Answer:
[tex]x=9[/tex] and [tex]y=13[/tex]
Step-by-step explanation:
We have been given a system of equations. We are asked to solve our given system algebraically.
[tex]5x-3y=6...(1)[/tex]
[tex]6x-4y=2...(2)[/tex]
From equation (1), we will get:
[tex]x=\frac{6+3y}{5}[/tex]
Upon substituting this value in equation (2), we will get:
[tex]6(\frac{6+3y}{5})-4y=2[/tex]
[tex]\frac{36+18y}{5}-4y=2[/tex]
[tex]\frac{36+18y}{5}-\frac{5*4y}{5}=2[/tex]
[tex]\frac{36+18y-20y}{5}=2[/tex]
[tex]\frac{36-2y}{5}=2[/tex]
[tex]\frac{36-2y}{5}*5=2*5[/tex]
[tex]36-2y=10[/tex]
[tex]36-36-2y=10-36[/tex]
[tex]-2y=-26[/tex]
[tex]\frac{-2y}{-2}=\frac{-26}{-2}[/tex]
[tex]y=13[/tex]
Upon substituting [tex]y=13[/tex] in equation (1), we will get:
[tex]5x-3(13)=6[/tex]
[tex]5x-39=6[/tex]
[tex]5x-39+39=6+39[/tex]
[tex]5x=45[/tex]
[tex]\frac{5x}{5}=\frac{45}{5}[/tex]
[tex]x=9[/tex]
