Find the values of $a$ and $b$ for which the following system of equations has infinitely many solutions:
$(2 a - 1) x - 3 y = 5$
$3 x + (b - 2) y = 3$

Open in App

Solution

Condition for infinitely many solutions is:

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Given,

$(2 a - 1) x - 3 y = 5$

$3 x + (b - 2) y = 3$

$a_{1} = 2 a - 1, a_{2} = 3$

$b_{1} = - 3, b_{2} = b - 2$

$c_{1} = 5, c_{2} = 3$

$\frac{2 a - 1}{3} = \frac{3}{b - 2} = \frac{5}{3}$

solving $\frac{2 a - 1}{3} = \frac{5}{3}$

we get,

$a = 3$

solving $\frac{3}{b - 2} = \frac{5}{3}$

we get,

$b = \frac{19}{5}$

Suggest Corrections

Similar questions

(2 a - 1) x - 3 y = 5

3 x + (b - 2) y = 3

(i) (2 a - 1) x - 3 y = 5 3 x + (b - 2) y = 3 (i i) 2 x - (2 a + 5) y = 5 (2 b + 1) x - 9 y = 15 (i i i) (a - 1) x + 3 y = 2 6 x + (1 - 2 b) y = 6 (i v) 3 x + 4 y = 12 (a + b) x + 2 (a - b) y = 5 a - 1 (v) 2 x + 3 y = 7 (a - b) x + (a + b) y = 3 a + b - 2 (v i) 2 x + 3 y - 7 = 0 (a - 1) x + (a + 1) y = (3 a - 1) (v i i) 2 x + 3 y = 7 (a - 1) x + (a + 2) y = 3 a

(v i i i) x + 2 y = 1 (a - b) x + (a + b) y = a + b - 2

(i x) 2 x + 3 y = 7 2 a x + a y = 28 - b y

Find the values of a and b for which the following system of equations has infinitely many solutions:

(i) (2a - 1) x - 3y = 5

3x + (b - 2) y = 3

(ii) 2x - (2a + 5) y = 5

(2b + 1) x - 9y = 15

(iii) (a - 1) x + 3y = 2

6x + (1 - 2b) y = 6

(iv) 3x + 4y = 12

(a + b) x + 2 (a - b) y = 5a - 1

(v) 2x + 3y = 7

(a - b) x + (a + b) y = 3a + b - 2

(vi) 2x + 3y - 7 = 0

(a - 1) x + (a + 1) y = (3a - 1)

(vii) 2x + 3y = 7

(a - 1) x + (a + 2) y = 3a

(viii) x + 2y = 1

(a - b) x + (a + b) y = a + b - 2

(ix) 2x + 3y = 7

2ax + ay = 28 - by

(2 a - 1) x + 3 y = 5, 3 x + (b - 1) y = 2 .

2 x + 3 y = k (k - 1) x + (k + 2) y = 3 k

Join BYJU'S Learning Program