Answer:
[tex]x = 3[/tex]
Step-by-step explanation:
Expand and rewrite both equations to make y the subject:
[tex]\begin{aligned}4(x-2y)-(5x+3y) & =30\\\implies 4x-8y-5x-3y & =30\\4x-5x-8y-3y & =30\\-x-11y & =30\\x+11y & = -30\\ 11y & = -x-30\\ y & = \dfrac{-x-30}{11}\end{aligned}[/tex]
[tex]\begin{aligned}3(3x+7y)-2(x+9y) & = 12\\\implies 9x+21y-2x-18y & = 12 \\9x-2x+21y-18y & = 12\\7x+3y & = 12 \\3y & = -7x+12\\ y & = -\dfrac{7}{3}x+4\end{aligned}[/tex]
Now solve for [tex]x[/tex] by substitution:
[tex]\begin{aligned}y & = y\\\\\implies \dfrac{-x-30}{11} & = -\dfrac{7}{3}x+4\\\\-x-30 & = 11\left(-\dfrac{7}{3}x+4\right)\\\\-x-30 & = -\dfrac{77}{3}x+44\\\\-x+\dfrac{77}{3}x & =44+30\\\\\dfrac{74}{3}x & = 74\\\\74x & = 222\\\\x & = \dfrac{222}{74}\\\\x & =3 \end{aligned}[/tex]
