Systems with Three Linear Equations

When working with systems of equations, you can solve for one variable at a time. So, if a third linear equation comes along (bringing, of course, its variable z), well, three’s a crowd. However, you can easily deal with all the variables as long as you address each in turn.

You solve systems of three (or more) linear equations using the elimination method:

1. Starting with three equations, eliminate one variable to create two equations with the two remaining variables.

   Pair the first equation with the second, the second with the third, or the first with the third to eliminate one of the variables. Then choose a different pairing and eliminate the same variable.

2. From those two new equations, eliminate a second variable so you can solve for the one that remains.

3. Substitute back into the other equations to find the values of the other variables.

   Plug the first variable you solved for into one of the two-variable equations you found in Step 1. Then solve for the third variable by plugging the known values into one of the original equations.

Sample question

1. Find the common solution of the system of equations x + 5y – 2z = 2, 4x + 3y + 2z = 2, and 3x – 3y – 5z = 38.

   x= 4, y = –2, z = –4 — also written as the ordered triple (4, –2, –4). You can choose to eliminate any of the three variables, but there’s usually a good-better-best-worse-worst decision that can be made.

   In this problem, the best choice is to eliminate the x variable. The x variable has the only coefficient of 1 in all of the equations. You look for a 1 or –1 or for multiples of the same number in the coefficients of a single variable.

   Do two pairings of elimination. Multiply the first equation by –4 and add it to the second equation:

   

   For the second pairing, multiply the first equation by –3 and add it to the third equation:

   

   Then add the two equations that result (after multiplying the second equation by –10 so you can eliminate the z’s):

   

   Divide each side of the equation by 163 to get y = –2. Replace the y in –18y + z = 32 with the –2, and you get –18(–2) + z = 32; 36 + z = 32; z = –4.

   Now take the values for y and z and put them into any of the original equations to solve for x. You get x + 5(–2) – 2(–4) = 2; x – 10 + 8 = 2; x – 2 = 2; x = 4.

Practice questions

1. Find the common solution of the system of equations 3x + 4y – z = 7, 2x – 3y + 3z = 5, and x + 5y – 2z = 0.

2. Find the common solution of the system of equations 8x + 3y – 2z = –2, x – 3y + 4z = –13, and 6x + 4y – z = –3.

Following are answers to the practice questions:

1. The answer is x = 4, y = –2, z = –3.

   Eliminate x’s by multiplying the third equation by –3 and adding it to the first equation; you get –11y + 5z = 7. Then eliminate x’s in another combination by multiplying the original third equation by –2 and adding it to the second equation; you get –13y + 7z = 5. Use Cramer’s rule on these two resulting equations:

   
   

   Now substitute –2 for y and –3 for z in the original third equation to solve for x. You get x + 5(–2) – 2(–3) = 0; x – 10 + 6 = 0; x – 4 = 0; x = 4.

2. The answer is x = –1, y = 0, z = –3.

   Eliminate z’s by multiplying the first equation by 2 and adding it to the second equation to get 17x + 3y = –17. Then eliminate z’s in another combination by multiplying the third equation by 4 and adding it to the second equation; you get 25x + 13y = –25. Use Cramer’s rule on these two resulting equations:

   
   

   Now substitute x = –1 and y = 0 into the original third equation to get 6(–1) + 4(0) – z = –3; –6 – z = –3; –z = 3; z = –3.