Programmed tutorial: Elimination Method for Linear Systems: Example 3

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Use the Elimination Method to solve

Equation 1

Equation 2

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Solving is easiest if we clear the denominators. In equation 1, what is the least common denominator?

Multiply both sides of equation 1 by 30 to clear the denominators.

Distribute the 30.

Do the multiplication.

5w + 3z = 120

In equation 2, what is the least common denominator?

Multiply both sides of equation 2 by 10 to clear the denominators.

Distribute the 10.

Do the multiplication.

5w – 2z = 20

Now we have these two equations to solve simultaneously:
5w + 3z = 120 Equation 3
5w – 2z = 20 Equation 4
How do we proceed?

We can multiply equation 4 by (-1) and add to eliminate w.

Do the multiplication.

5w + 3z = 120
–5w + 2z = –20

Add these equations.

We get 5z = 100

Solve for z.

z = 20

Use this value of z in either equation to find w.

Using equation 3, 5w + 3z = 120,
and substituting z = 20, we get
5w + 3(20) = 120

Do the multiplication.

5w + 60 = 120

Subtract 60 from both sides.

5w = 60

Divide by 5.

w = 12

Summarize our results.

The solution to this linear system is
w = 12, and z = 20

Check w = 12 and z = 20 in the original equations.

In equation 1:

becomes

2 + 2 = 4; Checks.

In equation 2:

becomes

6 – 4 = 2; Checks.

The end. If you found this helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor