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Another method of solving a system of two linear equations in two variables is called the method of elimination by addition . It is similar to the method of elimination by substitution in that the process eliminates one equation and one variable. The method of elimination by addition makes use of the following two properties.
To solve a system of two linear equations in two variables by addition,
Solve
$\{\begin{array}{rrr}\hfill x-y=2& \hfill & \hfill \left(1\right)\\ \hfill 3x+y=14& \hfill & \hfill \left(2\right)\end{array}$
Step 1: Both equations appear in the proper form.
Step 2: The coefficients of
$y$ are already opposites, 1 and
$-1,$ so there is no need for a multiplication.
Step 3: Add the equations.
$\frac{\begin{array}{c}x-y=2\\ 3x+y=14\end{array}}{4x+0=16}$
Step 4: Solve the equation
$4x=16.$
$4x=16$
$x=4$
The problem is not solved yet; we still need the value of
$y$ .
Step 5: Substitute
$x=4$ into either of the original equations. We will use equation 1.
$\begin{array}{rrrrr}\hfill 4-y& \hfill =& \hfill 2& \hfill & \hfill \text{Solve\hspace{0.17em}for\hspace{0.17em}}y.\\ \hfill -y& \hfill =& \hfill -2& \hfill & \hfill \\ \hfill y& \hfill =& \hfill 2& \hfill & \hfill \end{array}$
We now have
$x=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2.$
Step 6: Substitute
$x=4$ and
$y=2$ into both the original equations for a check.
$\begin{array}{rrrrrrrrrrrr}\hfill (1)& \hfill & \hfill x-y& \hfill =& \hfill 2& \hfill & \hfill (2)& \hfill & \hfill 3x+y& \hfill =& \hfill 14& \hfill \\ \hfill & \hfill & \hfill 4-2& \hfill =& \hfill 2& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}& \hfill & \hfill & \hfill 3(4)+2& \hfill =& \hfill 14& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill 2& \hfill =& \hfill 2& \hfill \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}& \hfill & \hfill & \hfill 12+2& \hfill =& \hfill 14& \hfill \text{Is\hspace{0.17em}this\hspace{0.17em}correct?}\\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill 14& \hfill =& \hfill 14& \hfill \text{Yes,\hspace{0.17em}this\hspace{0.17em}is\hspace{0.17em}correct}\text{.}\end{array}$
Step 7: The solution is
$\left(4,2\right).$
The two lines of this system intersect at
$\left(4,2\right).$
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