Q6. - Algebra 2

Question

Solve each system of equations by using elimination
$\begin{array}{c}2r-3s=11\\ 2r+2s=6\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations is $\left(4 , - 1\right)$.

Step-1 – Apply the elimination method of solving equations

The algebraic method of elimination involves adding or subtracting the equations to eliminate one of the variables and forming new equation that is true. Sometimes, direct addition or subtraction of equations does not eliminate the variable then one equation requires formation of equivalent equation through multiplication so that one of the two variables has the same or opposite coefficient in both the equations. Multiplying the equation by a nonzero number, resulting new equation has same set of solutions. 

Step-2 – Adding/Subtracting the equations

Since, coefficient of r is already same so subtract both the equations as shown and solve.

$\begin{array}{ccccccc}& 2r& -& 3s& =& & 11\\ & 2r& +& 2s& =& & 6\\ -& & -& & & -& \\ & 0& -& 5s& =& & 5\end{array}$

Simplify it further as

$\begin{array}{c}-5s=5\\ s=-1\end{array}$

Thus, the value of s is $-1$.

Step-3 – Substitute the value of variable

To find the value of r, substitute $s=- 1$ in the equation $2r-3s=11$ and then solve as shown.

$\begin{array}{c}2r-3s=11\\ 2r-3\left(-1\right)=11\\ 2r+3=11\\ 2r=8\end{array}$

Simplify it further as

$\begin{array}{c}2r=8\\ r=4\end{array}$

Thus, the value of r is $4$.

Hence, the solution of the provided system of equations is $\left(4 , - 1\right)$.