Solve through substitution–3x+3y=4y=x+3Can you please give me point of intersection

–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 [tex]\mathrm{a}_{1} x+\mathrm{b}_{1} y+\mathrm{c}_{1}=0[/tex] and [tex]\mathrm{a}_{2} x+\mathrm{b}_{2} y+\mathrm{c}_{2}=0[/tex] are two equation, then if, [tex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}[/tex] then the given system of equation has no solution.In this problem, [tex]\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}[/tex][tex]\text { In our case also } \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}[/tex] 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.