Solution: Suppose the cost of 1 pen and 1 pencil are Rs.$x$ and Rs.$y$ respectively. Then according to the question $\begin{array}{ll} 5 x+8 y=120 \ldots \ldots \text { (i) } \\ 8 x+5 y=153 \ldots...
tables and 
chairs, together, cost 
and tables and chairs cost 
. Find the cost of 
chairs and 
table.Let’s assume the cost of $1$ table is a and cost of $1$ chair is b. Then, according to the question $4a+3b=2250$ … (a) $3a+4b=1950$ … (b) On multiplying (a) with $3$ and (b) with $4$, We get,...
bags and pens together cost 
whereas bags and pens together cost 
. Find the total cost of bag and 
pens.Let the cost of a bag and a pen be a and b, respectively. Then, according to the question $3a+4b=257$ … (a) $4a+3b=324$ … (b) On multiplying equation (a) by $3$ and (b) by $4$, We get, $9a+12b=770$...
, 
Given system of equations are: $2x+3y–7=0$ $(F+1)x+(2F-1)y–(4F+1)=0$ The above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$ Here,...
, 
The given system of equations is: $2x+(F-2)y–F=0$ $6x+(2F-1)y–(2F+5)=0$ The above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$...
, 
The given system of equations is: $Fx+3y–(2F+1)=0$ $2(F+1)x+9y–(7F+1)=0$ The above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$...
, 
The given system of equations is: $a+(F+1)y–4=0$ $(F+1)a+9y–(5F+2)=0$ The above equations are of the form ${{a}_{1}}a+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}y-{{c}_{2}}=0$...
, 
Given system of equations is: $2a+3y–2=0$ $(F+2)a+(2F+1)y–2(F-1)=0$ Above equations are of the form ${{a}_{1}}a+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}y-{{c}_{2}}=0$...
, 
The given system of equations is: $2x–3y–7=0$ $(F+2)x–(2F+1)y–3(2F-1)=0$ The above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$...
, 
Given equations are: $8a+5y–9=0$ $Fa+10y–18=0$ The above equations are of the form ${{a}_{1}}a+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}y-{{c}_{2}}=0$ Now, ${{a}_{1}}=8,{{b}_{1}}=5,{{c}_{1}}=-9$...
, 
Given equations are: $Fa–2y+6=0$ $4a–3y+9=0$ The above equations are of the form ${{a}_{1}}a+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}y-{{c}_{2}}=0$ Now,${{a}_{1}}=F,{{b}_{2}}=-2,{{c}_{2}}=9$...
, 
The Given equations are: $4a+5y–3=0$ $ka+15y–9=0$ Above equations are of the form ${{a}_{1}}a+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}y-{{c}_{2}}=0$ Here, ${{a}_{1}}=4,{{b}_{1}}=5,{{c}_{1}}=-3$...
, 
The Given equations are: $2a+3y-5=0$ $6a+Fy-15=0$ Above equations are of the form ${{a}_{1}}a+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}y-{{c}_{2}}=0$ Now, ${{a}_{1}}=2,{{b}_{1}}=3,{{c}_{1}}=-5$...
, 
Given equations are: $a+2b–3=0$ $5a+kb+7=0$ Above equations are of the form ${{a}_{1}}a+{{b}_{1}}b-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}b-{{c}_{2}}=0$ Now, ${{a}_{1}}=1,{{b}_{1}}=2,{{c}_{1}}=-3$...
, 
Given equations are: $4a–5b–k=0$ $2a–3b–12=0$ Above equations are of the form ${{a}_{1}}a+{{b}_{1}}b-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}b-{{c}_{2}}=0$ Now, ${{a}_{1}}=4,{{b}_{1}}=5,{{c}_{1}}=-k$...
, 
Given equations are: $4a+kb+8=0$ $2a+2b+2=0$ Above equations are of the form ${{a}_{1}}a+{{b}_{1}}b-{{c}_{1}}=0$ ${{a}_{2}}a+{{b}_{2}}b-{{c}_{2}}=0$ Now, ${{a}_{1}}=4,{{b}_{1}}=k,{{c}_{1}}=8$...
