NCERT Solutions for Exercise 3.4 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Q1(i) Solve the following pair of linear equations by the elimination method and the substitution method :

$x+y=5\text{and}2x-3y=4$

Answer:

Elimination Method:

Given, equations

$$\begin{gathered} x+y=5............(1)\text{and} \\ 2x-3y=4........(2) \end{gathered}$$

Now, multiplying (1) by 3 we get

$3x+3y=15............(3)$

Now, adding (2) and (3), we get

$2x-3y+3x+3y=4+15$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value in (1), we get

$\frac{19}{5}+y=5$

$\Rightarrow y=5-\frac{19}{5}$

$\Rightarrow y=\frac{6}{5}$

Hence,

$x=\frac{19}{5}andy=\frac{6}{5}$

Substitution method :

Given, equations

$$\begin{gathered} x+y=5............(1)\text{and} \\ 2x-3y=4........(2) \end{gathered}$$

Now, from (1) we have,

$y=5-x.......(3)$

Substituting this value in (2)

$2x-3(5-x)=4$

$\Rightarrow 2x-15+3x=4$

$\Rightarrow 5x=19$

$\Rightarrow x=\frac{19}{5}$

Substituting this value of x in (3)

$\Rightarrow y=5-x=5-\frac{19}{5}=\frac{6}{5}$

Hence,

$x=\frac{19}{5}andy=\frac{6}{5}$

Q1(ii) Solve the following pair of linear equations by the elimination method and the substitution method :

$3x+4y=10\text{and}2x-2y=2$

Answer:

Elimination Method:

Given, equations

$$\begin{gathered} 3x+4y=10............(1)\text{and} \\ 2x-2y=2..............(2) \end{gathered}$$

Now, multiplying (2) by 2 we get

$4x-4y=4............(3)$

Now, adding (1) and (3), we get

$3x+4y+4x-4y=10+4$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Putting this value in (2) we get

$2(2)-2y=2$

$\Rightarrow 2y=2$

$\Rightarrow y=1$

Hence,

$x=2andy=1$

Substitution method :

Given, equations

$$\begin{gathered} 3x+4y=10............(1)\text{and} \\ 2x-2y=2..............(2) \end{gathered}$$

Now, from (2) we have,

$y=\frac{2x-2}{2}=x-1.......(3)$

Substituting this value in (1)

$3x+4(x-1)=10$

$\Rightarrow 3x+4x-4=10$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=x-1=2-1=1$

Hence, $x=2andy=1$

Q1(iii} Solve the following pair of linear equations by the elimination method and the substitution method: (iii) $3x-5y-4=0\text{and}9x=2y+7$

Answer:

Elimination Method:

Given, equations

$$\begin{gathered} 3x-5y-4=0..........(1)\text{and} \\ 9x=2y+7 \end{gathered}$$

$\Rightarrow 9x-2y-7=0........(2)$

Now, multiplying (1) by 3 we get

$9x-15y-12=0............(3)$

Now, subtracting (3) from (2), we get

$9x-2y-7-9x+15y+12=0$

$\Rightarrow 13y+5=0$

$\Rightarrow y=\frac{-5}{13}$

Putting this value in (1), we get

$3x-5(\frac{-5}{13})-4=0$

$\Rightarrow 3x=4-\frac{25}{13}$

$\Rightarrow 3x=\frac{27}{13}$

$\Rightarrow x=\frac{9}{13}$

Hence,

$x=\frac{9}{13}andy=-\frac{5}{13}$

Substitution method :

Given, equations

$$\begin{gathered} 3x-5y-4=0..........(1)\text{and} \\ 9x=2y+7 \end{gathered}$$

$\Rightarrow 9x-2y-7=0........(2)$

Now, from (2) we have,

$y=\frac{9x-7}{2}.......(3)$

Substituting this value in (1)

$3x-5\left(\frac{9x-7}{2}\right)-4=0$

$\Rightarrow 6x-45x+35-8=0$

$\Rightarrow -39x+27=0$

$\Rightarrow x=\frac{27}{39}=\frac{9}{13}$

Substituting this value of x in (3)

$\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}$

Hence, $x=\frac{9}{13}andy=-\frac{5}{13}$

Q1(iv) Solve the following pair of linear equations by the elimination method and the substitution method :(iv) $\frac{x}{2}+\frac{2y}{3}=-1\text{and}x-\frac{y}{3}=3$

