Solving a System of Equations by Elimination Involving Fractions

You just ordered the steps for solving a system of equations using elimination. Try using these steps when the system has equations containing fractions.

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the least common denominator (LCD) of all the fractions in the equation.

Solve the system by elimination: $\{\begin{array}{c}\hfill x+\frac{1}{2}y=6\hfill \\ \hfill \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}\hfill \end{array}$

Step 1 - Check to see if each equation is in standard form $Ax+By=C$. Both equations are in standard form.

$x+\frac{1}{2}y=6$

$\frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}$

Step 2 - To clear the fractions, multiply each equation by its LCD (Least Common Denominator).

$2(x+\frac{1}{2}y=6)$

$6(\frac{3}{2}x+\frac{2}{3}y=\frac{17}{2})$

Step 3 - Simplify the equations.

$2x+y=12$

$9x+4y=51$

Step 4 - Eliminate one of the variables and simplify the equation.

$-4(2x+y=12)=-8x-4y=-48$

Step 5 - Add the two equations to eliminate $y$.

$\begin{array}{cc}\hfill -8x-4y& =-48\hfill \\ \hfill 9x+4y& =51\hfill \\ \hfill x& =3\hfill \end{array}$

Step 6 - Solve for $y$ by substituting $x$ back into one of the equations.

$\begin{array}{cc}\hfill 2x+y& =12\hfill \\ \hfill 2(3)+y& =12\hfill \\ \hfill 6+y& =12\hfill \\ \hfill y& =6\hfill \end{array}$

Step 7 - Write the solution as an ordered pair.

The ordered pair is $(3,6)$.

Step 8 - Check that the ordered pair is a solution to both original equations.

$\begin{array}{cc}\hfill & \hfill \frac{3}{2}x+\frac{2}{3}y=\frac{17}{2}\\ \hfill x+\frac{1}{2}y\stackrel{?}{=}6& \hfill \frac{3}{2}(3)+\frac{2}{3}(6)\stackrel{?}{=}\frac{17}{2}\\ \hfill 3+\frac{1}{2}(6)\stackrel{?}{=}6& \hfill \frac{9}{2}+4\stackrel{?}{=}\frac{17}{2}\\ \hfill 3+3\stackrel{?}{=}6& \hfill \frac{9}{2}+\frac{8}{2}\stackrel{?}{=}\frac{17}{2}\\ \hfill 6=6✓& \hfill \frac{17}{2}=\frac{17}{2}✓\end{array}$

The solution is $(3,6)$.