Solve a System of Three Linear Equations Using Elimination
If your finite math instructor asks you to solve a system of linear equations, one approach is to use elimination. For example, if you’re asked to solve a system of three linear equations in three unknowns, elimination is the best way to do this.
To solve the problem, you have to pick which variable to eliminate first. For example, the following system has three variables.
A convenient variable to eliminate is z. You see that opposite z terms appear in the first and second equations. And to deal with the last equation, you use a multiplier of 4, multiplied through the second equation, making that equation’s z term the opposite of the z term in the last equation.
First, add the first and second equations together:
Next, multiply the middle equation by 4 and add it to the last equation:
You’ve created two new equations that have just two variables: 5x – 4y = 22 and 7x – 14y = 56. Because the second equation has all numbers divisible by 7, you divide each term by 7 to make the numbers smaller, giving you x – 2y = 8.
Now write the two new equations, one under the other:
If you multiply the terms in the bottom equation by –2, you can eliminate the y terms when you add the two equations together.
Dividing both sides by 3, you get that x = 2. Substituting 2 for x in the equation 5x – 4y = 22, you get that y = –3. And substituting both the x and y values into the first original equation, you have 4(2) – (–3) + z = 10, simplifying to 11 + z = 10 or z = –1. The solution, as a point, is (2, –3, –1).