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Pair of Linear Equations in Two Variables - Solved Examples

Exercise 3.4

Exercise 3.2

NCERT Exercise 3.3



1
Solve the following pairs of linear equations by the substitution method:
(i)
$x + y = 14$
$x - y = 4$

(ii)
$s - t = 3$
$\dfrac{s}{3} + \dfrac{t}{2} = 6$

(iii)
$3x - y = 3$
$9x - 3y = 9$

(iv)
$0.2x + 0.3y = 1.3$
$0.4x + 0.5y = 2.3$

(v)
$\sqrt{2}x + \sqrt{3}y = 0$
$\sqrt{3}x - \sqrt{8}y = 0$

(iv)
$\dfrac{3}{2}x - \dfrac{5}{3}y = -2$
$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$



Solution

(i)
$x + y = 14$ -----(i)
$x - y = 4$ -----(ii)

From (ii), $x = y + 4$

Putting this in equation (i), we get
$4 + y + y = 14 ⇒ 2y = 10 ⇒ y = 5$

Putting value of y in equation (i), we get,
$x + 5 = 14 ⇒ x = 14 – 5 = 9$
Therefore, $x = 9$ and $y = 5$


(ii)
$s - t = 3$ -----(i)
$\dfrac{s}{3} + \dfrac{t}{2} = 6$ -----(ii)

From (i), $s = t + 3$

Putting this in equation (ii), we get
$\dfrac{t + 3}{3} + \dfrac{t}{2} = 6$
$\dfrac{6 + 2t + 3}{6} = 6$
$⇒ 5t + 6 = 36 ⇒ 5t = 30 ⇒ t = 6$

Putting value of t in equation (i), we get,
$s - 6 = 3 ⇒ s = 9$
Therefore, $s = 9$ and $t = 6$


(iii)
$3x - y = 3$ -----(i)
$9x - 3y = 9$ -----(ii)

From (i), $3x - y = 3 ⇒ y = 3x - 3$

Putting this in equation (ii), we get
$9x - 3(3x - 3) = 9 ⇒ 9 = 9$
Hence, this pair of linear equations have infinitely many solutions


(iv)
$0.2x + 0.3y = 1.3$ -----(i)
$0.4x + 0.5y = 2.3$ -----(ii)

From (i), $0.2x + 0.3y = 1.3$
$⇒ x = \dfrac{1.3 - 0.3y}{0.2}$
Putting this in equation (ii), we get
$0.4× \dfrac{1.3 - 0.3y}{0.2} + 0.5y = 2.3$
$2(1.3 - 0.3y) + 0.5y = 2.3$
$2.6 - 0.6y + 0.5y = 2.3 ⇒ y = 3$
Putting value of y in equation (i), we get,
$0.2x + 0.3 × 3 = 1.3 ⇒ x = 2$
Therefore, $x = 2$ and $y = 3$


(v)
$\sqrt{2}x + \sqrt{3}y = 0$ -----(i)
$\sqrt{3}x - \sqrt{8}y = 0$ -----(ii)

From (i), $y = - \dfrac{\sqrt{2}x}{\sqrt{3}}$

Putting this in equation (ii), we get
$\sqrt{3}x - \sqrt{8}$ × $\dfrac{-\sqrt{2}x}{\sqrt{3}} = 0$
$⇒ 3x + 4x = 0 ⇒ x = 0$

Putting value of x in equation (i), we get,
$\sqrt{3} × 0 - \sqrt{8}y = 0 ⇒ y = 0$

Therefore, $x = 0$ and $y = 0$


(vi)
$\dfrac{3}{2}x - \dfrac{5}{3}y = -2$ -----(i)
$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$ -----(ii)

From (ii), $x = 3× \left(\dfrac{13}{6} - \dfrac{y}{2}\right)$

$⇒ x = \left(\dfrac{13}{2} - \dfrac{3y}{2}\right)$
Putting this in equation (i), we get
$\dfrac{3}{2} × \left(\dfrac{13}{2} - \dfrac{3y}{2}\right) - \dfrac{5}{3}y = -2$
$⇒ \left(\dfrac{39}{4} - \dfrac{9y}{4}\right) - \dfrac{5y}{3} = -2$
$⇒ \dfrac{-27y - 20y}{12} = -2 - \dfrac{39}{4} $
$⇒ \dfrac{-47y}{12} = \dfrac{-47}{4} ⇒ y = 3$

Putting this in equation (ii), we get
$\dfrac{x}{3} + \dfrac{3}{2} = \dfrac{13}{6} ⇒ x = 2$ -----(ii)

Therefore, $x = 2$ and $y = 3$



2
Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.


Solution

$2x + 3y = 11$ ----- (i)
$2x − 4y = −24$ ----- (ii)

Using equation (ii), we get
$2x = −24 + 4y$ $⇒ x = −12 + 2y$

Putting this in equation (i), we get
$2 (−12 + 2y) + 3y = 11$
$⇒ −24 + 4y + 3y = 11$
$⇒ 7y = 35 ⇒ y = 5$

Putting value of y in equation (i), we get
$2x + 3 (5) = 11$
$⇒ 2x + 15 = 11$
$⇒ 2x = 11 – 15 = −4 ⇒ x = −2$
Therefore, $x = −2$ and $y = 5$

Putting values of x and y in y = mx + 3, we get
$5 = m (−2) + 3$
$⇒ 5 = −2m + 3$
$⇒ −2m = 2 ⇒ m = −1$



3
Form a pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii)The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes $\dfrac{9}{11}$, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes $\dfrac{5}{6}$. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?


