}+2^{0}=9^{0} \\...

### The larger of the two supplementary angles exceeds the smaller by . Find them

Solution: Suppose that $x$ and $y$ be the supplementary angles, where $x>y$ According to the given condition $x+y=180^{0}\dots \dots(i)$ And $x-y=18^{0}\dots \dots(ii)$ Adding equation(i) and...

### 90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution. Find the quantity of each type of acid to be mixed to form the mixture.

Solution: Suppose that $x$ litres and $y$ litres be respectively the amount of $90 \%$ and $97 \%$ pure acid solutions. According to the given condition $\begin{array}{l} 0.90 x+0.97 y=21 \times...

### A jeweler has bars of 18 -carat gold and 12 -carat gold. How much of each must be melted together to obtain a bar of 16 -carat gold, weighing ? (Given: Pure gold is 24 -carat).

Solution: Let ussuppose that $x \mathrm{~g}$ and y $\mathrm{g}$ be the weight of 18 -carat and 12 - carat gold respectively. According to the given condition $\begin{array}{l} \frac{18...

### A chemist has one solution containing acid and a second one containing acid. How much of each should be used to make 10 litres of a acid solution?

Solution: Suppose that $x$ litres and y litres be the amount of acids from $50 \%$ and $25 \%$ acid solutions respectively. According to the question $50 \%$ of $x+25 \%$ of $y=40 \%$ of 10...

### A lending library has fixed charge for the first three days and an additional charge for each day thereafter. Mona paid for a book kept for 7 days, while Tanvy paid for the book she kept for 5 days find the fixed charge and the charge for each extra day.

Solution: Suppose that the fixed charge be Rs.$x$ and the charge for each extra day be Rs.$y$. In case of Mona, according to the question $x+4 y=27\dots \dots(i)$ In case of Tanvy, according to the...

### On selling a tea-set at loss and a lemon-set at gain, a shopkeeper gains Rs. 7 . However, if he sells the tea-set at gain and the lemon-set at gain, he gains Rs. 14 . Find the price of the tea-set and that of the lemon-set paid by the shopkeeper.

Solution: Suppose that the actual price of the tea and lemon set be Rs.$x$ and Rs.$y$ respectively. When gain is Rs.7, then $\begin{array}{l} \frac{y}{100} \times 15-\frac{x}{100} \times 5=7 \\...

### The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.

Solution: Suppose the woman's present age be $\mathrm{x}$ years. and her daughter's present age be $y$ years. Therefore, we have: $\begin{array}{l} x=3 y+3 \\ \Rightarrow x-3 y=3\dots \dots(i)...

### If twice the son’s age in years is added to the mother’s age, the sum is 70 years. But, if twice the mother’s age is added to the son’s age, the sum is 95 years. Find the age of the mother and that of the son.

Solution: Suppose the mother's present age be $\mathrm{x}$ years. and her son's present age be $y$ years. Therefore, we have: $x+2 y=70\dots \dots(i)$ And, $2 x+y=95\dots \dots(ii)$ On multiplying...

### The present age of a man is 2 years more than five times the age of his son. Two years hence, the man’s age will be 8 years more than three times the age of his son. Find their present ages.

Solution: Suppose that the man's present age be $x$ years. and his son's present age be $y$ years. As per the question, we have: Two years ago: Age of the man $=$ Five times the age of the son...

### Five years hence, a man’s age will be three times the sum of the ages of his son. Five years ago, the man was seven times as old as his son. Find their present ages.

Solution: Let us suppose the present age of the man be $\mathrm{x}$ years and that of his son be $y$ years. After 5 years man's age $=x+5$ After 5 years ago son's age $=y+5$ According to the...

### A railway half ticket costs half the full fare and the reservation charge is the some on half ticket as on full ticket. One reserved first class ticket from Mumbai to Delhi costs โน4150 while one full and one half reserved first class ticket cost โน 6255. What is the basic first class full fare and what is the reservation charge?

Solution: Let us suppose the basic first class full fare be Rs.$x$ and the reservation charge be Rs.$y$. Case 1: One reservation first class full ticket cost Rs.4,150 $x+y=4150\dots \dots(i)$ Case...

### The area of a rectangle gets reduced by 67 square meters, when its length is increased by and the breadth is decreased by . If the length is reduced by and breadth is increased by , the area is increased by 89 square meters, Find the dimension of the rectangle.

Solution: Let us suppose the length and the breadth of the rectangle be $x \mathrm{~m}$ and $\mathrm{y} \mathrm{m}$, respectively. Case 1: When length is increased by $3 \mathrm{~m}$ and the breadth...

### The area of a rectangle gets reduced by , when its length is rcduccd by and its breadth is increased by . If we increase the length by and breadth by , the area is increased by . Find the length and the breadth of the rectangle.

Solution: Let us suppose the length and the breadth of the rectangle be $\mathrm{x} \mathrm{m}$ and $\mathrm{y} \mathrm{m}$, respectively. $\therefore$ Area of the rectangle $=(x y)$ sq.m Case 1:...

### The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 meters and the breadth is decreased by 2 meters, the area remains the same. Find the length and the breadth of the room.

Solution: Let us suppose the length of the room be $\mathrm{x}$ meters and he breadth of the room be y meters. Therefore, we have: Area of the room $=x y$ As per the question, we have: $x=y+3$...

### 2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.

Solution: Let one man alone can finish the work in $\mathrm{x}$ days and one boy alone can finish it in $y$ days. $\therefore$ One man's one day's work $=\frac{1}{x}$ And, one boy's one day's work...

### A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Solution: Let us suppose the speed of the boat in still water be $\mathrm{x} \mathrm{km} / \mathrm{h}$ and the speed of the stream be y $\mathrm{km} / \mathrm{h}$. Therefore we have Speed upstream...

