Find out the value of a for which the following equations has a unique solution$a+2b=3$, $5a+kb+7=0$

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Find out the value of a for which the following equations has a unique solution $a+2b=3$ , $5a+kb+7=0$

Find out the value of a for which the following equations has a unique solution $a+2b=3$ , $5a+kb+7=0$

Given equations are:

$a+2b-3=0$

$5a+kb+7=0$

Above equations are of the form

${{a}_{1}}a+{{b}_{1}}b-{{c}_{1}}=0$

${{a}_{2}}a+{{b}_{2}}b-{{c}_{2}}=0$

Now, ${{a}_{1}}=1,{{b}_{1}}=2,{{c}_{1}}=-3$

${{a}_{2}}=5,{{b}_{2}}=k,{{c}_{2}}=7$

According to the question,

For unique solution, the condition is

${{a}_{1}}/{{a}_{2}}\ne {{b}_{1}}/{{b}_{2}}$

$1/5\ne 2/k$

$\Rightarrow k\ne 10$

Therefore, the given equations will have unique solution for all real values of k other than $10$ .

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