, 
Given equations are: $ax+2y–5=0$ $3x+y–1=0$ Above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$ Now, ${{a}_{1}}=k,{{b}_{1}}=2,{{c}_{1}}=-5$...
, 
Given system of equations are: $x–2y–8=0$ $5x–10y–10=0$ Above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$...
,
Given system of equations is: $3x–5y–20=0$ $6x–10y–40=0$ Above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$...
, 
Given system of equations are: $2x + y – 5 = 0$ $4x + 2y – 10 = 0$ Above equations are of the form ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$ Therefore,...
, 
Given system of equations is: $x-3y-3=0$ $3x-9y-2=0$ Above equations are in the form of ${{a}_{1}}x+{{b}_{1}}y-{{c}_{1}}=0$ ${{a}_{2}}x+{{b}_{2}}y-{{c}_{2}}=0$ Here,...
apart from each other on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet in 
and if they move in opposite directions, they meet in 
. Find the speeds of the cars.Let the speed of the car leaving A to be 'x' km/h and the speed of the car leaving B to be 'y' km/h. Total distance=80 km Also 1 hour 20 mins= 4/3 hour When an automobile goes in the same direction...
partly by train and partly by car. If he covers the 
by train and the rest by car, it takes him 
and 
. But, if he travels 
by train and the rest by car, he takes half an hour longer. Find the speed of the train and the speed of the car.Let’s suppose speed of the train be $Akm/hr$ speed of the car $=Bkm/hr$ There are two parts Part 1: When the women travels $400km$ by train and the rest by car. Part 2: When Riya travels $200km$ by...
to his home partly by train and partly by car. It takes if he travels 
by train and the rest by car. He takes 
more if he travels 
by train and the rest by car. Find the speed of the train and car respectively.Let’s assume, The speed of the train be $Ckm/hr$ The speed of the car $=Dkm/hr$ From the question, it’s understood that there are two parts # Part 1: When Ramesh travels $160Km$ by train and the...
Let’s assume C to be the speed of the stream. So, we know that Speed of boat in downstream $=(5+C)$ and, Speed of boat in upstream $=(5–C)$ It is given that, The distance in one way is $40km$. And,...
an hour faster, he takes 1 hour less. But, if he walks 1km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.Let the actual speed of the man be $Ckm/hr$ and D be the actual time taken by him in hours. So, we know that Distance covered = speed C distance ⇒ Distance$=C\times D=CD$ …………………………. (i) First...
, thereby making a profit of 
on the table and 
on the chair. If she had taken a profit of on the table and on the chair she would have got 
. Find the cost price of each.Let the cost price of one table and one chair be ₹ a and ₹ b, respectively. The selling price of the table, when it’s sold at a profit of $10\%=₹a+10a/100=₹110a/100$ The selling price of the chair,...
books and 
pens together cost 
whereas books and pens together cost 
. Find the total cost of book and pens.Let’s assume the cost of a book and a pen are ₹ a and ₹ b, respectively. According to the question $5a+7b=79$… (a) $7a+5b=77$… (b) multiplying equation (a)by $5$ and (b)by $7$, We get,...
upstream and 
downstream in 
. It goes 
upstream and 
downstream in 
hours. Find the speed of the boat in still water and also speed of the stream.Let’s assume, The speed of the boat in still water as $Ckm/hr$ And, The speed of the stream as $Dkm/hr$ We know that, Speed of the boat in upstream $=(C–D)km/hr$ Speed of the boat in downstream...
in number. If she has more pencils and less pens, then the number of pencils would become times the number of pens. Find the original number of pens and pencils.Let’s assume the number of pens and pencils are a and b, respectively. Forming equations according to the question, we have $a+b=40$… (a) $(b+5)=4(a-5)$ $b+5=4a–20$ $5+20=4a–b$ $4a –b=25$… (b)...