### A sailor goes 8km downstream in 420 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current .

Solution: Let us suppose the speed of the sailor in still water be $\mathrm{x} \mathrm{km} / \mathrm{h}$ and that of the current $\mathrm{y} \mathrm{km} / \mathrm{h}$. Speed downstream...

### Places A and B are 160 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.

Solution: Let us suppose the speed of the car A and B be x $\mathrm{km} / \mathrm{h}$ and y $\mathrm{km} / \mathrm{h}$ respectively. Let $\mathrm{x}>\mathrm{y}$. Case-1: When they travel in the...

### Abdul travelled 300km by train and 200km by taxi taking 5 hours and 30 minutes. But, if he travels 260 km by train and 240 km by he takes 6 minutes longer. Find the speed A taxi, of the train and that of taxi.

Solution: Let us suppose that the speed of the train and taxi be $\mathrm{x} \mathrm{km} / \mathrm{h}$ and $\mathrm{y} \mathrm{km} / \mathrm{h}$ respectively. Then according to the question...

### A train covered a certain distance at a uniform speed. If the train had been faster, it would have taken 3 hours less than the scheduled time. And, if the train were slower by 4 , it would have taken 3 hours more than the scheduled time. Find the length of the journey.

Solution: Let us suppose that the original speed be $\mathrm{x} \mathrm{kmph}$ and let the time taken to complete the journey be $y$ hours. $\therefore$ Length of the whole journey $=(\mathrm{xy})...

### Points A and B are apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.

Solution: Let us suppose that $\mathrm{X}$ and $\mathrm{Y}$ be the cars starting from points $\mathrm{A}$ and $\mathrm{B}$, respectively and let their speeds be $\mathrm{x}$ $\mathrm{km} /...

### A man sold a chair and a table together for Rs. 1520 , thereby making a profit of on chair and on table. By selling them together for Rs. 1535, he would have made a profit of on the chair and on the table. Find the cost price of each.

Solution: Let us suppose the cost price of the chair and table be Rs.x and Rs.y respectively. Then according to the question The selling price of chair $+$ Selling price of table $=1520$...

### The monthly incomes of A and B are in the ratio of and their monthly expenditures are in the ratio of . If each saves Rs. 9000 per month, find the monthly income of each.

Solution: Let us suppose the monthly income of $\mathrm{A}$ and $\mathrm{B}$ are Rs.x and Rs.y respectively. Therefore $\begin{array}{l} \frac{x}{y}=\frac{5}{4} \\ \Rightarrow y=\frac{4 x}{5}\dots...

### A man invested an amount at per annum simple interest and another amount at per annum simple interest. He received an annual interest of Rs. 1350 . But, if he had interchanged the amounts invested, he would have received Rs. 45 less. What amounts did he invest at different rates?

Solution: Let us suppose that the amounts invested at $10 \%$ and $8 \%$ be Rs.x and Rs.y respectively. Therefore $\begin{array}{l} \frac{x \times 10 \times 1}{100}=\frac{y \times 8 \times...

### A part of monthly hostel charges in a college are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs. 4550 as hostel charges whereas a student , who takes food for 30 days, pays Rs. 5200 as hostel charges. Find the fixed charges and the cost of the food per day.

Solution: Let us suppose the fixed charges be Rs.x and the cost of food per day be Rs.y. Therefore $\begin{array}{l} x+25 y=4500\dots \dots(i) \\ x+30 y=5200\dots \dots(ii) \end{array}$ Subtracting...

### Taxi charges in a city consist of fixed charges per day and the remaining depending upon the distance travelled in kilometers. If a person travels , he pays Rs. 1330 , and for travelling , he pays Rs. 1490 . Find the fixed charges per day and the rate per km.

Solution: Let us suppose the fixed charges be Rs.x and rate per km be Rs.y. Therefore $\begin{array}{l} x+80 y=1330\dots \dots(i) \\ x+90 y=1490\dots \dots(ii) \end{array}$ Subtracting equation(i)...

### There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.

Solution: Let us suppose the no. of students in classroom $\mathrm{A}$ be $\mathrm{x}$ Let's suppose the no. of students in classroom B be y. If 10 students are transferred from A to B, then we...

### The sum of two numbers is 16 and the sum of their reciprocals is . Find the numbers.

Solution: Let us supposse the larger number be $\mathrm{x}$ and the smaller number be y. Therefore, we have: $x+y=16\dots \dots(i)$ And, $\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\dots \dots(ii)$...

### The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3 . They are in the ratio of . Determine the fraction.

Solution: Let us suppose the required fraction be $\frac{x}{y}$. According to the question $\begin{array}{l} x+y=4+2 x \\ \Rightarrow y-x=4\dots \dots(i) \end{array}$ After changing the numerator...

### Find a fraction which becomes when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes when 7 is subtracted from the numerator and 2 is subtracted from the denominator.

Solution: Let us suppose the required fraction be $\frac{x}{y}$. Therefore, we have: $\begin{array}{l} \frac{x-1}{y+2}=\frac{1}{2} \\ \Rightarrow 2(x-1)=1(y+2) \\ \Rightarrow 2 x-2=y+2 \\...

### The denominator of a fraction is greater than its numerator by 11 . If 8 is added to both its numerator and denominator, it becomes . Find the fraction.

Solution: Let us suppose the required fraction be $\frac{x}{y}$. Therefore, we have: $\begin{array}{l} y=x+11 \\ \Rightarrow y-x=11\dots \dots(i) \end{array}$ Again, $\frac{x+8}{y+8}=\frac{3}{4}$...

### If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the denominator, it reduces to . Find the fraction.

Solution: Let us suppose the required fraction be $\frac{x}{y}$. Therefore, we have: $\begin{array}{l} \frac{x+2}{y}=\frac{1}{2} \\ \Rightarrow 2(x+2)=y \\ \Rightarrow 2 x+4=y \\ \Rightarrow 2...

