Question

Solve the following systems of equations by the method of cross-multiplication:
$(a+2b)x+(2a-b)y=2$
$(a-2b)x+(2a+b)y=3$

Answer 1

$(a+2b)x+(2a-b)y=2⟶\left(i\right)(a-2b)x+(2a+b)y=3⟶\left(ii\right)$

Adding the two we get $2ax+4ay=5⟶\left(iii\right)$

and subtracting $\left(ii\right)$from $\left(i\right),$ we get $,$

$4bx-2by=-1⟶\left(iv\right)$

Multiplying $\left(iii\right)$ by $2b$ and $\left(iv\right)$ by $a,$ we get

$4abx+8aby=10b⟶\left(v\right)$

$4abx-2aby=-a⟶\left(vi\right)$

Subtracting $\left(vi\right)$ from $\left(v\right),$ we have $,$

$10aby=10b+a\Rightarrow y=\frac{10b+a}{10ab}\Rightarrow \frac{1}{a}+\frac{1}{10b}$

Multiplying $\left(vi\right)$ by $4$ and adding $\left(v\right),$ we get $,$

$20abx=10b-4a$

$x=\frac{10b-4a}{20ab}$$=\frac{1}{2a}-\frac{1}{5b}$

$\therefore x=\frac{1}{2a}-\frac{1}{5b}$ and $y=\frac{1}{a}+\frac{1}{10b}$