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The equation $ax-(a+b)y=1$ and $(a-b)x+ay=5$ have a unique solution:

1. for all values of $a$ and $b$
2. only when $a=b$
3. only when $a^{2}:b^{2}=1:2$
4. only when $a=0$ and $b=0$

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The pair of linear equations represented by these lines $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$

1. If  $\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$ then the pair of linear equations has exactly one solution.
2. If $\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$ then the pair of linear equations has infinitely many solutions.
3. If $\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$then the pair of linear equations has no solution.

Given that, the equation $ax-(a+b)y=1$ and $(a-b)x+ay=5$

$\implies ax-(a+b)y-1 = 0,\;(a-b)x+ay – 5 = 0$

For unique solution$:\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$

$\implies \dfrac{a}{a-b} \neq \dfrac{-(a+b)}{a}$

$\implies a^{2} + (a-b)(a+b) \neq 0$

$\implies a^{2} + (a^{2} – b^{2}) \neq 0$

$\implies 2a^{2} – b^{2} \neq 0$

$\implies 2a^{2} \neq b^{2}$

$\implies \dfrac{a^{2}}{b^{2}} \neq \dfrac{1}{2}$

$\implies a^{2} :b^{2} \neq 1:2$

by (5.8k points) 5 56 296
edited
+1

Nice explanation, but in 3rd last $eq^{n}$ you have made a calculation mistake, according to me it should be

=> $a^{2}$ + (a – b)(a + b) $\neq$ 0

=> $a^{2}$ + ($a^{2}$ – $b^{2}$) $\neq$ 0

=> $2a^{2}$ $\neq$ $b^{2}$

Answer would have Option C. only when $a^{2}$: $b^{2}$ = 1 : 2  If the question had given “NOT have a unique solution”
Correct me if I am wrong.

0

Thank you for correcting me.

NOT have a unique solution”  means infinitely many solutions or no solution.

I think options is not correct.