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Pair of Linear Equations in Two Variables - Solved Examples

Exercise 3.5

Exercise 3.3

NCERT Exercise 3.4



1
Solve the following pair of linear equations by the elimination method and the substitution method:

(i) $x + y = 5$ and $2x – 3y = 4$

(ii) $3x + 4y = 10$ and $2x – 2y = 2$

(iii) $3x − 5y – 4 = 0$ and $9x = 2y + 7$

(iv) $ \dfrac{x}{2} + \dfrac{2y}{3} = -1$ and $ x - \dfrac{y}{3} = 3$


Solution

(i)
$x + y = 5$ -----(i)

$2x – 3y = 4$ -----(ii)

Elimination method:
Multiplying equation (i) by 2, we get equation (iii)
$2x + 2y = 10$ -----(iii)

Subtracting equation (ii) from (iii), we get
$5y = 6 ⇒ y = \dfrac{6}{5} $

Putting value of y in (i), we get

$x + \dfrac{6}{5} = 5 ⇒ x = 5 − \dfrac{6}{5} = \dfrac{19}{5}$
Therefore, $x = \dfrac{19}{5}$ and $y = \dfrac{6}{5}$

Substitution method:
From equation (i), we get
$ x = 5 - y $

Putting this in equation (ii), we get
$ 2(5 - y) - 3y = 4$
$ ⇒ 10 - 5y = 4 ⇒ y = \dfrac{6}{5}$

Putting value of y in (i), we get
$ x + \dfrac{6}{5} = 5$
$ \dfrac{5x + 6}{5} = 5 ⇒ x = \dfrac{19}{5} $

Therefore, $x = \dfrac{19}{5}$ and $y = \dfrac{6}{5}$


(ii)
$3x + 4y = 10$ -----(i)

$2x – 2y = 2$ -----(ii)

Elimination method:
Multiplying equation (ii) by 2, we get equation (iii)
$4x - 4y = 4$ -----(iii)

Adding equation (i) and (iii), we get
$7x = 14 ⇒ x = 2 $

Putting value of x in (i), we get

$ 3 × 2 + 4y = 10 ⇒ y = 1$
Therefore, $x = 2$ and $y = 1$

Substitution method:
From equation (ii), we get
$ 2x = 2y + 2 ⇒ x = y + 1$

Putting this in equation (i), we get
$3(y + 1) + 4y = 10$
$ ⇒ 7y = 7 ⇒ y = 1$

Putting value of y in (ii), we get
$ 2x - 2 × 1 = 2$
$ ⇒ 2x - 2 = 2 ⇒ x = 2 $

Therefore, $x = 2$ and $y = 1$


(iii)
$3x − 5y – 4 = 0$ -----(i)

$9x = 2y + 7$ -----(ii)

Elimination method:
Multiplying equation (i) by 3, we get equation (iii)
$9x - 15y - 12 = 0$ -----(iii)

Subtracting equation (ii) and (iii), we get
$ -13y - 5 = 0 ⇒ y = \dfrac{-5}{13} $

Putting value of y in (i), we get

$ 3x -5 \dfrac{-5}{13} - 4 = 0 $
$⇒ 3x = 4 - \dfrac{25}{13} = \dfrac{27}{13}$
$⇒ x = \dfrac{9}{13}$
Therefore, $x = \dfrac{9}{13}$ and $y = \dfrac{-5}{13}$

Substitution method:
From equation (i), we get
$ 3x = 5y + 4 ⇒ x = \dfrac{5y + 4}{3}$

Putting this in equation (ii), we get
$9 \dfrac{5y + 4}{3} - 2y = 7$
$ ⇒ 12 + 15y - 2y = 7 ⇒ y = \dfrac{-5}{13}$

Putting value of y in (i), we get
$ 3x - 5 \dfrac{-5}{13} = 4$
$ ⇒ 3x = 4 - \dfrac{25}{13} = \dfrac{27}{13} $

$ ⇒ x = \dfrac{9}{13} $

Therefore, $x = \dfrac{9}{13}$ and $y = \dfrac{-5}{13}$




(iv)
$ \dfrac{x}{2} + \dfrac{2y}{3} = -1$ ----- (i)

$ x - \dfrac{y}{3} = 3$ -----(ii)

Elimination method:
Multiplying equation (ii) by 2, we get (iii)
$ 2x - \dfrac{2y}{3} = 3$ -----(iii)

