Solve for $a$ and $b$ $2 (a + b) - (a - b) = 6, 4 (a - b) = 2 (a + b) - 9$

a = \frac{3}{4}, b = 1 \frac{3}{4}

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a = \frac{1}{2}, b = 1 \frac{1}{2}

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a = 1 \frac{3}{4}, b = \frac{3}{4}

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a = \frac{3}{2}, b = \frac{3}{4}

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Solution

The correct option is C $a = \frac{3}{4}, b = 1 \frac{3}{4}$
Given,
$2 (a + b) - (a - b) = 6$
$4 (a - b) = 2 (a + b) - 9$
$2 (a + b) - (a - b) = 6$
$2 a + 2 b - a + b = 6$
$a + 3 b = 6$ ........ $(1)$
$4 (a - b) = 2 (a + b) - 9$
$4 a - 4 b = 2 a + 2 b - 9$
$2 a - 6 b = - 9$ ...... $(2)$
On adding equations $(1)$ and $(2)$ , we get
$2 a + 6 b = 12$
$2 a - 6 b = - 9$
__________________
$4 a = 3$
$a = \frac{3}{4}$
Putting the value of $a$ in eq $(1)$ , we get
$\frac{3}{4} + 3 b = 6$
$3 b = 6 - \frac{3}{4}$
$b = \frac{21}{4} \times \frac{1}{3} = \frac{21}{12} = 1 \frac{9}{12} = 1 \frac{3}{4}$
Hence, $a = \frac{3}{4}$ and $b = 1 \frac{3}{4}$

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