upstream and 
downstream in hours. In 
hours, it can go 
upstream and 
downstream. Determine the speed of the stream and that of the boat in still water.Let’s assume, The speed of the boat in still water as $Ckm/hr$ And, The speed of the stream as $Dkm/hr$ We know that, Speed of the boat in upstream $=(C–D)km/hr$ Speed of the boat in downstream...
downstream in minutes and returns in hour. Determine the speed of the sailor in still water and the speed of the current.Let’s assume, The speed of the sailor in still water as $Ckm/hr$ And, The speed of the current as $Dkm/hr$ We know that, Speed of the sailor in upstream $=(C–D)km/hr$ Speed of the sailor in...
audio cassettes and videocassettes cost 
, while audio cassettes and videocassettes cost 
. Find the cost of audio cassettes and a video cassette.Let’s assume the cost of an audio cassette and that of a video cassette be ₹a and ₹b, respectively. Then forming equations according to the question, we have $7a+3b=1110$…(a) $5a+4b=1350$… (b) On...
pens and 
pencils together cost 
and pens and pencils cost 
. Find the cost of pen and pencil.Let’s assume the cost of a pen and pencil be ₹ a and ₹ b respectively. Then, forming equations according to the question $5a+6b=9$…(a) $3a+2b=5$…(b) On multiplying equation (a)by $2$ and equation...
. apart on a highway. P car starts from P and another car starts from B simultaneously. If the D travel in the same direction, the D meet in 
, but if the D travel towards each other, the D meet in one hour. Find the speed of two cars.Let’s consider the car starting from point P as C and its speed as C km/hr. The car starts from point B as D and its speed as D km/hr. There are two cases in the question: Case i: Car C and D are...
times the sum of its digits. If 
is added to the number, the digits are interchanged. Find the number.Let’s assume the digit at unit’s place is a and at ten’s place is b. Thus from the question, the number we need to find is $10b+a$. From the question since the number is $4$ times the sum of the two...
. If the digits differ by , find the number.Let’s assume the digit at unit’s place is a and ten’s place is b. Thus from the question, the number we need to find is $10b+a$. From the question since the two digits of the number are differing bb...
and the difference between their square is 
. Find the numbers.Let the two numbers be a and b and assume that a is greater than or equal to b. According to question, we can write the sum of the two numbers as $a+b=1000$ ……….. (i) And the difference between the...
. If the two digits differ by , find the number. How many such numbers can be found?Let’s the digit at unit’s place be a and ten’s place be b. According to question, the two digits of the number are differing by $2$. Thus, we can write $a-b=\pm 2$………….. (i) Now, on reversing the...
. The number obtained by interchanging the order of digits of the given number, it exceeds the given number by 
. Find the number.Let the digits at unit’s place be a and ten’s place be v, respectively. Thus, the number we need to find is $10b+a$. As per the given question, the sum of the two digit number is $15$. Thus, we...
Let the digit at unit’s place be a and ten’s place be b. Thus, the number to be found is $10b+a$. From given question, the sum of two digit number is equal to $5$. Thus we can write equation...
. If the number is subtracted from the one obtained by interchanging the digits, we get the value 
. Find the number?Let the digit at the unit’s place be a and at ten’s place be b. Then the required number is $10b+a$. According to the given question, The sum of the two digit number is $13$, So, $a+b=13$………… (i) On...
. If the sum of two numbers is four times their difference, find the numbersLet’s assume the two numbers to be ‘a’ and ‘b’. Let’s consider that, ‘a’ is greater than or equal to ‘b’. Now, according to the question The sum of the two numbers, $a+b=8$…………. (i) Also, sum is...
