# How do you solve by elimination with fractions?

Sep 29, 2014

A good start would be to multiply each equation by the least common multiple (LCM) of the denominator. If you multiply all parts of the equation by the LCM, it does not affect the solution. This would eliminate the fractions and you could go forth and solve the system by elimination.

Here is an example:

$- \frac{1}{10} x + \frac{1}{2} y = \frac{4}{5}$

$\frac{1}{7} x + \frac{1}{3} y = - \frac{2}{21}$

Step 1: Multiply Equation 1 by the LCM which 10.

$10 \left(- \frac{1}{10} x + \frac{1}{2} y\right) = \left(\frac{4}{5}\right) 10$

$10 \left(- \frac{1}{10} x\right) + 10 \left(\frac{1}{2} y\right) = \left(\frac{4}{5}\right) 10$

$- x + 5 y = 8$

Step 2: Multiply Equation 2 by the LCM which is 21.

$21 \left(\frac{1}{7} x + \frac{1}{3} y\right) = \left(- \frac{2}{21}\right) 21$

$21 \left(\frac{1}{7} x\right) + 21 \left(\frac{1}{3} y\right) = \left(- \frac{2}{21}\right) 21$

$3 x + 7 y = - 2$

Step 3: Place the new equations together to create a new system:

$- x + 5 y = 8$
$3 x + 7 y = - 2$

Step 4: To solve by elimination, multiply first equation by 3 (this will help to eliminate the $x$ variable).

$3 \left(- x + 5 y\right) = \left(8\right) 3$
$3 x + 7 y = - 2$

becomes

$- 3 x + 15 y = 24$
$3 x + 7 y = - 2$

Step 5: Add the two equations to eliminate the x variable:
$- 3 x + 15 y = 24$
$3 x + 7 y = - 2$

becomes $22 y = 22$

Step 6: Solve for $y$:
$22 y = 22$

$\frac{22 y}{22} = \frac{22}{22}$

and thus

$y = 1$

Step 7: substitute $y$ into one of the equations to solve for $x$.

$- x + 5 y = 8$
$- x + 5 \left(1\right) = 8$
$- x + 5 = 8$
$- x + 5 - 5 = 8 - 5$
$- x = 3$
$\frac{- x}{-} 1 = \frac{3}{-} 1$

$x = - 3$

Step 8: Write the solution to the system as a coordinate.

$x = - 3 , y = 1$

$\left(x , y\right) = \left(- 3 , 1\right)$