Question

Solve for a and b:

2(a + b) – (a – b) = 6, 4(a – b) = 2(a + b) – 9

(a) \(a = \frac 34, b= 1\frac 34\)

(b) \(a = \frac 12,b= 1\frac 12\)

(c) \(a = 1\frac 34, b= \frac 34\)

(d) \(a = \frac 32, b= \frac 34\)

Answer 1

Correct option is (a) \(a = \frac 34, b= 1\frac 34\)

Given,

2(a + b) − (a − b) = 6

4(a − b) = 2(a + b) − 9

2(a + b) − (a − b) = 6

2a + 2b − a + b = 6

a + 3b = 6     ........ (1)

4(a − b) = 2(a + b) − 9

4a − 4b = 2a + 2b − 9

2a − 6b = −9    ...... (2)

On adding equations (1) and (2), we get

2a + 6b = 12

2a − 6b = −9

4a = 3

a = \(\frac 34\)

Putting the value of a in eq (1), we get

\(\frac 34\) + 3b = 6

3b = 6 - \(\frac 34\)

b = \(\frac{21} 4 \times \frac13\)

\(= \frac{21}{12}\)

\(= 1 \frac9{12}\)

\(= 1\frac 34\)

Hence, a = \(\frac 34\) and b = \( 1\frac 34\).

Answer 2

The correct answer is: (a) a = 3/4, b = 1³/₄