Solution

(i) Let first number be $x$ and second number be $y$. Let $x$ be greater than $y$
According to given conditions, we have
$x – y = 26$ ----- (i)
$x = 3y$ ----- (ii)

Putting equation (ii) in (i), we get
$3y – y = 26$ $⇒ 2y = 26$ $⇒ y = 13$

Putting value of $y$ in equation (ii), we get

$x = 3y = 3 × 13 = 39$

Therefore, the two numbers are 13 and 39.


(ii) Let smaller angle be $x$ and let larger angle be $y$
According to given conditions, we have
y = x + 18 ----- (i)
As the angles are supplementary angles, $x + y = 180°$---- (ii)

Putting (i) in equation (ii), we get
$x + x + 18 = 180$
$⇒ 2x = 180 – 18 = 162$ $⇒ x = 81°$
Putting value of $x$ in equation (i), we get

$y = x + 18 = 81 + 18 = 99$
Therefore, two angles are 81° and 99°.


(iii) Let cost of each bat be Rs $x$ and let cost of each ball be Rs $y$

According to given conditions, we have
$7x + 6y = 3800$ ----- (i)
and
$3x + 5y = 1750$ ----- (ii)

From equation (i), we get
$7x = 3800 − 6y ⇒ x =$ $\dfrac{3800 - 6y}{7}$

Putting this in equation (ii), we get
$3 × \dfrac{3800 - 6y}{7} + 5y = 1750$

$⇒ \dfrac{11400 - 18y}{7} + 5y = 1750$

$⇒ \dfrac{-18y}{7} + 5y = 1750 - \dfrac{11400}{7}$

$⇒ \dfrac{35y - 18y}{7} = \dfrac{12250 - 11400}{7}$

$⇒ 17y = 850 ⇒ y = 50$

Putting value of y in (ii), we get
$3x + 250 = 1750$
$⇒ 3x = 1500 ⇒ x = 500$

Therefore, cost of each bat is Rs 500 and cost of each ball is Rs 50


(iv) Let the fixed charge be Rs $x$ and let charge for every km be Rs $y$

According to given conditions, we have
$x + 10y = 105$ ----- (i)
$x + 15y = 155$ ----- (ii)

Using equation (i), we get
$x = 105 − 10y$

Putting this in equation (ii), we get
$105 − 10y + 15y = 155$
$⇒ 5y = 50 ⇒ y = 10$

Putting value of $y$ in equation (i), we get
$x + 10 (10) = 105$
$⇒ x = 105 – 100 = 5$

Therefore, fixed charge = Rs 5 and charge per km = Rs 10

So, to travel a distance of 25 km, person will have to pay = Rs ($x + 25y$) = Rs (5 + 25 × 10) = Rs 255


(v) Let numerator be $x$ and let denominator be $y$

According to given conditions, we have
$\dfrac{x + 2}{y + 2} = \dfrac{9}{11}$ ----- (i)
$\dfrac{x + 3}{y + 3} = \dfrac{5}{6}$ ----- (ii)

Using equation (i), we get
$11(x + 2) = 9(y + 2)$
$⇒ 11x + 22 = 9y + 18$
$⇒ 11x = 9y – 4$ $⇒ x = \dfrac{9y - 4}{11}$

Putting value of x in equation (ii), we get
$6 × \left(\dfrac{9y - 4}{11} + 3 \right)$ = $5 (y + 3)$

$\dfrac{54y - 24}{11} + 18 $ = $5y + 15$

$\dfrac{54y}{11} - \dfrac{24}{11} + 18 $ = $5y + 15$

$\dfrac{54y}{11} - 5y$ = $15 - \dfrac{24}{11} - 18$

$\dfrac{54y - 55y}{11}$ = $\dfrac{24 - 33}{11}$

$⇒ y = 9$

Putting value of y in (i), we get
$\dfrac{x + 2}{9 + 2} = \dfrac{9}{11}$ ----- (i)
$⇒ x + 2 = 9 ⇒ x = 7$

Therefore, the fraction is $\dfrac{7}{9}$


(vi) Let present age of Jacob be $x$ years and that of his son be $y$ years

According to given conditions, we have
Five years hence
$(x + 5) = 3 (y + 5)$ ----- (i)
Five years ago
$(x − 5) = 7 (y − 5)$ ----- (ii)

From equation (i), we get
$x + 5 = 3y + 15$
$⇒ x = 10 + 3y$

Putting value of $x$ in equation (ii) we get
$10 + 3y – 5 = 7y − 35$
$⇒ −4y = −40 ⇒ y = 10$ years

Putting value of y in equation (i), we get
$x + 5 = 3 (10 + 5) = 45$
$⇒ x = 45 – 5 = 40$ years

Therefore, present age of Jacob is 40 years and present age of Jacob’s son is 10 years