Q41. - Algebra 1

Question

Use elimination to solve each system of equations.

$\begin{array}{l}\mathbf{8}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{18}\\ \mathbf{6}\mathbf{x}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{6}\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}$.

Step 1. State the concept of elimination method to solve system of equations.

The elimination method is the method in which addition or subtraction method is used to get an equation in the one variable. 

If the coefficients are opposite, use the addition and when the coefficients are same, use the subtraction.

Step 2. Add or subtract the equations to eliminate one variable.

The given system of equation is: 

$8x-5y=18$             … (1)

$6x+6y=-6$             … (2)

Multiply equation (1) by 6 and equation (2) by 5.

$\begin{array}{l}6\times (8x-5y)=6\left(18\right)\\ 48x-30y=108\left(3\right)\end{array}$

 

 $\begin{array}{l}5\times (6x+6y)=5(-6)\\ 30x+30y=-30\left(4\right)\end{array}$ 

Note that both the equations (3) and (4) have $30y$ in opposite signs, therefore, variable $y$ can be eliminated if both equations are added.

 $\begin{array}{l}48x-30y=108\\ \underset{¯}{(+)30x+30y=-30}\\ 78x=78\text{ Divide both sides by 78}\\ x=1\end{array}$

Step 3. Solve for y .

Substitute 1 for $x$ into the equation (2) and solve for $y$.

 $\begin{array}{c}6\left(1\right)+6y=-6\\ 6+6y-6=-6-6\\ 6y=-12\\ y=-2\end{array}$ 

The solution is $(1,-2)$.