Pair of Linear Equations in Two Variables

### Solve:

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,...

### 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...

### 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: $(-y)^{2} y=y^{3}>0(\because y>0)$...

### 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...

### 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 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: $\begin{array}{l} \because 2^{x+y}=2^{x-y}=\sqrt{8} \\ \therefore \mathrm{x}+\mathrm{y}=\mathrm{x}-\mathrm{y} \\ \Rightarrow \mathrm{y}=0 \end{array}$

### 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,...

### 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, $\frac{x+8}{y+8}=\frac{3}{4}$...

### 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 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...

### 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 $=(10 \mathrm{x}+\mathrm{y})$ $x+y=12\dots \dots(i)$...

### 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 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 $2 x+3 y=7$ $\Rightarrow 2 x+3 y-7=0\dots \dots(i)$ and $(a+b+1) x-(a+2 b+2) y=4(a+b)+1$ $(a+b+1) x-(a+2 b+2) y-[4(a+b)+1]=0\dots \dots(i)$...

### 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 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,...

### 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: ,

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:

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 : ,

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 : ,

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 : ,

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 : ,

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 : ,

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 : ,

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 : , , 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 : ,

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 : ,

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 : ,

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 : ,

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: ,

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...