Question

Solve
$2x+7y-5=0$
$-3x+8y=-11$

Answer 1

The given system of equations is
$2x+7y-5=0$
$-3x+8y=-11$

By cross multiplication method, we know that, for a system of linear equations in $x$ and $y$, of the form $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we have:

$\frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Hence, for $2x+7y-5=0$ and $-3x+8y+11=0$, we have:

$\frac{x}{\left(7\right)\left(11\right)-\left(8\right)(-5)}=\frac{y}{(-5)(-3)-\left(2\right)\left(11\right)}=\frac{1}{\left(2\right)\left(8\right)-(-3)\left(7\right)}$

$\Rightarrow \frac{x}{77+40}=\frac{y}{15-22}=\frac{1}{16+21}$

$\Rightarrow \frac{x}{117}=\frac{y}{-7}=\frac{1}{37}$

On comparing $\frac{x}{117}$ with $\frac{1}{37}$, we get:

$x=\frac{117}{37}$

And on comparing $\frac{y}{-7}$ with $\frac{1}{37}$, we get:

$y=-\frac{7}{37}$

Hence, the solution of the given system of equations is $(\frac{117}{37},-\frac{7}{37})$.