# Solving by Elimination

## Key Questions

• 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.

$x = - 3 , y = 1$

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

• You follow a sequence of steps.

In general, the steps are:

1. Enter the equations.
2. Multiply each equation by a number to get the lowest common multiple for one of the variables.
3. Add or subtract the two equations to eliminate that variable .
4. Substitute that variable into one of the equations and solve for the other variable.

EXAMPLE:

How do you use the elimination method to solve $2 x + 3 y = 7 , 3 x + 4 y = 10$?

Solution:

Step 1. Enter the equations.

[1] $2 x + 3 y = 7$
[2] $3 x + 4 y = 10$

Step 2. Find the lowest common multiple.

Multiply Equation 1 by $3$ and Equation 2 by $2$.

[3] $6 x + 9 y = 21$
[4] $6 x + 8 y = 20$

Step 3. Subtract Equation 4 from Equation 3.

[5] $y = 1$

Step 4. Substitute Equation 5 in Equation 1.

$2 x + 3 y = 7$
$2 x + 3 = 7$
$2 x = 4$

$x = 2$

Check: Substitute the values of $x$ and $y$ in Equation 2.

If you use one equation to get the second variable, use the other equation for the check.

$3 x + 4 y = 10$
3×2+4×1=10
$6 + 4 = 10$
$10 = 10$

It checks!

The solution is correct.