Adding (iii) and (i), we get
$ ⇒ \frac{5}{2}x = 5 ⇒ x = 2$
Putting value of $x$ in (ii), we get

$ 2 - \dfrac{y}{3} = 3 ⇒ y = -3$
Therefore, x = 2 and y = −3

Substitution method:
From equation (ii), we get
$ x = \dfrac{y}{3} + 3 ⇒ x = \dfrac{y + 9}{3}$

Putting this in equation (i), we get
$ \dfrac{y + 9}{6} + \dfrac{2y}{3} = -1$

$ \dfrac{y + 9 + 4y}{6} = -1$

$ 5y + 9 = -6 ⇒ y = -3$

Putting value of y in (i), we get
$ \dfrac{x}{2} + \dfrac{2 × -3}{3} = -1$
$ \dfrac{x}{2} -2 = -1 ⇒ x = 2 $

Therefore, x = 2 and y = −3



2
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 ½ if we only add 1 to the denominator. What is the fraction?

(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?

(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.

(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.

(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.


Solution

(i) Let numerator be x and denominator be y
According to given condition, we have
$\dfrac{x + 1}{y - 1} = 1$
$\dfrac{x}{y + 1} = \dfrac{1}{2}$
$⇒ x + 1 = y – 1$ and $2x = y + 1$
$⇒ x – y = −2$ ----- (i) and $2x – y = 1$ ----- (ii)

Multiplying equation (i) by 2 we get (iii)
$2x − 2y = −4$ ----- (iii)
Subtracting equation (ii) from (iii), we get
$−y = −5 ⇒ y = 5$

Putting value of y in (i), we get
$x – 5 = −2 ⇒ x = −2 + 5 = 3$

Therefore, the fraction is $\dfrac{x}{y} = \dfrac{3}{5}$

(ii) Let present age of Nuri be x years and that of Sonu be y years
5 years ago, age of Nuri = $(x – 5)$ years and age of Sonu = $(y – 5)$ years
According to given condition, we have
$(x − 5) = 3 (y − 5)$
$⇒ x – 5 = 3y – 15$
$⇒ x − 3y = −10$ ----- (i)
10 years later, age of Nuri = $(x + 10)$ years and age of Sonu = $(y + 10)$ years
According to given condition, we have
$(x + 10) = 2 (y + 10)$
$⇒ x + 10 = 2y + 20$
$⇒ x − 2y = 10$ ----- (ii)

Subtracting equation (i) from (ii), we get
$y = 10 − (−10) = 20$ years
Putting value of y in (i), we get
$x – 3 (20) = −10$
$⇒ x – 60 = −10 ⇒ x = 50$ years

Therefore, present age of Nuri is 50 years and present age of Sonu is 20 years

(iii) Let digit at ten’s place be $x$ and digit at one’s place be $y$
According to given condition, we have
$x + y = 9$ ----- (i)
$9 (10x + y) = 2 (10y + x)$
$⇒ 90x + 9y = 20y + 2x$
$⇒ 88x = 11y ⇒ 8x = y$
$⇒ 8x – y = 0$ ----- (ii)

Adding (i) and (ii), we get
$9x = 9 ⇒ x = 1$

Putting value of $x$ in (i), we get
$1 + y = 9 ⇒ y = 9 – 1 = 8$
Therefore, the number is $10x + y = 10 (1) + 8 = 10 + 8 = 18$

(iv) Let number of Rs 100 notes be $x$ and let number of Rs 50 notes be $y$
According to given conditions, we have
$x + y = 25$ ----- (i)
$100x + 50y = 2000$
$⇒ 2x + y = 40$ ----- (ii)

Subtracting (ii) from (i), we get
$−x = −15⇒ x = 15$

Putting value of $x$ in (i), we get
$15 + y = 25 ⇒ y = 25 – 15 = 10$

Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10

(v) Let the fixed charge for 3 days be Rs $x$
Let additional charge for each day thereafter be Rs $y$

According to given condition, we have
$x + 4y = 27$ ----- (i)
$x + 2y = 21$ ----- (ii)

Subtracting (ii) from (i), we get
$2y = 6 ⇒ y = 3$

Putting value of $y$ in (i), we get
$x + 4 (3) = 27 ⇒ x = 27 – 12 = 15$
Therefore, fixed charge for 3 days is Rs 15 and additional charge for each day after 3 days is Rs 3