Solve:
ax + by = a - b
bx - ay = a + b

x = 2, y = -1

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x = -2, y = 1

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x = 1, y = - 1

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Insufficient data

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None of the above

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Solution

The correct option is C x = 1, y = - 1
$a x + b y = a - b$ and $b x - a y = a + b$

Here, $\frac{a_{1}}{a_{2}} = \frac{a}{b} a n d \frac{b_{1}}{b_{2}} = - \frac{b}{a} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Therefore, a unique solution exists.

Writing the coefficients in the following array

$\begin{matrix} b & a - b a & b - a & a + b b & - a \end{matrix}$

we get, $\frac{x}{b (a + b) + a (a - b)} = \frac{y}{b (a - b) - a (a + b)}$

$= \frac{- 1}{- a^{2} - b^{2}}$

$\Rightarrow \frac{x}{b^{2} + a^{2}} = \frac{y}{- b^{2} - a^{2}} = \frac{1}{a^{2} + b^{2}}$

$∴ x = 1$ and $y = - 1$

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Similar questions

Solve each of the following systems of eqautions by the method of cross-multiplication:

ax+by=a-b

bx-ay=a+b

\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b), (x \neq \frac{1}{a}, \frac{1}{b})

a x + b y = (a - b), b x - a y = (a + b) .

a x + b y = a - b

b x - a y = a + b

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