For what value of p will the following pair of linear equation have infinitely many solution- (p-3) x+3y=p; px+py=12.

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Solution

The equations $(p - 3) x + 3 y - p = 0 p x + p y - 12 = 0$
For infinite many solution
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Here $a_{1} = p - 3, b_{1} = 3, c_{1} - p a_{2} = p, b_{2} = p, c_{2} = - 12$
$\frac{p - 3}{p} = \frac{3}{p} = \frac{- 1}{- 12}$
Solving $\frac{3}{p} = \frac{- 1}{- 12}$
$⟹ p^{2} = 36 ⟹ p = \pm 6$
Now solving $\frac{p - 3}{p} = \frac{- 1}{- 12}$
$⟹ p - 3 = 3 ⟹ p = 6$
Hence the value of $p = 6$ .

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Find the values of p and q for the following pair of equations 2x + 3y = 7 and 2px + py = 28 – qy, if the pair of equations has infinitely many solutions.

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For what value of p will the following pair of linear equation have infinitely many solution- (p-3) x+3y=p; px+py=12.

The equations (p−3)x+3y−p=0px+py−12=0For infinite many solutiona1a2=b1b2=c1c2Here a1=p−3,b1=3,c1−pa2=p,b2=p,c2=−12p−3p=3p=−1−12Solving 3p=−1−12⟹p2=36⟹p=±6Now solving p−3p=−1−12⟹p−3=3⟹p=6Hence the value of p=6.