Question

Find the value of p and q for which the following system of linear equations has infinite number of solutions
2x+3y=9. And. (p+q)x+(2p-q)y=3(p+q+1)

Answer 1

$Given \: system \:of \: Linear \: equations :$

$2x + 3y - 9 = 0 \: --(1) \:and \:$

$(p+q)x+(2p-q)y-3(p+q+1) = 0 \: --(2)$

$Compare \:above \: equations \: with$

$a_{1}x +b_{1}y +c_{1} = 0 \: and$

$a_{2}x +b_{2}y +c_{2} = 0, we \:get$

$a_{1} = 2 , b_{1} = 3, \: c_{1} = -9$

$a_{2} = p+q , b_{2} = (2p-q) , \: c_{2} = -(3(p+q+1)$

/* It is given that , equations has infinitely many solutions */

$\therefore \frac{a_{1}}{a_{2}} = \frac{a_{1}}{a_{2}} = \frac{a_{1}}{a_{2}}$

$\implies \frac{2}{p+q} = \frac{3}{2p-q}$

$\implies 2(2p-q) =3(p+q)$

$\implies 4p - 2q = 3p + 3q$

$\implies 4p - 3p = 3q + 2q$

$\implies p = 5q \: --(3)$

$and \: \frac{2}{p+q} = \frac{-9}{-3(p+q+1)}$

$\implies \frac{2}{p+q} = \frac{3}{(p+q+1)}$

$\implies 2(p+q+1) = 3(p+q)$

$\implies 2p+2q+2= 3p+3q$

$\implies 2p-3p= 3q- 2q - 2$

$\implies -p = q - 2$

$\implies - 5q = q - 2 \: [ From \: (3) ]$

$\implies -5q - q = -2$

$\implies -6q = -2$

$\implies q = \frac{-2}{-6}$

$\implies q = \frac{1}{3}\:--(4)$

$Put \: value \: of \: q \: in \: equation \:(3) , we$

$get$

$p = \frac{5}{3}\: --(5)$

Therefore.,

$\red{ Value \:of \: p } \green { = \frac{5}{3}}$

$\red{ Value \:of \: q } \green { = \frac{1}{3}}$

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