. If is added in the denominator, the fraction becomes . Find the fraction so obtained.Let the numerator be A and the denominator be B. So, the required fraction is $A/B$. ATQ, The sum of the numerator and denominator of the fraction is $12$. $A+B=12$ ⇒ $A+B–12=0$ ATQ, If $3$ is added...
and numerator is increased by , a fraction becomes . It also becomes 
if we only increase the denominator by . So now find the fraction?Let the numerator of the fraction to be A and the denominator of the fraction to be B. So, the required fraction is $A/B$. ATQ, If denominator is decreased by $1$ and numerator is increased by $1$,...
is subtracted from both its numerator and denominator then fraction becomes 
. If numerator and denominator are added by , it becomes 
. Find the fraction.Let the numerator of the fraction to be A and the denominator of the fraction to be B. So, the required fraction is $A/B$. ATQ, Thus, the equation so formed is, $(A–1)/(B−1)=1/3$ ⇒ $3(A–1)=(B–1)$ ⇒...
more than the numerator. If the denominator is eight times the numerator then the numerator is lessen by and denominator is increased by . Find the original fraction calculated.Let the numerator of the fraction to be A and the denominator of the fraction to be B. So, fraction is $A/B$. The numerator of the fraction is $4$ less the denominator. Thus, the equation so formed...
is added to both numerator and the denominator then the fraction becomes 
. If is added to both the numerator and the denominator it becomes 
. Find the fraction.Let’s assume the numerator of the fraction to be A and the denominator of the fraction to be B. So, the required fraction is $A/B$. ATQ , the equation so formed is, $A+2B+2=9/11$ ⇒ $11\left( A+2...
Let’s the present ages of the father as a years and that of his son’s age as b years. According to the question,, The present age of father is three years more than three times the age of the son....
Let’s the present ages of the father as a years and that of his son’s age as b years. According to the question,, After 10 years, the age of father will be $(a+10)$ years and son’s age will be...
Let the present ages of the father as a years and that of his son’s age as b years. According to question After 6 years, the man’s age will be $(a+6)$ years and son’s age will be $(b+6)$ years. So,...
Assuming that, the present age of $A=a$ The present age of $B=b$ The present age of $F=z$ The present age of $S=t$ Now from the questions A is elder to b by 2 years. ⇒ $a=b+2$ F is twice as old as...
Let the present ages of A be a years and that of B be b years According to the question, After 10 years, A’s age will be $(a+10)$ years and B’s age will be $(b+10)$ years. Now, the relation between...
Let’s assume the present ages of the father as a years and that of his son’s age as b years. From the question it’s given that, Father is 3 times as old as his son. (Present) So, the equation formed...
, 
Taking LCM for both the given equations, we have $(2B+3A)/AB=9/AB ⇒3A+2B=9$………. (i) now, $(4B+9A)/AB=21/AB ⇒9A+4B=21$………(ii) Performing substraction between (ii) & (i) $9A+4B-2\left( 3A+2B...
, 
The given pair of equations are: \[~\left( A+B \right)/AB=2\Rightarrow 1/B+1/A=2\ldots \ldots .\text{ }\left( i \right)\] \[\left( AB \right)/AB=6\Rightarrow 1/B1/A=6\ldots \ldots \ldots \left( ii...
, 
Assume $1/\sqrt{A}=u$and $1/\sqrt{B}=v$ So, the given equations becomes $2u+3v=2$………………….. (i) $4u-9v=-1$ ………………………. (ii) Multiplying (ii) by $3$ and Adding equation (i) and (ii)A$3$ we get,...
, 
Let $1/A=u$ and $1/B=v$ So, the given equations becomes $2u+3v=13$………………….. (i) $5u–4v=-2$ ………………………. (ii) Adding equation (i) and (ii) we get, $2u+3v+5u–4v=13–2$ ⇒ $7u–v=11$ ⇒ $v=7u–11$…….. (iii)...
, 
Taking $1/A=u$, the given equation becomes $4u+5B=7$…………………….. (i) $3u+4B=5$…………………….. (ii) Subtracting (ii) from (i), we get $4u+5B–(3u+4B)=7–5$ ⇒ $u+B=2$ ⇒ $u=2–B$……………………… (iii) Using (iii) in...
, 
Taking $1/A=u$, the given equation becomes $4u+3B=14$…………………….. (i) $3u–4B=23$…………………….. (ii) Adding (i) and (ii), we get $4u+3B+3u–4B=14+23$ ⇒ $7u–B=37$ ⇒ $B=7u–37$……………………… (iii) Using (iii) in...
