Solve this system of equations byusing the elimination method.6x » 5y = 33x – 8y = –15=72

Given: The system of equations below
[tex]\begin{gathered} equation1:6x-5y=3 \\ equation2:3x-8y=-15 \end{gathered}[/tex]To Determine: The solution of the equations
Solution
Step 1: Eliminate x by multiplying equation 1 by 1 and equation 2 by 2
[tex]\begin{gathered} 1\times(6x-5y=3)\rightarrow6x-5y=3 \\ 2\times(3x-8y=-15)\rightarrow6x-16y=-30 \end{gathered}[/tex]Step 2: Subtract derived equation from equation 2 from the derived equation from equation 1
[tex]\begin{gathered} (6x-5y=3)-(6x-16y=-30) \\ 6x-6x-5y--16y=3--30 \\ -5y+16y=3+30 \\ 11y=33 \\ y=\frac{33}{11} \\ y=3 \end{gathered}[/tex]Step 3: Substitute the value of y into equation 1
[tex]\begin{gathered} 6x-5y=3 \\ 6x-5(3)=3 \\ 6x-15=3 \\ 6x=3+15 \\ 6x=18 \\ x=\frac{18}{6} \\ x=3 \end{gathered}[/tex]Hence, x = 3, y = 3
(3,3)