Answer:

Elimination Method:

Given, equations

$$\begin{gathered} \frac{x}{2}+\frac{2y}{3}=-1........(1)\text{and} \\ x-\frac{y}{3}=3............(2) \end{gathered}$$

Now, multiplying (2) by 2, we get

$2x-\frac{2y}{3}=6............(3)$

Now, adding (1) and (3), we get

$\frac{x}{2}+\frac{2y}{3}+2x-\frac{2y}{3}=-1+6$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Putting this value in (2), we get

$2-\frac{y}{3}=3$

$\Rightarrow \frac{y}{3}=-1$

$\Rightarrow y=-3$

Hence,

$x=2andy=-3$

Substitution method :

Given, equations

$$\begin{gathered} \frac{x}{2}+\frac{2y}{3}=-1........(1)\text{and} \\ x-\frac{y}{3}=3............(2) \end{gathered}$$

Now, from (2) we have,

$y=3(x-3)......(3)$

Substituting this value in (1)

$\frac{x}{2}+\frac{2(3(x-3))}{3}=-1$

$\Rightarrow \frac{x}{2}+2x-6=-1$

$\Rightarrow \frac{5x}{2}=5$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=3(x-3)=3(2-1)=-3$

Hence, $x=2andy=-3$

Q2(i) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

Answer:

Let the numerator of the fraction be x, and the denominator is y,

Now, according to the question,

$\frac{x+1}{y-1}=1$

$\Rightarrow x+1=y-1$

$\Rightarrow x-y=-2.........(1)$

Also,

$\frac{x}{y+1}=\frac{1}{2}$

$\Rightarrow 2x=y+1$

$\Rightarrow 2x-y=1..........(2)$

Now, subtracting (1) from (2), we get

$x=3$

Putting this value in (1)

$3-y=-2$

$\Rightarrow y=5$

Hence, $x=3andy=5$

And the fraction is: $\frac{3}{5}$

Q2(ii) Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Answer:

Let the age of Nuri be x and the age of Sonu be y.

Now, according to the question

$x-5=3(y-5)$

$\Rightarrow x-5=3y-15$

$\Rightarrow x-3y=-10.........(1)$

Also,

$x+10=2(y+10)$

$\Rightarrow x+10=2y+20$

$\Rightarrow x-2y=10........(2)$

Now, subtracting (1) from (2), we get

$y=20$

Putting this value in (2)

$x-2(20)=10$

$\Rightarrow x=50$

Hence, the age of Nuri is 50 and the age of Nuri is 20.

Q2(iii) Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer:

Let the unit digit of the number be x and the 10's digit be y.

Now, according to the question,

$x+y=9.......(1)$

Also

$9(10y+x)=2(10x+y)$

$\Rightarrow 90y+9x=20x+2y$

$\Rightarrow 88y-11x=0$

$\Rightarrow 8y-x=0.........(2)$

Now adding (1) and (2), we get,

$\Rightarrow 9y=9$

$\Rightarrow y=1$

Now putting this value in (1)

$x+1=9$

$\Rightarrow x=8$

Hence, the number is 18.

Q2(iv) Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Answer:

Let the number of Rs 50 notes be x and the number of Rs 100 notes be y.

Now, according to the question,

$x+y=25..........(1)$

And

$50x+100y=2000$

$\Rightarrow x+2y=40.............(2)$

Now, subtracting (1) from (2), we get

$y=15$

Putting this value in (1).

$x+15=25$

$\Rightarrow x=10$

Hence, Meena received 10, 50 Rs notes and 15, 100 Rs notes.

Q2(v) Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method : (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

Let the fixed charge be x, and per day charge is y.

Now, according to the question,

$x+4y=27...........(1)$

And

$x+2y=21...........(2)$

Now, Subtracting (2) from (1). We get,

$4y-2y=27-21$

$\Rightarrow 2y=6$

$\Rightarrow y=3$

Putting this in (1)

$x+4(3)=27$

$\Rightarrow x=27-12=15$

Hence, the fixed charge is 15 Rs and the per-day charge is 3 Rs.

Also Read:

• Pair of Linear Equations in Two Variables Exercise 3.1
  
• Pair of Linear Equations in Two Variables Exercise 3.2