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Question

For what value of $$p$$ will the following pair of linear equations have infinitely many solutions:
$$(p-3)x+3y=p$$
$$px+py=12$$


Solution

Given equation are
$$(p-3)x+3y=p$$
$$px+py=12$$
The pair of linear equation has infinitely many solutions.
$$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$
$$\dfrac{p-3}{p}=\dfrac{3}{p}=\dfrac{p}{12}$$
$$\dfrac{a_1}{a_2}=\dfrac{c_1}{c_2}$$
$$\dfrac{p-3}{p}=\dfrac{p}{12}$$
$$(p-3)12=p^2$$
$$12p-36=p^2$$
$$p^2-12p-36=0$$
$$p^2-6p-6p-36=0$$
$$p(p-6)-6(p-6)=0$$
$$p-6=0$$
$$\therefore p=6$$.

1194068_1286474_ans_2eaff0d8f77a46af92293a8aef775225.jpg

Mathematics

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