Find the values of 'a' and 'b' for which the system of equations 3x + y = 3 and (a - b)x + (a + b)y = 3a + b - 3 has infinite solutions.

Question

Accepted Answer

For a system of two linear equations to have infinite solutions, the two equations must be dependent, i.e., one equation is a scalar multiple of the other.

Given equations:
1) 3x + y = 3
2) (a - b)x + (a + b)y = 3a + b - 3

For infinite solutions, the ratios of coefficients of x, y, and the constants must be equal:

(3) / (a - b) = (1) / (a + b) = (3) / (3a + b - 3)

Let this common ratio be k.

From the first two ratios:
3 / (a - b) = 1 / (a + b)
Cross-multiplied:
3(a + b) = a - b
3a + 3b = a - b
Bring all terms to one side:
3a + 3b - a + b = 0
(3a - a) + (3b + b) = 0
2a + 4b = 0
Divide by 2:
a + 2b = 0
=> a = -2b

Now, from the first and third ratios:
3 / (a - b) = 3 / (3a + b - 3)
Cross-multiplied:
3(3a + b - 3) = 3(a - b)
Divide both sides by 3:
3a + b - 3 = a - b
Bring all terms to one side:
3a + b - 3 - a + b = 0
(3a - a) + (b + b) - 3 = 0
2a + 2b - 3 = 0
Divide by 2:
a + b - 3/2 = 0
=> a + b = 3/2

Recall from above: a = -2b
Substitute into a + b = 3/2:
-2b + b = 3/2
-b = 3/2
b = -3/2

Then a = -2b = -2 * (-3/2) = 3

Therefore, a = 3 and b = -3/2.

Hence, option 3 is correct.

Answer

Answer

Answer