### The sum of the numerator and denominator of a fraction is 8 . If 3 is added to both of the numerator and the denominator, the fraction becomes . Find the fraction.

Solution: Let's suppose the required fraction be $\frac{x}{y}$. Therefore, we have: $x+y=8\dots \dots(i)$ And, $\frac{x+3}{y+3}=\frac{3}{4}$ $\begin{array}{l} \Rightarrow 4(x+3)=3(y+3) \\...

### The sum of a two-digit number and the number obtained by reversing the order of its digits is 121 , and the two digits differ by 3 . Find the number.

Solution: Let us say that $\mathrm{x}$ be the ones digit and $\mathrm{y}$ be the tens digit. Therefore Two digit number before reversing $=10 \mathrm{y}+\mathrm{x}$ Two digit number after reversing...

### A two-digit number is such that the product of its digits is 18 . When 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution: Let us suppose the tens and the units digits of the required number be $\mathrm{x}$ and $\mathrm{y}$, respectively. Therefore, we have: $xy=18\dots \dots(i)$ The required number $=(10...

### A two-digit number is such that the product of its digits is 35 . If 18 is added to the number, the digits interchange their places. Find the number.

Solution: Let us suppose the tens and the units digits of the required no. be $\mathrm{x}$ and $\mathrm{y}$, respectively. Therefore, we have: $x y=35\dots \dots(i)$ Required number $=(10...

### A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.

Solution: It is known that: Dividend $=$ Divisor $\times$ Quotient $+$ Remainder Let the tens and the units digits of the required no. be $\mathrm{x}$ and $\mathrm{y}$, respectively. Required number...

### A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution: Let's suppose the tens and the units digits of the required number be $\mathrm{x}$ and $\mathrm{y}$, respectively. Required number $=(10 x+y)$ $\begin{array}{l} 10 x+y=4(x+y)+3 \\...

### The sum of the digits of a two-digit number is 15 . The number obtained by interchanging the digits exceeds the given number by Find the number.

Solution: Let's suppose that the tens and the units digits of the required number be $\mathrm{x}$ and $\mathrm{y}$, respectively. Required number $=(10 \mathrm{x}+\mathrm{y})$ $x+y=15\dots \dots(i)$...

### A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

Solution: Let us suppose that the tens and the units digits of the required number be $\mathrm{x}$ and $\mathrm{y}$, respectively. Required number $=(10 \mathrm{x}+\mathrm{y})$ $10 x+y=7(x+y)$ $10...

### The sum of the digits of a two-digit number is 12 . The number obtained by interchanging its digits exceeds the given number by 18 . Find the number.

Solution: Let us suppose that the tens and the units digits of the required no. be $\mathrm{x}$ and $\mathrm{y}$, respectively. Required number $=(10 \mathrm{x}+\mathrm{y})$ $x+y=12\dots \dots(i)$...

### The difference between two numbers is 14 and the difference between their squares is 448 . Find the numbers.

Solution: Let us suppose that the larger number be $\mathrm{x}$ and the smaller number be $\mathrm{y}$. Therefore, we have: $\begin{array}{l} x-y=14 \text { or } x=14+y\dots \dots(i) \\...

### If 2 is added to each of two given numbers, their ratio becomes 1: 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5: 11. Find the numbers.

Solution: Let us say that the required numbers be $\mathrm{x}$ and $\mathrm{y}$. So now, we have: $\frac{x+2}{y+2}=\frac{1}{2}$ By cross multiplication, we get: $\begin{array}{l} 2 x+4=y+2 \\...

### If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.

Solution: It is known that: Dividend $=$ Divisor $\times$ Quotient $+$ Remainder Let's assume the larger no. be $\mathrm{x}$ and the smaller be y. Then, we have: $\begin{array}{l} 3 x=y \times 4+8...

### If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.

Solution: Let's suppose that the greater number be $\mathrm{x}$ and the smaller number be y. Therefore, we have: $\begin{array}{l} 25 x-45=y \text { or } 2 x-y=45\dots (i) \\ 2 y-21=x \text { or...

### Find the numbers such that the sum of thrice the first and the second is 142 , and four times the first exceeds the second by 138 .

Solution: Let us assume that the first number be $\mathrm{x}$ and the second number be $\mathrm{y}$. Therefore, we have: $3 \mathrm{x}+\mathrm{y}=142\dots (i)$ $4 \mathrm{x}-\mathrm{y}=138\dots...

### Find two numbers such that the sum of twice the first and thrice the second is 92 , and four times the first exceeds seven times the second by 2 .

Solution: Let's suppose that the first number be $\mathrm{x}$ and the second number be $\mathrm{y}$. Therefore, we have: $\begin{array}{l} 2 x+3 y=92\dots (i) \\ 4 x-7 y=2\dots (ii) \end{array}$ On...

### The sum of two numbers is 137 and their differences are 43 . Find the numbers.

Solution: Let us assume that the larger number be $\mathrm{x}$ and the smaller number be $\mathrm{y}$. Then, we have: $\begin{array}{l} x+y=137\dots(i) \\ x-y=43 \dots(ii) \end{array}$ On adding...

### A lady has only 50 -paisa coins and 25 -paisa coins in her purse. If she has 50 coins in all totaling Rs.19.50, how many coins of each kind does she have?

Solution: Let's assume that $x$ and $y$ be the number of 50 -paisa and 25 -paisa coins respectively. Then $x+y=50\dots (i)$ $0.5 x+0.25 y=19.50\dots(ii)$ Multiplying equation(ii) by 2 and...

### 23 spoons and 17 forks cost Rs.1770, while 17 spoons and 23 forks cost Rs.1830. Find the cost of each spoon and that of a fork.

