Unit 13.2 Solving Systems of Equations Using Elimination
Elimination Method
The elimination method is another effective technique used to solve systems of linear equations. This method involves adding and multiplying the equations to eliminate one of the variables, making it easier to solve for the remaining variable. Once one variable is eliminated, the resulting equation can be solved for the other variable. Then, the value of this variable can be substituted back into one of the original equations to find the value of the eliminated variable.
- Align the Equations:
- Ensure that the equations are written in standard form Ax + By = C
- Multiply (if necessary):
- If the coefficients of the variable to be eliminated are not opposites or identical, multiply one or both equations by a suitable number to make the coefficients of one of the variables either equal but opposites.
- Add or Subtract the Equations:
- Add the equations to eliminate one variable. The result should be an equation with a single variable.
- Solve for the Remaining Variable:
- Solve the equation obtained in step 3 to find the value of the remaining variable.
- Substitute Back to Find the Other Variable:
- Substitute the value of the solved variable back into one of the original equations to find the value of the eliminated variable.
- Check the Solution:
- Substitute the values of both variables into both original equations to ensure they satisfy the equations.
Systems of Equations Elimination Method
Solve the following systems of equations
2x + 3y = 6 (eq. 1)
4x – y = 5 (eq. 2)
Since there is no variable we can get rid of directly by adding, we need to multiply. We can choose to get rid of either x or y. Since y is negative already in equation 2, we will get rid of y
Multiply the whole “equation 2” by 3 so that the “y” can be eliminated
3 ( 4x – y = 5 ) (eq. 2)
12x – 3y = 15 (eq. 2a)
Add equation 1 and equation 2a together to eliminate y
2x + 3y = 6 (eq. 1)
+ 12x – 3y = 15 (eq. 2a)
Add the x, y and constant values straight down
2x + 12x = 14x
3y + (-3y) = 0y
6 + 15 = 21
So our new equation is
14x = 21
x = [latex]\frac{21}{14}[/latex] = [latex]\frac{3}{2}[/latex]
Substitute x into one of the original equations to solve for y
Equation 1:
2([latex]\frac{3}{2}[/latex]) + 3y = 6
3 + 3y = 6
3y = 3
y = 1
Our solution values are ([latex]\frac{3}{2}[/latex], 1)
Checking our answers with the original equations
4([latex]\frac{3}{2}[/latex]) – 1 = 5
True. The solutions are valid