Question

Solve the following pair of linear equations by the substitution and cross-multiplication methods :

$8x+5y=9$; $3x+2y=4$

Answer 1

Substitution method:

$8x+5y=9$

$\therefore y=\frac{9-8x}{5}$ ....(1)

$3x+2y=4$ ...(2)

Substitute eq(1) in eq (2), we get

$3x+2\left(\frac{9-8x}{5}\right)=4$

Multiply by $5$ on both sides, we get

$15x+18-16x=20$

$\Rightarrow -x=2$

$\Rightarrow x=-2$

Substitute $x=-2$ in eq (1), we get

$y=\frac{9-8(-2)}{5}$

$\Rightarrow y=\frac{25}{5}=5$

So, $(x,y)=(-2,5)$

Cross-multiplication method:

$8x+5y=9$ ...(1)

So, $a_1=8,b_1=5,c_1=-9$

$3x+2y=4$ ...(2)

So, $a_2=3,b_2=2,c_2=-4$

$\therefore x=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}$

$\Rightarrow x=\frac{5(-4)-2(-9)}{8\left(2\right)-3\left(5\right)}$$=\frac{-20+18}{16-15}=-2$

$y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}$

$\Rightarrow y=\frac{-9\left(3\right)-(-4)8}{8\left(2\right)-3\left(5\right)}$ $=\frac{-27+32}{16-15}=5$

So, $(x,y)=(-2,5)$