Solution: Let us assume that the cost of a spoon be Rs.x and that of a fork be Rs.y. Then $\begin{array}{l} 23 \mathrm{x}+17 \mathrm{y}=1770\dots \dots(i) \\ 17 \mathrm{x}+23 \mathrm{y}=1830\dots...

### 5 chairs and 4 tables together cost โน5600, while 4 chairs and 3 tables together cost โน 4340 . Find the cost of each chair and that of each table.

Solution: $\begin{array}{l} 5 x+4 y=5600\dots \dots(i) \\ 4 x+3 y=4340\dots \dots(ii) \end{array}$ On multiplying equation(i) by 3 and equation(ii) by 4 , we obtain $\begin{array}{l} 15 x-16...

### Find the value of k for which the system of equations has a non-zero solution.

Solution: Given system of equations: $\begin{array}{l} 5 x-3 y=0\dots \dots(i) \\ 2 x+k y=0\dots \dots(ii) \end{array}$ The given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0, a_{2}...

### Find the value of k for which the system of equations has no solution.

Solution: We can write the given system of equations as $\begin{array}{l} \mathrm{kx}+3 \mathrm{y}+3-\mathrm{k}=0 \dots \dots(i)\\ 12 \mathrm{x}+\mathrm{ky}-\mathrm{k}=0\dots \dots(ii) \end{array}$...

### Find the value of k for which the system of equations has no solution.

Solution: Given system of equations: $3 x-y-5=0\dots \dots(i)$ And, $6 \mathrm{x}-2 \mathrm{y}+\mathrm{k}=0\dots \dots(ii)$ The given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0,...

### Find the value of k for which the system of equations has no solution.

Solution: Given system of equations: $\begin{array}{l} \mathrm{kx}+3 \mathrm{y}=3 \\ \mathrm{kx}+3 \mathrm{y}-3=0\dots \dots(i) \\ 12 \mathrm{x}+\mathrm{ky}=6 \\ 12 \mathrm{x}+\mathrm{ky}-6=0\dots...

### Find the value of k for which the system of equations

, has a non-zero solution.

Solution: Given system of equations: $\begin{array}{l} 8 x+5 y=9 \\ 8 x+5 y-9=0\dots \dots(i) \\ k x+10 y=15 \\ k x+10 y-15=0\dots \dots(ii) \end{array}$ The given equations are of the following...

### Find the values of a and b for which the system of linear equations has an infinite number of solutions:

,

Solution: We can write the given system of equations as $\begin{array}{l} 2 \mathrm{x}+3 \mathrm{y}-7=0\dots \dots(i) \\ 2 \mathrm{ax}+(\mathrm{a}+\mathrm{b}) \mathrm{y}-28=0\dots \dots(ii)...

### Find the values of a and b for which the system of linear equations has an infinite number of solutions:

,

Solution: We can write the given system of equations as $\begin{array}{l} 2 x+3 y-7=0\dots \dots(i) \\ (a+b) x+(2 a-b) y-21=0\dots \dots(ii) \end{array}$ The given system is of the form: $a_{1}...

### Find the values of a and b for which the system of linear equations has an infinite number of solutions:

,

Solution: We can write the given system of equations as $2 x+3 y=7$ $\Rightarrow 2 x+3 y-7=0\dots \dots(i)$ and $(a+b+1) x-(a+2 b+2) y=4(a+b)+1$ $(a+b+1) x-(a+2 b+2) y-[4(a+b)+1]=0\dots \dots(i)$...

### Find the values of a and b for which the system of linear equations has an infinite number of solutions:

,

Solution: We can write the given system of equations as $\begin{array}{l} 2 x-3 y=7 \\ \Rightarrow 2 x-3 y-7=0\dots \dots(i) \end{array}$ and $(a+b) x-(a+b-3) y=4 a+b$ $\Rightarrow(a+b) x-(a+b-3)...

### Find the values of a and b for which the system of linear equations has an infinite number of solutions:

,

Solution: We can write the given system of equations as $\begin{array}{l} (2 a-1) x+3 y=5 \\ \Rightarrow(2 a-1) x+3 y-5=0\dots \dots(i) \end{array}$ and $3 x+(b-1) y=2$ $\Rightarrow 3...

### Find the values of a and b for which the system of linear equations has an infinite number of solutions:

,

Solution: We can write the given system of equations as $\begin{array}{l} (a-1) x+3 y=2 \\ \Rightarrow(a-1) x+3 y-2=0\dots \dots(i) \\ \text { and } 6 x+(1-2 b) y=6 \end{array}$ $\Rightarrow 6...

### Find the value of k for which the system of linear equations has a unique solution;

,

Solution: Given system of equations can be written as $\begin{array}{l} (\mathrm{k}-3) \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0 \\ \mathrm{kx}+\mathrm{ky}-12=0 \end{array}$ The given system is of the...

### Find the value of k for which the system of linear equations has an infinite number of solutions:

Solution: Given system of equations: $\begin{array}{l} (\mathrm{k}-1) \mathrm{x}-\mathrm{y}=5 \\ \Rightarrow(\mathrm{k}-1) \mathrm{x}-\mathrm{y}-5=0\dots \dots(i) \end{array}$ And, $(k+1) x+(1-k)...

### Find the value of k for which the system of linear equations has an infinite number of solutions:

Solution: Given system of equations: $\begin{array}{l} 5 x+2 y=2 k \\ \Rightarrow 5 x+2 y-2 k=0\dots \dots(i) \end{array}$ And, $2(\mathrm{k}+1) \mathrm{x}+\mathrm{ky}=(3 \mathrm{k}+4)$ $\Rightarrow...

### Find the value of k for which the system of linear equations has an infinite number of solutions:

Solution: The system of equations: $\begin{array}{l} \mathrm{kx}+3 \mathrm{y}=(2 \mathrm{k}+1) \\ \Rightarrow \mathrm{kx}+3 \mathrm{y}-(2 \mathrm{k}+1)=0\dots \dots(i) \end{array}$ And,...

