Given:
6x + 8y = 8
5x + 4y = 3
Let's solve the system of equations using elimination method.
Take the followings steps:
Step 1:
Multiply each equation by any value that makes the coefficient of y opposite.
Multiply equation 2 by (-2):
6x + 8y = 8
-2(5x + 4y = 3)
6x + 8y = 8
-10x - 8y = -6
Add the two equations:
6x + 8y = 8
+ -10x - 8y = -6
_______________
-4x + 0 = 2
-4x = 2
Divide both sides by -4:
[tex]\begin{gathered} \frac{-4x}{-4}=\frac{2}{-4} \\ \\ x=-\frac{1}{2} \end{gathered}[/tex]Substitute -½ for x in either of the equations.
Take equation 1:
[tex]\begin{gathered} 6x+8y=8 \\ \\ 6(-\frac{1}{2})+8y=8 \\ \\ -3+8y=8 \end{gathered}[/tex]Add 3 to both sides:
[tex]\begin{gathered} 3-3+8y=8+3 \\ \\ 8y=11 \end{gathered}[/tex]Divide both sides by 8:
[tex]\begin{gathered} \frac{8y}{8}=\frac{11}{8} \\ \\ y=\frac{11}{8} \end{gathered}[/tex]Thus, we have the solutions:
[tex](x,y)\Longrightarrow(-\frac{1}{2},\frac{11}{8})[/tex]ANSWER:
[tex](-\frac{1}{2},\frac{11}{8})[/tex]