Q25. - Algebra 2

Question

Solve each system of equations by using either substitution or elimination.

$\begin{array}{c}25.r+4s=-8\\ 3r+2s=6\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\left(4 , - 3\right)$.

Step-1 – Apply the substitution method of solving equations

The algebraic method of substitution involves solving the one of the two equations for one variable in terms of other variable and then substituting the expression so formed for the variable in the second equation.

Step-2 – Solving one equation for r in terms of s

To solve the equation $r+4s=-8$ for r in terms of s, subtract 4s from both sides as shown below.

$\begin{array}{c}r+4s=-8\\ r+4s-4s=-8-4s\\ r=-8-4s\end{array}$

Step-3 – Substitute the expression

Now, substitute $r=-8-4s$in the equation $3r+2s=6$ and solve for s.

$\begin{array}{c}3r+2s=6\\ 3\left(-8-4s\right)+2s=6\\ -24-12s+2s=6\\ -10s=30\end{array}$

Simplify it further as

$\begin{array}{c}-10s=30\\ s=-3\end{array}$

Thus, the value of s is $-3$.

Step-4 – Substitute the value of variable

To find the value of r, substitute $s=-3$ in the equation $r+4s=-8$ and then solve for r as shown.

$\begin{array}{c}r+4s=-8\\ r+4\left(-3\right)=-8\\ r-12=-8\\ r=4\end{array}$

Thus, the value of r is $4$.

Hence, the solution of the provided system of equations is $\left(4 , - 3\right)$.