Solve for x and y: 6(ax + by) = 3a + 2b, 6(bx - ay) = 3b - 2a - Vidyadip

Question

Solve for $x$ and $y$:
$6(ax + by) = 3a + 2b,$
$6(bx - ay) = 3b - 2a$

✓

Answer

$6(ax + by) = 3a + 2b$
$6ax + 6bx = 3a + 2b ...(1)$
$6(bx - ay) = 3b - 2a$
$6bx - 6ay = 3b - 2a ...(2)$
$6ax + 6bx = 3a + 2b ...(1)$
$6bx - 6ay = 3b - 2a ...(2)$
Multiplying (1) by by a and (2) by b
$6 a^2 x+6 b^2 x=3 a^2+2 a b \ldots(3)$
$ 6 a^2 x-6 b^2 x=3 b^2-2 a b \ldots(4)$
Adding (3) and (4), we get
$ 6 a^2 x+6 b^2 x=3 a^2+3 b^2 $
$ 6\left(a^2+b^2\right) x=3\left(a^2+b^2\right)$
$\text{x}=\frac{3\big(\text{a}^2+\text{b}^2\big)}{6\big(\text{a}^2+\text{b}^2\big)}=\frac{3}{6}=\frac{1}{2}$
Substituting $\text{x}=\frac{1}{2}$ in (1), we get
$\text{6a}\times\frac{1}{2}+\text{6by}=\text{3a}+\text{2b}$
$\text{3a}+\text{6by}=\text{3a}+\text{2b}$
$\text{6by}=\text{3a}+\text{2b}-\text{3a}$
$\text{6by}=\text{2b}$
$\text{y}=\frac{\text{2b}}{\text{6b}}=\frac{1}{3}$
Hence, the solution is $\text{x}=\frac{1}{2},\ \text{y}=\frac{1}{3}$

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