### Find the value of k for which the system of linear equations has an infinite number of solutions:

Solution: The system of equations: $2 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}=\mathrm{k}$ $\Rightarrow 2 \mathrm{x}+(\mathrm{k}-2) \mathrm{y}-\mathrm{k}=0\dots(i)$ And, $6 \mathrm{x}+(2 \mathrm{k}-1)...

### Find the value of k for which the system of linear equations has an infinite number of solutions:

Solution: The given system of equations: $\begin{array}{l} 2 \mathrm{x}+3 \mathrm{y}=7 \\ \Rightarrow 2 \mathrm{x}+3 \mathrm{y}-7=0\dots \dots(i) \end{array}$ And, $(\mathrm{k}-1)...

### For what value of k, the system of equations

Have (i) a unique solution, (ii) no solution? Also, show that there is no value of for which the given system of equation has infinitely namely solutions

Solution: The given system of equations: $\begin{array}{l} x+2 y=3 \\ \Rightarrow x+2 y-3=0\dots \dots(i) \end{array}$ And, $5 x+k y+7=0\dots \dots(ii)$ The equations are of the following form:...

### For what value of k, the system of equations

has (i) a unique solution, (ii) no solution?

Solution: The given system of equations are: $\begin{array}{l} x+2 y=5 \\ \Rightarrow x+2 y-5=0\dots \dots(i) \\ 3 x+k y+15=0\dots \dots(ii) \end{array}$ The equations are of the forms: $a_{1}...

### For what value of k, the system of equations

has (i) a unique solution, (ii) no solution?

Solution: Given system of equations: $\begin{array}{l} \mathrm{kx}+2 \mathrm{y}=5 \\ \Rightarrow \mathrm{kx}+2 \mathrm{y}-5=0\dots \dots(i) \\ 3 \mathrm{x}-4 \mathrm{y}=10 \\ \Rightarrow 3...

### Show that the system of equations

has no solution.

Solution: We can write the system of equations as $6 x+5 y-11=0\dots \dots(i)$ $\Rightarrow 9 \mathrm{x}+\frac{15}{2} \mathrm{y}-21=0\dots \dots(ii)$ The given system is of the form $a_{1} x+b_{1}...

### Show that the system equations

has an infinite number of solutions

Solution: The system of equations: $\begin{array}{l} 2 x-3 y=5 \\ \Rightarrow 2 x-3 y-5=0\dots \dots(i) \\ 6 x-9 y=15 \\ \Rightarrow 6 x-9 y-15=0\dots \dots(ii) \end{array}$ The equations are of the...

### Find the value of k for which the system of equations has a unique solution:

Solution: The system of given equations: $\mathrm{kx}+3 \mathrm{y}=(\mathrm{k}-3)$ $\Rightarrow \mathrm{kx}+3 \mathrm{y}-(\mathrm{k}-3)-0 \quad \ldots .(\mathrm{i})$ And, $12...

### Find the value of k for which the system of equations has a unique solution:

Solution: The system of given equations are $\begin{array}{l} 4 x-5 y=k \\ \Rightarrow 4 x-5 y-k=0 \quad \text {....(i) } \end{array}$ And, $2 x-3 y=12$ $\Rightarrow 2 x-3 y-12=0\dots \dots(ii)$ The...

### Find the value of k for which the system of equations has a unique solution:

Solution: The system of equations given are $\begin{array}{l} 4 x+k y+8=0 \\ x+y+1=0 \end{array}$ This given system is of the form: $\begin{array}{l} a_{1} x+b_{1} y+c_{1}=0 \\ a_{2} x+b_{2}...

### Find the value of k for which the system of equations has a unique solution:

Solution: The system of equations given are $\begin{array}{l} 5 x-7 y-5=0 \\ 2 x+k y-1=0 \end{array}$ This given system is of the form: $a_{1} x+b_{1} y+c_{1}=0$ $a_{2} x+b_{2} y+c_{2}=0$ where,...

### Find the value of k for which the system of equations has a unique solution:

Solution: The system of equations given are $\begin{array}{l} x-k y-2=0 \\ 3 x+2 y+5=0 \end{array}$ This system of equations given is of the form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2}...

### Find the value of k for which the system of equations has a unique solution:

Solution: The system of equations given are $\begin{array}{l} 2 x+3 y-5=0 \\ k x-6 y-8=0 \end{array}$ The given system is of the form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ where,...

### Show that the following system of equations has a unique solution: , Also, find the solution of the given system of equations.

Solution: The system of equations given is: $\begin{array}{l} \frac{x}{3}+\frac{y}{2}=3 \\ \Rightarrow \frac{2 x+3 y}{6}=3 \\ 2 x+3 y=18 \\ \Rightarrow 2 x+3 y-18=0\dots \dots(i) \end{array}$ and...

### Show that the following system of equations has a unique solution:

, Also, find the solution of the given system of equations.

Solution: The given system of eq. is: $\begin{array}{l} 2 x-3 y-17=0\dots \dots(i) \\ 4 x+y-13=\dots \dots(ii) \end{array}$ The given eq. are of the form $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2}...

### Show that the following system of equations has a unique solution:

,

. Also, find the solution of the given system of equations.

Solution: The given system of equations is $3 x+5 y=12$ $5 x+3 y=4$ These equations are of the forms: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+$ where, $a_{1}=3, b_{1}=5$ $a_{1} x+b_{1}...

### Solve the system of equations by using the method of cross multiplication:

,

, where and

Solution: On substituting $\frac{1}{x}=u$ and $\frac{1}{y}=v$ in the equations given, we obtain $\mathrm{au}-\mathrm{bv}+0=0$ $a b^{2} u+a^{2} b v-\left(a^{2}+b^{2}\right)=0$ Here, $a_{1}=a,...

