Question

Solve each system by substitution. Check your solution.(I'll send the image)

Answer 1

`\begin{gathered} x=6 \\ y=4 \end{gathered}`

Step-by-step explanation

`\begin{gathered} x+2y=14\text{ Equation(1)} \\ y=3x-14\text{ Equation(2)} \end{gathered}`

Step 1

isolate x in equation (1) and replace in equation (2)

`\begin{gathered} x+2y=14\text{ Equation(1)} \\ subtract\text{ 2y in both sides} \\ x+2y-2y=14-2y \\ x=14-2y \\ \text{now, replace in equation (2)} \\ y=3x-14\text{ Equation(2)} \\ y=3(14-2y)-14 \\ y=42-6y-14 \\ y=28-6y \\ \text{subtract y in both sides} \\ y-y=28-6y-y \\ 0=28-7y \\ \text{subtract 28 in both sides} \\ 0-28=28-7y-28 \\ -28=-7y \\ \text{divide both sides by -7} \\ (-28)/(-7)=(-7y)/(-7) \\ 4=y \\ y=4 \end{gathered}`

Step 2

now,replace the y value in equation (1) to find x

`\begin{gathered} x+2y=14\text{ Equation(1)} \\ x+2\cdot4=14\text{ } \\ x+8=14 \\ \text{subtract 8 in both sides} \\ x+8-8=14-8 \\ x=6 \end{gathered}`

I hope this helps you