Question

Solve the system of equations by substitution.

Answer 1

Answer:

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$x=3,\:y=-1$

Step-by-step explanation:

$3x - 4y = 13$

$5x + 4y = 11$

isolate x for 3x-4y=13

$\mathrm{Subsititute\:}x=\frac{13+4y}{3}$

$\begin{bmatrix}5\cdot \frac{13+4y}{3}+4y=11\end{bmatrix}$

$\frac{65+32y}{3}=11$

now isolate y for $\frac{65+32y}{3}=11$

$\frac{65+32y}{3}=11$

$65+32y=33$

$32y=-32$

Divide both sides by 32

$\frac{32y}{32}=\frac{-32}{32}$

$y=-1$

$\mathrm{For\:}x=\frac{13+4y}{3}$

$\mathrm{Subsititute\:}y=-1$

$x=\frac{13+4\left(-1\right)}{3}$

  $=\frac{13-4\cdot \:1}{3}$

  $=\frac{9}{3}$

$\mathrm{Divide\:the\:numbers:}\:\frac{9}{3}=3$

  $=3$

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$x=3,\:y=-1$

Answer 2

You can see how I get the answer in the picture below. Hope it can help you.