### Solve the system of equations by using the method of cross multiplication:

Solution: We can write the given equation as: $\begin{array}{l} 2 a x+3 b y-(a+2 b)=0\dots \dots(i) \\ 3 a x+2 b y-(2 a+b)=0\dots \dots(i) \end{array}$ Here, $a_{1}=2 \mathrm{a}, \mathrm{b}_{1}=3...

### Solve the system of equations by using the method of cross multiplication:

,

Solution: We can write the given equations as: $\begin{array}{l} \frac{a x}{b}-\frac{b y}{a}-(a+b)=0\dots \dots(i) \\ a x-b y-2 a b=0\dots \dots(ii) \end{array}$ Here, $a_{1}=\frac{a}{b},...

### Solve the system of equations by using the method of cross multiplication:

,

Solution: Taking $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, the given equations become: $5 u-2 v+1=0\dots \dots(i)$ $15 u+7 v-10=0 \quad \ldots \ldots($ ii $)$ Here, $\mathrm{a}_{1}=5,...

### Solve the system of equations by using the method of cross multiplication:

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Solution: Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the given equations become: $\begin{array}{l} \mathrm{u}+\mathrm{v}=7 \\ 2 \mathrm{u}+3 \mathrm{v}=17 \end{array}$ We can write the given...

### Solve the system of equations by using the method of cross multiplication:

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Solution: The given equations may be written as: $\begin{array}{l} \frac{x}{6}+\frac{y}{15}-4=0\dots \dots(i) \\ \frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0\dots \dots(ii) \end{array}$ Here...

### Solve the system of equations by using the method of cross multiplication:

Solution: The given equations may be written as: $\begin{array}{l} 7 x-2 y-3=0\dots \dots(i) \\ 11 x-\frac{3}{2} y-8=0\dots \dots(ii) \end{array}$ Here $a_{1}=7, b_{1}=-2, c_{1}=-3, a_{2}=11,...

### Solve the system of equations by using the method of cross multiplication:

Solution: The given equations may be written as: $2 x+y-35=0\dots \dots(i)$ $3 \mathrm{x}+4 \mathrm{y}-65=0 \quad \ldots \ldots(ii)$ Here $a_{1}=2, b_{1}=1, c_{1}=-35, a_{2}=3, b_{2}=4$ and...

### Solve the system of equations by using the method of cross multiplication:

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Solution: The given equations may be written as: $\begin{array}{l} 2 x+5 y-1=0\dots \dots(i) \\ 2 x+3 y-3=0\dots \dots(ii) \end{array}$ Here $a_{1}=2, b_{1}=5, c_{1}=-1, a_{2}=2, b_{2}=3$ and...

### Solve the system of equations by using the method of cross multiplication:

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Solution: The given equations are: $3 x+2 y+25=0\dots \dots(i)$ $2 \mathrm{x}+\mathrm{y}+10=0 \quad \ldots \ldots(ii)$ Here $\mathrm{a}_{1}=3, \mathrm{~b}_{1}=2, \mathrm{c}_{1}=25, \mathrm{a}_{2}=2,...

### Solve the system of equations by using the method of cross multiplication:

Solution: The given equations are: $\begin{array}{l} 6 x-5 y-16=0\dots \dots(i) \\ 7 x-13 y+10=0\dots \dots(ii) \end{array}$ Here $a_{1}=6, b_{1}=-5, c_{1}=-16, a_{2}=7, b_{2}=-13$ and $c_{2}=10$ On...

### Solve the system of equations by using the method of cross multiplication:

Solution: The given equations are: $\begin{array}{ll} 3 \mathrm{x}-2 \mathrm{y}+3=0 & \ldots \ldots (i)\\ 4 \mathrm{x}+3 \mathrm{y}-47=0 & \ldots \ldots (ii) \end{array}$ Here $a_{1}=3,...

### Solve the system of equations by using the method of cross multiplication:

Solution: The given equations are: $\begin{array}{l} x+2 y+1=0\dots \dots (i) \\ 2 x-3 y-12=0\dots \dots(ii) \end{array}$ Here $a_{1}=1, b_{1}=2, c_{1}=1, a_{2}=2, b_{2}=-3$ and $c_{2}=-12$ On cross...

### Solve for x and y :

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Solution: The given equations are $\begin{array}{l} \frac{x}{a}+\frac{y}{b}=\mathrm{a}+\mathrm{b}\dots \dots(i) \\ \frac{x}{a^{2}}+\frac{y}{b^{2}}=2\dots \dots(ii) \end{array}$ Multiplying...

### Solve for x and y :

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Solution: The given equations are $\begin{array}{l} a^{2} x+b^{2} y=c^{2}\dots \dots (i) \\ b^{2} x+a^{2} y=d^{2} \dots \dots(ii) \end{array}$ Multiplying equation(i) by $\mathrm{a}^{2}$ and...

### Solve for x and y :

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Solution: The given equations are $\begin{array}{l} \mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b} \quad\{\ldots \ldots(i) \\ \mathrm{ax}-\mathrm{by}=\mathrm{a}^{2}-\mathrm{b}^{2}+\ldots \ldots(ii) \\...

### Solve for x and y :

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Solution: The given eq. are: $\frac{b x}{a}+\frac{a y}{b}=\mathrm{a}^{2}+\mathrm{b}^{2}$ By taking LCM, we obtain: $\begin{array}{l} \frac{b^{2} x+a^{2} y}{a b}=a^{2}+b^{2} \\ \Rightarrow b^{2}...

### Solve for x and y :

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Solution: The given equations are: $\frac{b x}{a}-\frac{a y}{b}+a+b=0$ By taking LCM, we obtain: $b^{2} x-a^{2} y=-a^{2} b-b^{2} a\dots \dots(i)$ and $b x-a y+2 a b=0$ $b x-a y=-2 a b\dots...

