Equation
[tex]\begin{gathered} 4x+4y=48 \\ 3x=2y+6 \end{gathered}[/tex]
Procedure
By subtracting the value of x for the second equation and replacing in the first equation
[tex]4(\frac{2y+6}{3})+4y=48[/tex][tex]\begin{gathered} \frac{8y}{3}+8+4y=48 \\ \frac{20y}{3}=48-8 \\ 20y=3(40) \\ y=\frac{120}{20} \\ y=6 \end{gathered}[/tex]Now, for x
[tex]\begin{gathered} x=\frac{2y+6}{3} \\ x=\frac{2\cdot6+6}{3} \\ x=\frac{12+6}{3} \\ x=\frac{18}{3} \\ x=6 \end{gathered}[/tex]The solutions to system equations are:
x= 6 and y= 6
