Solving Equations – Steps for Elimination Method
For examples on solving equations with steps for the elimination method, please look at the examples after the quiz below.
Solving Equations Quiz with Steps & Examples
Instructions: Exercises on solving equations using the steps for the elimination method are included in this quiz. Answer each question and check your answers
[WpProQuiz 9]
Solving Equations – Elimination Method Example
Solving equations with the steps for the elimination method will really be useful for your algebra exam.
Systems of equations are also called simultaneous equations.
We need to use the steps for elimination method when the coefficients on the variables in the system of equations are not the same.
You will remember from our post on algebraic terms that coefficients are the number part of a term.
Look at the example below.
Solving Equations – Steps for Elimination Method
Look at the example below on solving equations and then study the steps that follow on how to carry out the elimination method.
Example: Solve the system of equations for x and y.
8x – 3y = 30
3x + y = 7
Answer: x = 3 and y = –2
STEP 1: Check the equations
Check to see if the equations you are solving have a term with a common coefficient.
Our first equation has 8x and the second equation has 3x, so the coefficients are 8 and 3.
The first equation has 3y and the second equation has y, so the coefficients are 3 and 1.
By comparing the equations in this way, we can see that we do not have any terms with common coefficients.
STEP 2: Check the algebraic terms
Then determine which equation and which term to perform multiplication on.
We can perform multiplication on the “y” variable in the second equation.
That is because the first equation has 3y and the second equation has y.
So. we can multiply the y in the second equation by 3 in order to get 3y.
STEP 3: Perform multiplication
Perform multiplication on the equation you chose in step 2.
Our second equation was: 3x + y = 7
Multiply each term in the second equation by 3 as show below.
(3x × 3) + (y × 3) = (7 × 3)
9x + 3y = 21
You have now created a term (3y) that has a coefficient in common with a term in the other equation.
STEP 4: Perform the elimination method
Perform the elimination method by adding or subtracting the new equation from step 3.
In this problem, we need to add the equations as show below:
8x – 3y = 30
+ (9x + 3y = 21)
17x = 51
17x ÷ 17 = 51 ÷ 17
x = 3
STEP 5: Substitute the value
Now substitute x with its value to solve for y.
We know from the previous step that x = 3.
So, take the first equation and substitute this value.
x = 3
8x – 3y = 30
(8 × 3) – 3y = 30
24 – 3y = 30
24 – 3y + 3y = 30 + 3y
24 – 3y + 3y = 30 + 3y
24 = 30 + 3y
24 – 30 = 30 – 30 + 3y
24 – 30 = 30 – 30 + 3y
24 – 30 = 3y
–6 = 3y
–6 ÷ 3 = 3y ÷ 3
–2 = y
y = –2
So, as you can see from the example above, you can use inverse operations in conjunction with the elimination method in order to add or subtract systems of equations that have common terms.
More Algebra Practice
If you need more algebra practice, try our other free algebra practice problems.
For exercises on solving equations using the steps for the substitution method, go to the next exercise:
