Question

Solve through substitution

Answer 1

–3x + 3y - 4 =0 and - x + y -3 = 0 do not have any solution which means two lines are parallel and there will be no intersection point.

Solution:

Need to determine solution of following system of equations

–3x + 3y = 4  

y = x + 3  

Let's modify given equation in standard form

–3x + 3y - 4 =0        ------- (1)

- x + y -3 = 0          ------- (2)

Lets first analyze whether given system of equation is having solution or not.

If $\mathrm{a}_{1} x+\mathrm{b}_{1} y+\mathrm{c}_{1}=0$ and $\mathrm{a}_{2} x+\mathrm{b}_{2} y+\mathrm{c}_{2}=0$ are two equation, then if, $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ then the given system of equation has no solution.

In this problem,

$\begin{array}{l}{a_{1}=-3, b_{1}=3 \text { and } c_{1}=-4} \\\\ {a_{2}=-1, b_{2}=1 \text { and } c_{2}=-3} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-3}{-1}=3} \\\\ {\frac{b_{1}}{b_{2}}=\frac{3}{1}=3} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-4}{-3}=\frac{4}{3}}\end{array}$

$\text { In our case also } \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

So equations –3x + 3y - 4 =0 and - x + y -3 = 0 do not have any solution which means two lines are parallel and there will be no intersection point.