Pair of Linear Equations in Two Variables - Solved Examples

Exercise 3.6

Exercise 3.4

Hence the solution for the given equations can be written as
$\dfrac{x}{(1)(-8) - (2)(-5)}$ = $\dfrac{y}{(-5)(3) - (-8)(2)}$ = $\dfrac{1}{(2)(2) - (1)(3)}$

$\dfrac{x}{-8 + 10}$ = $\dfrac{y}{-15 + 16}$ = $\dfrac{1}{4 - 3}$

$\dfrac{x}{2} = \dfrac{y}{1} = \dfrac{1}{1}$

$⇒ x = 2$ and $y = 1$


(iii)
$3x − 5y = 20$
$6x − 10y = 40$

Comparing equation
$3x - 5y – 20 = 0$ with $a_1x + b_1y + c_1 = 0$ and
$6x - 10y – 40 = 0$ with $a_2x + b_2y + c_2 = 0$, we get,
$a_1 = 3$, $b_1 = -5$, $c_1 = -20$ and $a_2 = 6$, $b_2 = -10$, $c_2 = -40$

In this case, $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ = $\dfrac{c_1}{c_2}$ which means that the two lines coincide with each other.
Hence there are infinitely many solutions.



(iv)
$x − 3y – 7 = 0$
$3x − 3y – 15 = 0$

Comparing equation
$x - 3y – 7 = 0$ with $a_1x + b_1y + c_1 = 0$ and
$3x - 3y – 15 = 0$ with $a_2x + b_2y + c_2 = 0$, we get,
$a_1 = 1, b_1 = -3, c_1 = -7$ and $a_2 = 3, b_2 = -3, c_2 = -15$

In this case, $\dfrac{a_1}{a_2} ≠ \dfrac{b_1}{b_2}$ which means that there is a unique solution for the given equations

Hence the solution for the given equations can be written as
$\dfrac{x}{(-3)(-15) - (-3)(-7)}$ = $\dfrac{y}{(-7)(3) - (-15)(1)}$ = $\dfrac{1}{(-3)(1) - (-3)(3)}$

$\dfrac{x}{45 - 21}$ = $\dfrac{y}{-21 + 15}$ = $\dfrac{1}{-3 + 9}$

$\dfrac{x}{24} = \dfrac{y}{-6} = \dfrac{1}{6}$

$⇒ x = 4$ and $y = -1$

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
$2x + 3y = 7$
$(a − b) x$ + $(a + b) y$ = $3a + b – 2$

(ii) For which value of k will the following pair of linear equations have no solution?
$3x + y = 1$
$(2k − 1) x + (k − 1) y = 2k + 1$

(i)
$2x + 3y = 7$
$(a − b) x$ + $(a + b) y$ = $3a + b – 2$

Comparing equation
$2x + 3y – 7 = 0$ with $a_1x + b_1y + c_1 = 0$ and
$(a - b)x$ + $(a + b)y$ – $3a - b + 2 = 0$ with $a_2x + b_2y + c_2 = 0$, we get,
$a_1 = 2, b_1 = 3, c_1 = -7$ and $a_2 = (a - b)$, $b_2 = (a + b)$, $c_2 = – 3a - b + 2$

For linear equations to have infinite solutions, they should meet criterion that $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ = $\dfrac{c_1}{c_2}$
$⇒ \dfrac{2}{a - b}$ = $\dfrac{3}{a + b}$ = $\dfrac{-7}{2 - b - 3a}$

$⇒ \dfrac{2}{a - b}$ = $\dfrac{3}{a + b}$ and $\dfrac{-7}{2 - b - 3a}$

$⇒ 2a + 2b = 3a - 3b$ and $6 - 3b - 9a = -7a -7b$
$⇒ a = 5b$ -----(i) and $-2a = -4b - 6$ -----(ii)

Putting (i) in (ii)
$-2 (5b) = -4b - 6$ ⇒ $-10b + 4b = -6$ ⇒ $b = 1$

Putting value of b in (i), we get
$a = 5 × 1 ⇒ 5$

Therefore, $a = 5$ and $b = 1$

(ii)
$3x + y = 1$
$(2k − 1) x + (k − 1) y = 2k + 1$

Comparing equation
$3x + y – 1 = 0$ with $a_1x + b_1y + c_1 = 0$ and
$(2k - 1)x$ + $(k - 1)y$ – $2k - 1 = 0$ with $a_2x + b_2y + c_2 = 0$, we get,
$a_1 = 3, b_1 = 1, c_1 = -1$ and $a_2 = (2k - 1)$, $b_2 = (k - 1)$, $c_2 = – 2k - 1$

Linear Equations have no solutions if $\dfrac{a_1}{a_2}$ = $\dfrac{b_1}{b_2}$ ≠ $\dfrac{c_1}{c_2}$
$⇒ \dfrac{3}{2k - 1}$ = $\dfrac{1}{k - 1}$ ≠ $\dfrac{-1}{-2k - 1}$

