Question

Show that the following system of equations has a unique solution and solve it:
$$\begin{gathered} 2x-3y=17 \\ 4x+y=13 \end{gathered}$$

Answer 1

The system of equations can be written as
$$\begin{gathered} 2x-3y-17=0.....\left(i\right) \\ 4x+y-13=0.....\left(ii\right) \end{gathered}$$
The given equations are of the form
$a_{1}x+b_{1}y+c_1=0\mathrm{and}{a}_{2}x+b_{2}y+c_2=0$
where $a_1=2,b_1=-3,c_1=-17\mathrm{and}{a}_2=4,b_2=1,c_2=-13$
Now,
$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\mathrm{and}\frac{b_1}{b_2}=\frac{-3}{1}=-3$
Since, $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, therefore the system of equations has unique solution.
Using cross multiplication method, we have
$$\begin{gathered} \frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1} \\ \Rightarrow \frac{x}{-3\left(-13\right)-1\times \left(-17\right)}=\frac{y}{-17\times 4-\left(-13\right)\times 2}=\frac{1}{2\times 1-4\times \left(-3\right)} \\ \Rightarrow \frac{x}{39+17}=\frac{y}{-68+26}=\frac{1}{2+12} \\ \Rightarrow \frac{x}{56}=\frac{y}{-42}=\frac{1}{14} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{56}{14},y=\frac{-42}{14} \\ \Rightarrow x=4,y=-3 \end{gathered}$$
Hence, $x=4\mathrm{and}y=-3$.