2.6.2 Writing a New System to Solve a Given System

Solving a System of Equations by Elimination Using Multiplication

Let’s look at an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Solve the system by elimination: $\{\begin{array}{c}4x-3y=9\hfill \\ 7x+2y=-6\hfill \end{array}$

Solution

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them. Both equations are in standard form.

$4x-3y=9$

$7x+2y=-6$

Step 2 - Check to see if the coefficients of one variable are opposites or equivalent. To get opposite coefficients of $y$, we will need to multiply the first equation by 2 and the second equation by 3. Multiply the first equation by 2 and the second equation by 3.

$2(4x-3y)=2(9)$

$3(7x+2y)=3(-6)$

Step 3 - Simplify

$8x-6y=18$

$21x+6y=-18$

Step 4 - Add the equations to eliminate one variable.

$\begin{array}{c}\hfill 8x-6y=18\\ \hfill 21x+6y=-18\\ \hfill 29x=0\end{array}$

Step 5 - Solve for the remaining variable.

$x=0$

Step 6 - Substitute the solution into one of the original equations. Then solve for the other variable.

$\begin{array}{ccc}\hfill 7x+2y& \hfill =\hfill & -6\hfill \\ \hfill 7(0)+2y& \hfill =\hfill & -6\hfill \\ \hfill 2y& \hfill =\hfill & -6\hfill \\ \hfill y& \hfill =\hfill & -3\hfill \end{array}$

Step 7 - Check that the ordered pair is a solution to both original equations.

$\begin{array}{cc}\hfill 4x-3y=9& \hfill 7x+2y=-6\\ \hfill 4(0)-3(-3)\stackrel{?}{=}9& \hfill 7(0)+2(-3)\stackrel{?}{=}-6\\ \hfill 9=9✓& \hfill -6=-6✓\end{array}$

The solution is $(0,-3)$.