Question

Solve the system of equations below. You may want to use the method by substitution starting with the second equation.

Answer 1

Given

The system of equation,

$\begin{gathered} 3x-2y=42 \\ 5x+10y=-10 \end{gathered}$

To find:

The solution to the system of equation.

Explanation:

It is given that,

$\begin{gathered} 3x-2y=42 \\ 5x+10y=-10 \end{gathered}$

That implies,

From the second equation,

$\begin{gathered} 5x+10y=-10 \\ \div5\Rightarrow x+2y=-2 \\ \Rightarrow x=-2y-2 \end{gathered}$

Substitute x in the first equaton,

$\begin{gathered} 3(-2-2y)-2y=42 \\ -6-6y-2y=42 \\ -8y=42+6 \\ -8y=48 \\ y=\frac{48}{-8} \\ y=-6 \end{gathered}$

Therefore,

$\begin{gathered} x=-2(-6)-2 \\ x=12-2 \\ x=10 \end{gathered}$

Hence, the solution is (10,-6).