Solving systems of linear equations by elimination  (Page 3/2)

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 $x=4,y=2.$

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 $\left(4,2\right).$