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).