, 
Let $1/A=u$ and $1/B=v$ So, the given equations becomes $u/5+v/6=12$…………………..(i) $u/3–3v/7=8$……………………….(ii) Taking LCM for both equations, we get $6u+5v=360$………. (iii) $7u–9v=168$……….. (iv)...
, 
Let $1/A=u$ and $1/B=v$ So, the given equations becomes $2u+5v=1$…………………..(i) $60u+40v=19$ ……………………….(ii) Multiplying equation (i) $8$ and (ii) $1$ we get, $16u+40v=8$ ………………………….. (iii)...
, 
Let $\frac{1}{u}=x$and $\frac{1}{v}=y$, then the given system of equation become $15x+2y=17$….(i) $x+y=\frac{36}{5}$….(ii) From (i) we get, $2y=17-15x$ $\Rightarrow y=\frac{17-15x}{2}$ Substituting...
, 
Let $1/A=u$ and $1/B=v$ So, the given equations becomes $15A+2B=17$ ………………………….. (i) $A+B=36/5$………………………. (ii) From equation (i) we get, $2B=17-15A$ =$B=(17-15A)/ 2$ …………………. (iii) Substituting...
, 
Let $1/A=u$ and $1/B=v$, So the given equations becomes, $u/2+v/3=2$ ………………(i) $u/3+v/2=13/6$ ……………(ii) From (i), we get $u/2+v/3=2$ ⇒ $3u+2v=12$ ⇒ $u=(12–2v)/3$ ………….(iii) Using (iii) in (ii)...
, 
The given pair of equations are: $1/(7A)+1/(6B)=3$………………………….. (i) $1/(2A)–1/(3B)=5$……………………………. (ii) Multiplying (ii) by $\frac{1}{2}$ we get, $1/(4A)–1/(6B)=52$……………………………. (iii) Now, solving...
, 
Now, let’s multiply LHS and RHS by $100$for both (i) and (ii) for making integral coefficients and constants. (i)$A100$ ⇒ $50A+70B=74$ ……………………….. (iii) (ii) $A100$ ⇒ $30A+50B=50$ …………………………… (iv)...
,
The given pair of equations are: $2A-(3/B)=9$……………………………. (i) $3A+(7/B)=2$…………………………… (ii) Substituting $1/B=u$ the above equations becomes, $2A–3u=9$ ………………………..(iii) $3A+7u=2$………………………..(iv) From...
, 
The given pair of equations are: $3A-(B+7)/11+2=10$……………….. (i) $2B+(A+11)/7=10$…………………….. (ii) From equation (i) $33A-B-7+22=(10\times 11)$ [After taking LCM] ⇒ $33A-B+15=110$ ⇒ $33A+15-110=B$ ⇒...
, 
The given pair of equations are: $\sqrt{2}A-\sqrt{3}B=0$……………………….. (i) $\sqrt{3}A-\sqrt{8}B=0$……………………….. (ii) From equation (i) $A=\sqrt{\left( 3/2 \right)}B$……………..(iii) Substituting this value...
, 
The given pair of equations are: $A+2B=3/2$ …………………. (i) $2A+B=3/2$…………………… (ii) Let us eliminate B from the given equations. The coefficients of B in the given equations are $2$ and $1$...
, 
The given pair of equations are: $A+B/2=4$ ……………………. (i) $2B+A/3=5$……………………. (ii) From (i) we get, $A+B/2=4$ ⇒ $2A+B=8$ [After taking LCM] ⇒ $B=8–2A$ …..(iv) From (ii) we get, $A+6B=15$ ……………… (iii)...
Taking $1/A=u$ Then the two equation becomes, $4u+3B=8$…………………… (i) $6u–4B=-5$……………………. (ii) From (i), we get $\begin{array}{l} 4u = 8 - 3B\\ u = (8 - 3B)/4 \end{array}$ …….. (iii) Substituting u in...
, 
The given pair of equations are: $A/3+B/4=11$…………………………. (i) $5A/6-B/3=−7$……………………………….. (ii) From (i), we get $A/3+B/4=11$ ⇒$4A+3B=\left( 11\times 12 \right)$ [After taking LCM] ⇒ $4A=132–3B$ ⇒...
, 
The given pair of equations are: $A/7+B/3=5$…………………………. (i) $A/2–B/9=6$………………………………..(ii) From (i), we get $A/7+B/3=5$ ⇒$3A+7B=\left( 5\times 21 \right)$ [After taking LCM] ⇒ $3A=105–7B$ ⇒...
, 
The given pair of equations are: $7(B+3)–2(A+2)=14$…………………………. (i) $4(B-2)+3(A-3)=2$……………………………….. (ii) From (i), we get $7B+21–2A–4=14$ $7B=14+4–21+2A$ ⇒ $B=(2A–3)/7$ From (ii), we get...
, 
The given pair of equations are: $A/2+B=0.8⇒A+2B=1.6$…… (a) $7/(A+B/2)=10⇒7=10(A+B/2)⇒7=10A+5B$ Let’s, multiply LHS and RHS of equation (a) by $10$ for easy calculation So, we finally get...
The given pair of equations are: $0.4A+0.3B=1.7$ $0.7A–0.2B=0.8$ Let’s, multiply LHS and RHS by $10$ to make the coefficients as an integer $4A+3B=17$ ……………………….. (i) $7A–2B=8$ …………………………… (ii) From...
The given equations are: $3A–7B+10=0$ …………………………. (a) $B–2A–3=0$ ……………………………….. (b) From (b) $B–2A–3=0$ $B=2A+3$ ……………………………… (c) Now, substitute B in equation (a) we get, ⇒ $3A–7(2A+3)+10=0$ ⇒...
, 
The given equations are: $11A+15B+23=0$ …………………………. (a) $7A–2B–20=0$ …………………………….. (b) From (b) $\begin{array}{l} 2A = 7A - 20\\ B = (7A - 20)/2 \end{array}$ ……………………………… (c) Now, substitute in...
and 
(ii) 
and 
Given, $2a–3b=6$ ……. (i) $a+b=1$……. (ii) For equation (i), ⇒ $b=(2a–6)/3$ When $a=3$, we have $b=(2(3)–6)/3=0$ When $a=0$, we have $b=(2(0)–6)/3=-2$ Thus, we have the following table giving points...
, 
is consistent or in-consistent using graphical methodGiven, $a–2b=2$……. (i) $4a–2b=5$……. (ii) For equation (i), ⇒ $b=(a–2)/2$ When $a=2$, we have $b=(2–2)/2=0$ When $a=0$, we have $b=(0–2)/2=-1$ Thus, we have the following table giving points on the...
, 
and 
(ii) 
, 
and 
Given, $2b–a=8$……. (i) $5b–a=14$……. (ii) $b–2a=1$……… (iii) For equation (i), ⇒ $b=(a+8)/2$ When $a=-4$, we have $b=(-4+8)/2=2$ When $a=0$, we have $b=(0+8)/2=4$ a $-4$ $0$ b $2$ $4$ Thus, we have...
Given, $3a–4b–1=0$……. (i) $2a–(8/3)b+5=0$……. (ii) From equation (i), ⇒ $b=(3a–1)/4$ When $a=-1$, we have $b=(3(-1)–1)/4=-1$ When $a=3$, we have $b=(3(3)–1)/4=2$ a $-1$ $3$ b $-1$ $2$ Thus, we have...
Given, $2b–a=9$……. (i) $6b–3a=21$……. (ii) For equation (i), ⇒ $b=(a+9)/2$ When $a=-3$, we get $b=(-3+9)/2=3$ When $a=-1$, we have $b=(-1+9)/2=4$ Thus, we have the following table giving points on...
Given, $a–2b=6$……. (i) $3a–6b=0$……. (ii) For equation (i), ⇒ $b=(a–6)/2$ When $a=6$, we have $b=(6–6)/2=0$ When $a=2$ we have $b=(2–6)/2=-2$ Thus, we have the following table giving points on the...