### Solve for x and y:

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Solution: The given equations are $\begin{array}{l} a x-b y=a^{2}+b^{2}\dots \dots(i) \\ x+y=2 a\dots \dots(ii) \end{array}$ From equation(ii) $y=2 a-x$ Substituting $\mathrm{y}=2...

### Solve for x and y :

Solution: The given equations are $\begin{array}{l} 6(a x+b y)=3 a+2 b \\ \Rightarrow 6 a x+6 b y=3 a+2 b\dots \dots(i) \end{array}$ and $6(b x-a y)=3 b-2 a$ $\Rightarrow 6 b x-6 a y=3 b-2 a\dots...

### Solve for x and y :

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Solution: The given eq. can be written as $\begin{array}{l} \frac{x}{a}-\frac{y}{b}=0\dots \dots(i) \\ a x+b y-u^{2}+b^{2}\dots \dots(ii) \end{array}$ From equation(i), $\mathrm{y}=\frac{b x}{a}$...

### Solve for x and y :

Solution: The given equations are $\begin{array}{l} \mathrm{px}+\mathrm{qy}=\mathrm{p}-\mathrm{q}\dots \dots(i) \\ \mathrm{qx}-\mathrm{py}=\mathrm{p}+\mathrm{q}\dots \dots(ii) \end{array}$...

### Solve for x and y :

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Solution: The given equations are: $\begin{array}{l} \frac{x}{a}+\frac{y}{b}=2 \\ \Rightarrow \frac{b x+a y}{a b}=2 \quad[\text { Taking LCM }] \\ \Rightarrow \mathrm{bx}+\mathrm{ay}=2 \mathrm{ab}...

### Solve for x and y :

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Solution: The given equations are $\begin{array}{l} \mathrm{x}+\mathrm{y}=\mathrm{a}+\mathrm{b}\dots \dots(i) \\ \mathrm{ax}-\mathrm{by}=\mathrm{a}^{2}-\mathrm{b}^{2}\dots \dots(ii) \end{array}$...

### Solve for x and y :

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Solution: The given equations can be written as $\begin{array}{l} \frac{3}{x}+\frac{6}{y}=7\dots \dots(i) \\ \frac{9}{x}+\frac{3}{y}=11\dots \dots(ii) \end{array}$ Multiplying equation(i) by 3 and...

### Solve for x and y :

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Solution: The given equations are $\begin{array}{l} \frac{2}{3 x+2 y}+\frac{3}{3 x-2 y}=\frac{17}{5}\dots \dots(i) \\ \frac{5}{3 x+2 y}+\frac{1}{3 x-2 y}=2\dots \dots(ii) \end{array}$ Substituting...

### Solve for x and y :

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where and

Solution: The given equations are $\begin{array}{l} \frac{1}{2(x+2 y)}+\frac{5}{3(3 x-2 y)}=-\frac{3}{2}\dots \dots(i) \\ \frac{1}{4(x+2 y)}-\frac{3}{5(3 x-2 y)}=\frac{61}{60}\dots \dots(ii)...

### Solve for x and y:

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Solution: The given equations are $\begin{array}{l} \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\dots \dots(i) \\ \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8} \end{array}$ $\frac{1}{3...

### Solve for x and y :

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Solution: The given equations can be written as $\begin{array}{l} \frac{5}{x}+\frac{2}{y}=6\dots \dots (i) \\ \frac{-5}{x}+\frac{4}{y}=-3\dots \dots(ii) \end{array}$ Adding equation(i) and...

### Solve for x and y : ,

Solution: The given equations are: $23 x-29 y=98 \quad \ldots .$ (i) $29 x-23 y=110 \ldots \ldots$ (ii) Adding equation(i) and equation(ii), we obtain: $\begin{array}{l} 52 x-52 y=208 \\ \Rightarrow...

### Solve for x and y:

217 x+131 y=913,

131 x+217 y=827

Solution: The given equations are: $\begin{aligned} 217 \mathrm{x}+131 \mathrm{y} =913\dots \dots(i) \\ 131 \mathrm{x}+217 \mathrm{y} =827 \ldots \text {....(ii) } \end{aligned}$ On adding...

### Solve for x and y :

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Solution: The given eq. are: $\begin{array}{l} 71 x+37 y=253\dots \dots(i) \\ 37 x+71 y=287 \ldots \ldots \text {...(ii) } \end{array}$ On adding equation(i) and equation(ii), we obtain:...

### Solve for x and y :

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, where

Solution: The given equations are $\frac{10}{x+y}+\frac{2}{x-y}=4 \quad \ldots \ldots$ (i) $\frac{15}{x+y}-\frac{9}{x-y}=-2\dots \dots(ii)$ Substituting $\frac{1}{x+y}=\mathrm{u}$ and...

### Solve for x and y :

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Solution: The given eq. are $\frac{44}{x+y}+\frac{30}{x-y}=10 \dots \dots(i)$ $\frac{55}{x+y}-\frac{40}{x-y}=13\dots \dots (ii)$ Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, we get: $44 u+30...

### Solve for x and y : , , where

Solution: The given eq. are $\frac{5}{x+1}+\frac{2}{y-1}=\frac{1}{2} \quad \ldots \ldots$ (i) $\frac{10}{x+1}-\frac{2}{y-1}=\frac{5}{2}\dots \dots (ii)$ On substituting $\frac{1}{x+1}=\mathrm{u}$...

### Solve for x and y :

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Solution: The given eq. are $\begin{array}{l} \frac{3}{x+y}+\frac{2}{x-y}=2ย \ldots \ldots \text { (i) } \\ \frac{9}{x+y}-\frac{4}{x-y}=1ย \ldots \ldots \text { (ii) } \end{array}$ Substituting...