$⇒ 3k - 3 = 2k - 1$
$⇒ k = 2$

Solve the following pair of linear equations by the substitution and cross-multiplication methods:
$8x + 5y = 9$
$3x + 2y = 4$

$8x + 5y = 9$ ----- (i)
$3x + 2y = 4$ ----- (ii)

Substitution Method
From equation (i),
$5y = 9 - 8x ⇒ y = \dfrac{9 - 8x}{5}$

Putting the value of $y$ in (ii), we get,
$3x + 2 × \dfrac{9 - 8x}{5} = 4$

$3x + \dfrac{18 - 16x}{5} = 4$

$15x + 18 - 16x = 20$

$x = -2$

Putting the value of x in (i),
$8(-2) + 5y = 9 ⇒ 5y = 25$ $⇒ y = 5$

Therefore, $x = -2$ and $y = 5$

Cross multiplication Method
$8x + 5y = 9$ ----- (i)
$3x + 2y = 4$ ----- (ii)

Hence the solution for the given equations can be written as
$\dfrac{x}{(5)(-4) - (-9)(2)}$ = $\dfrac{y}{(-9)(3) - (8)(-4)}$ = $\dfrac{1}{(8)(2) - (5)(3)}$

$\dfrac{x}{-20 + 18}$ = $\dfrac{y}{-27 + 32}$ = $\dfrac{1}{16 - 15}$

$\dfrac{x}{-2} = \dfrac{y}{5} = \dfrac{1}{1}$

$⇒ x = -2$ and $y = 5$

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

(i)Let fixed monthly charge be Rs $x$ and let charge of food for one day be Rs $y$
According to given conditions,
$x + 20y = 1000$ ----- (i)
and
$x + 26y = 1180$ ----- (ii)

Subtracting equation (i) from equation (ii), we get
$6y = 180 ⇒ y = 30$

Putting value of $y$ in (i), we get
$x + 20 (30) = 1000 ⇒ x = 1000 – 600 = 400$

Therefore, fixed monthly charges = Rs 400 and charges of food for one day = Rs 30

(ii) Let numerator be x and the denominator be y

According to given conditions,
$\dfrac{x - 1}{y}$ = $\dfrac{1}{3}$

$⇒ 3x – 3 = y$
$⇒ 3x - y = 3$ ----- (i)

$\dfrac{x}{y + 8}$ = $\dfrac{1}{4}$

$⇒ 4x = y + 8$
$⇒ 4x – y = 8$ ----- (ii)

Subtracting equation (i) from (ii), we get
$4x - y - (3x - y) = 8 - 3 $ $⇒ x = 5$

Putting value of $x$ in (i), we get
$3 (5) - y = 3$ $⇒ 15 - y = 3 ⇒ y = 12$

Therefore, numerator = 5, denominator = 12 and the fraction is $\dfrac{5}{12}$

(iii) Let number of correct answers be $x$ and let number of wrong answers be $y$

According to given conditions,
$3x - y = 40$ ----- (i)
and
$4x - 2y = 50$ ----- (ii)

From equation (i), $y = 3x - 40$
Putting this in (ii), we get
$4x - 2 (3x - 40) = 50$
$⇒ 4x - 6x + 80 = 50$
$⇒ -2x = -30 ⇒ x = 15$

Putting value of $x$ in (i), we get
$3 (15) - y = 40$
$⇒ 45 - y = 40$
$⇒ y = 45 - 40 = 5$

Therefore, number of correct answers = 15 and number of wrong answers = 5

Total questions = $x + y$ = 15 + 5 = 20

(iv)Let the speed of car which starts from A be x km/hr
Let speed of car which starts from B be y km/hr

According to given conditions (Assuming x > y)
$5x - 5y = 100$
$⇒ x - y = 20$ ----- (i)

$x + y = 100$ ----- (ii)

Adding (i) and (ii), we get
$2x = 120 ⇒ x = 60$ km/hr

Putting value of $x$ in (i), we get
$60 - y = 20$ $⇒ y = 60 - 20 = 40$ km/hr

Therefore, speed of car starting from A = 60 km/hr and Speed of car starting from point B = 40 km/hr

(v) Let length of rectangle be $x$ units and let breadth of rectangle be $y$ units

Area = xy square units.

According to given conditions,
$xy - 9 = (x − 5) (y + 3)$
$⇒ xy – 9 = xy + 3x − 5y – 15$ $⇒ 3x − 5y = 6$ ----- (i)

$xy + 67 = (x + 3) (y + 2)$
$⇒ xy + 67 = xy + 2x + 3y + 6$
$⇒ 2x + 3y = 61$ ----- (ii)

From equation (i),
$3x = 6 + 5y$
$⇒ x = \dfrac{6 + 5y}{3}$

Putting this in (ii), we get
$2 \dfrac{6 + 5y}{3} + 3y = 61$
$⇒ 12 + 10y + 9y = 183$
$⇒ 19y = 171 ⇒ y = 9$ units
Putting value of $y$ in (ii), we get
$2x + 3 (9) = 61$
$⇒ 2x = 61 – 27 = 34$
$⇒ x = 17$ units

Therefore, length = 17 units and breadth = 9 units