Given, $3a–5b=20$……. (i) $6a–10b=–40$……. (ii) From equation (i), ⇒ $b=(3a–20)/5$ When $a=5$, we have $b=(3(5)–20)/5=-1$ When $a=0$, we have $b=(3(0)–20)/5=-4$ Thus, we have the following table...
Given, $a–2b+11=0$……. (i) $3a–6b+33=0$……. (ii) From equation (i), ⇒ $b=(a+11)/2$ When $a=-1$, we get $b=(-1+11)/2=5$. When $a=-3$, we get $b=(-3+11)/2=4$. Thus, we have the following table giving...
Given, $3a+b=8$……. (i) $6a+2b=16$……. (ii) From equation (i), ⇒ $b=(8–3a)$ When $a=2$, we get $b=(8–3(2))=2$ When $a=3$, we get $b=(8–3(3))=-1$ Thus, we have the following table giving points on the...
Given, $a–2b=5$……. (i) $3a–6b=15$……. (ii) For equation (i), ⇒ $b=(a–5)/2$ When $a=3$, we get $b=(3–5)/2=-1$ When $a=5$, we get $b=(5–5)/2=0$ Thus, we have the following table giving points on the...
Given, $2a+3b=6$……. (i) $4a+6b=12$……. (ii) From equation (i), ⇒ $b=(6–2a)/3$ When $a=0$, we have $b=(6–2(0))/3=2$ When $a=3$, we have $b=(6–2(3))/3=0$ Thus, we have the following table giving points...
Given, $2a-3b+13=0$……. (i) $3a-2b+12=0$……. (ii) From equation (i), ⇒ $b=\left( 2a+13 \right)/3$ When $a=-5$, we get $b=\left( 2\left( -5 \right)+13 \right)/3=1$ When $a=-2$, we get $b=\left( 2\left(...
Given, $2a+3b=4$……. (i) $a–b+3=0$……. (ii) From equation (i), ⇒ $b=(4–2a)/3$ When $a=-1$, we get $b=(4–2(-1))/3=2$ When $a=2$, we get $b=(4–2(2))/3=0$ Thus, we have the following table giving Points...
Given, $a+b=4$……. (i) $2a–3b=3$……. (ii) From equation (i), ⇒ $b=(4–a)$ When $a=4$, we get $b=(4–4)=0$ When $a=2$, we get $b=(4–2)=2$ Thus, we have the following table giving Points on the line...
Given, $a–2b=6$……. (i) $3a–6b=0$……. (ii) From equation (i), ⇒ $b=(a–6)/2$ When $a=2$, we get $b=(2–6)/2=-2$ When $a=0$, we get $b=(0–6)/2=-3$ Thus, we have the following table giving Points on the...
Given, $a+b=6$……. (i) $a–b=2$……. (ii) From equation (i), ⇒ $b=(6–a)$ When $a=2$, we get $b=(6–2))=4$ When $a=3$, we get $b=(6–3)=3$ Thus, we have the following table giving Points on the line...
Solution: Given, $2a+b–3=0$……. (i) $2a–3b–7=0$……. (ii) From equation (i), ⇒$b=(3–2a)$ When a$=0$, we get b $=(3–2(0))=3$ When $a=1$, we get $b=(3–2(1))=1$ Thus, we have the following table giving...
Solution: Let the fraction be a/b According to the given information, $\left( a+1 \right)/\left( b-1 \right)\text{ }=\text{ }1$ $=>\text{ }a\text{ }\text{ }b\text{...
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7 By the method of elimination: $\begin{align} & \begin{array}{*{35}{l}} 3x\text{ }\text{ }5y\text{ }\text{ }4\text{ }=\text{ }0\text{ }\ldots...
Solution: Given, $3a+b+1=0$ ……. (i) $2a–3b+8=0$……. (ii) From equation (i), ⇒ $b=-(1+3a)$ When $a=0$, we have $b=-1$ When $a=-1$, we have $b=2$ a $-1$ $0$ b $2$ $-1$ Thus, we have the following table...