### Solve for x and y :

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Solution: The given eq. are $\frac{5}{x+y}-\frac{2}{x-y}=-1 \quad \ldots \ldots \text { (i) }$ $\frac{15}{x+y}-\frac{7}{x-y}=10\dots \dots(ii)$ Substituting $\frac{1}{x+y}=\mathrm{u}$ and...

### Solve for x and y:

x + y = 5xy,

3x + 2y = 13xy

Solution: The given eq. are: $x + y = 5xy \dots \dots(i)$ $3x + 2y = 13xy \dots \dots(ii)$ From equation (i), we have: $\begin{array}{l} \frac{x+y}{x y}=5 \\ \Rightarrow \frac{1}{y}+\frac{1}{x}=5...

### Solve for x and y :

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Solution: The given eq. are: $\begin{array}{l} 4 x+6 y=3 x y \quad \ldots \ldots(i) \\ 8 x+9 y=5 x y \quad \ldots \ldots(i i) \end{array}$ From eq. (i), we have: $\begin{array}{l} \frac{4 x+6 y}{x...

### Solve for x and y :

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Solution: The given eq. are: $\begin{array}{l} \frac{3}{x}+\frac{2}{y}=12 \ldots \ldots \ldots \text { (i) } \\ \frac{2}{x}+\frac{3}{y}=13 \ldots \ldots . \text { (ii) } \end{array}$ Multiplying...

### Solve for x and y :

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Solution: The given eq. are: $\begin{array}{l} \frac{5}{x}-\frac{3}{y}=1 \ldots \ldots(i) \\ \frac{3}{2 x}+\frac{2}{3 y}=5 \ldots \ldots (ii) \end{array}$ Putting $\frac{1}{x}=u$ and...

### Solve for x and y:

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Solution: The given eq. are: $\begin{array}{l} \frac{9}{x}-\frac{4}{y}=8 \ldots \ldots . .(i) \\ \frac{13}{x}+\frac{7}{y}=101 \ldots \ldots (ii) \end{array}$ Putting $\frac{1}{x}=u$ and...

### Solve for x and y :

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Solution: The given eq. are: $\begin{array}{l} \frac{3}{x}-\frac{1}{y}+9=0 \\ \Rightarrow \frac{3}{x}-\frac{1}{y}=-9 \quad \ldots \ldots(i) \\ \Rightarrow \frac{2}{x}-\frac{3}{y}=5 \ldots \ldots...

### Solve for x and y:

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Solution: The given eq. are: $\begin{array}{l} 2 x-\frac{3}{y}=9 \ldots \ldots \text { (i) } \\ 3 x+\frac{7}{y}=2 \ldots \ldots \text { (ii) } \end{array}$ Putting $\frac{1}{y}=\mathrm{v}$, we...

### Solve for x and y :

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Solution: The given eq. are: $\begin{array}{l} x+\frac{6}{y}=6 \quad \ldots \ldots \text {(i) } \\ 3 x-\frac{8}{y}=5 \ldots \ldots \text { (ii) } \end{array}$ Putting $\frac{1}{y}=\mathrm{v}$, we...

### Solve for x and y :

Solution: The given eq. are: $\begin{array}{l} \frac{5}{x}+6 y=13 \ldots \ldots \text { (i) } \\ \frac{3}{x}+4 y=7 \ldots \ldots \text { (ii) } \end{array}$ Putting $\frac{1}{x}=\mathrm{u}$, we...

### Solve for x and y :

Solution: The given eq. are: $\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}$ i.e., $\frac{x+y-8}{2}=\frac{3 x+y-12}{11}$ On cross multiplication, we obtain: $\begin{array}{l} 11 x+11 y-88=6...

### Solve for x and y:

6x + 5y = 7x + 3y + 1 = 2(x + 6y โ 1)

Solution: The given equations are: $\begin{array}{l} 6 x+5 y=7 x+3 y+1=2(x+6 y-1) \\ \Rightarrow 6 x+5 y=2(x+6 y-1) \\ \Rightarrow 6 x+5 y=2 x+12 y-2 \\ \Rightarrow 6 x-2 x+5 y-12 y=-2 \\...

### Solve for x and y:

7(y + 3) โ 2(x + 2) = 14,

4(y โ 2) + 3(x โ 3) = 2

Solution: The given eq. are: $7(y + 3) โ 2(x + 2) = 14$ $\Rightarrow 7y + 21 โ 2x โ 4 = 14$ $\Rightarrow -2x + 7y = -3 \dots \dots(i)$ and $4(y โ 2) + 3(x โ 3) = 2$ $\Rightarrow 4y โ 8 + 3x โ 9 = 2$...

### Solve for x and y:

0.3x + 0.5y = 0.5,

0.5x + 0.7y = 0.74

Solution: The given system of eq. is $0.3 \mathrm{x}+0.5 \mathrm{y}=0.5\dots \dots(i)$ $0.5 \mathrm{x}+0.7 \mathrm{y}=0.74 \dots \dots(ii)$ On multiplying equation(i) by 5 and equation(ii) by 3 and...

### Solve for x and y:

0.4x + 0.3y = 1.7,

0.7x โ 0.2y = 0.8.

Solution: The given system of eq. is $0.4x + 0.3y = 1.7 \dots \dots(i)$ $0.7x โ 0.2y = 0.8 \dots \dots(ii)$ On multiplying equation(i) by $0.2$ and equation(ii) by $0.3$ and adding them, we obtain...

### Solve for x and y :

Solution: The given eq. are: $\begin{array}{l} \frac{7-4 x}{3}=y \\ \Rightarrow 4 x+3 y=7 \dots \dots(i) \end{array}$ $\begin{array}{l} \text { and } 2 x+3 y+1=0 \\ \Rightarrow 2 x+3 y=-1 \ldots...