Question

Determine the value of a and b for which the following system of linear equations have infinite number of solutions 2 − ( − 4) = 2 + 1 4 − ( − 1) = 5 − 1

Answer 1

The correct question is,

Find the values of a and b for which the following  system of linear equations has an infinite number of solutions 2x−3y = 7; (a+b)x−(a+b−3)y = 4a+b

Given:

2x−3y = 7; (a+b)x−(a+b−3)y = 4a+b

To find:

The values of a and b for which the given system of linear equations has an infinite number of solutions.

Solution:

From given, we have the system of linear equations

2x − 3y = 7

(a+b)x − (a+b−3)y = 4a+b

So, we get,

a1 = 2, b1 = -3, c1 = 7

a2 = a + b, b2 = - (a + b - 3), c2 = 4a + b

For the system of linear equations to have infinite number of solutions, we have a condition:

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

$\Rightarrow \dfrac{2}{a + b}=\dfrac{-3}{- (a + b - 3)}=\dfrac{7}{4a + b}$

$\Rightarrow \dfrac{2}{a + b}=\dfrac{3}{a + b - 3}=\dfrac{7}{4a + b}$

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

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

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

2a + 2b - 6 = 3a + 3b

a + b = -6 ............(1)

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

$\dfrac{3}{a + b - 3}=\dfrac{7}{4a + b}$

3 (4a + b) = 7 (a + b - 3)

12a + 3b = 7a + 7b - 21

5a - 4b = -21 ...........(2)

solving equations (1) and (2), we get,

a = -5 and b = -1