Solution: Given, $a–2b=5$……. (i) $2a+3b=10$……. (ii) From equation (i) we get, ⇒ $b=(a–5)/2$ When we Put $b=0$, we get, $a=5$ When we Put $a=1$, we get, $b=-2$ Thus, we have the following table...
Solution: Given, $a+b=3$……. (i) $2a+5b=12$……. (ii) From equation (i), When we Put $b=0$, we get, $a=3$ When we Put $a=0$, we get, $b=3$ Thus, we have the following table giving Points on the line...
and route of train B is given by the equation 
. Represent the route of train A and B graphically.Solution- Given, linear equations represents the routes of train A and train B, $3x+4y–12=0$………………………….. (i) $6x+8y–48=0$ ………………………….. (ii) Represent these equations graphically, we need at least...
Let the present age of Ankur and his niece be x and y respectively. Seven years ago, Age of Ankur $=x–7$ and Age of his niece $=y–7$ According to question, $x–7=7(y–7)$ ⇒ $x–7y=-42$ ……… (i) After...
and a costs of Hoopla is 
. If she spent 
in the fair, represent the given situation algebraically as well as graphically.Let ‘a’ be the number of rides Radhika had on the giant wheel. And, let ‘b’ be the number of times she played Hoopla. According to question, we can write the equations. $a=(1/2)b$ ⇒ $a-2b=0$……. (i)...
Solution: Let the fixed charge of the car be ₹ x and, Let the variable charges of the car be ₹ y per km. So according to the question, we get \[2\] equations \[x\text{ }+\text{ }12y\text{ }=\text{...
Solution: We need to prove that ΔABC is right angled. \[\angle A\text{ }=\text{ }{{x}^{o}},\angle B\text{ }=\text{ }3{{x}^{o}}\]and \[\angle C\text{ }=\text{ }{{y}^{o}}\] Sum of the three angles in...
Solution: Let the total number of correct answers be x and the total number of incorrect answers be y. Their sum will give the total number of questions in the test = x + y According to the question...
Solution: The sum of the opposite angles of cyclic quadrilateral should be 180o. And, in the cyclic quadrilateral ABCD, Angles \[\angle A\] and \[\angle C\] & angles \[\angle B\] and \[\angle...
Solution: Given, \[\angle A\text{ }=\text{ }{{x}^{o}}\] \[\angle B\text{ }=\text{ }{{\left( 3x\text{ }\text{ }2 \right)}^{o}},\] \[\angle C\text{ }=\text{ }{{y}^{o}}\]Also given, \[\angle C\text{...
Solution: Let the time required for a man alone to finish the work be x days The time required for a boy alone to finish the work be y days. Then, The work done by a man in one day \[=\text{ }1/x\]...
Solution: Let he money with A be ₹ x and the money with B be ₹ y. According to the question, Case 1: If A gives ₹ \[30\]to B, then B will have twice the money left with A. Equation, \[y\text{...
Solution: Leave the pay alone signified by x and the consumption be meant by y. Then, at that point, from the inquiry we have The pay of X is ₹ 8x and the use of X is 19y. The pay of Y is ₹ 7x and...
Solution: Let the length and breadth of the rectangle be x units and y units. The area of rectangle = xy sq.units Length is increased by 3 m ⇒ The new length is \[x+3\] Breadth is reduced by 4 m...
Solution: Let the length and breadth of the rectangle be x units and y units respectively. Hence, the area of rectangle = xy sq.units Length is increased by 7 m ⇒ now, the new length is...
Solution: Let the length and breadth of the rectangle be x units and y units respectively. The area of rectangle = xy sq.units Case 1: Length is increased by 2 units ⇒ The new length is \[x+2\]units...
\[\mathbf{5}.\] \[\left( \mathbf{x}\text{ }+\text{ }\mathbf{y} \right)/\text{ }\mathbf{xy}\text{ }=\text{ }\mathbf{2}\] \[\left( \mathbf{x}\text{ }\text{ }\mathbf{y} \right)/\text{...