Question

For which values of p and q, will the following pair of linear equations have infinitely many solutions? 4x + 5y = 2 , (2p + 7q) x + (p + 8q) y = 2q-p + 1

Answer 1

$let4x+5y=2$

$4x+5y-2=0.......\left(i\right)$

$and(2p+7q)x+(p+8q)y=2q-p+1$

$(2p+7q)x+(p+8q)y-(2q-p+1)=0.......\left(ii\right)$

$Nowforinfinetelymanysolutions$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ $\therefore \frac{4}{(2p+7q)}=\frac{5}{(p+8q)}=\frac{-2}{-(2q-p+1)}$

$\left(i\right)4(p+8q)=5(2p+7q)$

$4p+32q=10p+35q$

$6p+3q=0$

$2p+q=0......\left(a\right)$

$\left(ii\right)5(2p-q+1)=2(p+8q)$

$10p-5q+5=2p+16q$

$8p-21q+5=0......\left(b\right)$

$fromeqn\left(a\right),q=-2p,substitutinf{}^{\prime }{q}^{\prime }ineqn\left(b\right)$

$8p-21(-2p)+5=0$

$8p+42P+5=0$

$50p=-5$

$\Rightarrow p=\frac{-1}{10}$

$nowq=-2p$

$=-2\frac{(-1)}{10}=\frac{1}{5}$

$\therefore p=\frac{-1}{10}andq=\frac{1}{5}$