# 8.2 Systems of Three Equations and Three Unknowns

- Page ID
- 1020

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Later, you will learn about matrices and how to row reduce which will allow you to solve systems of equations in a new way. In order to set you up so that using matrices is logical and helpful, it is important to first solve a few systems of three equations using a very specific type of variable elimination.

When solving systems, what are you allowed to do to each equation?

**Solving Systems of Equations with Three Unknowns**

A system of three equations with three unknowns represents three planes in three dimensional space. When solving the system, you are figuring out how the planes intersect. One way that three planes could intersect is in a point:

A system of equations that has at least one solution is called a **consistent system**.

It is also possible for two or more planes to be parallel or each pair of planes to intersect in a line. In either of these cases the three planes do not intersect at a single point and the system is said to have no solution. A system of equations with no solutions is called an **inconsistent** **system**. If the three planes intersect at a line or a plane, there are an **infinite number of solutions. **

The following system of equations has the solution (1,3,7) . You can verify this by substituting 1 for \(x\), 3 for \(y,\) and 7 for \(z\) into each equation.

\(\begin{aligned} x+2 y-z &=0 \\ 7 x-0 y+z &=14 \\ 0 x+y+z &=10 \end{aligned}\)

One thing to be mindful of when given a system of equations is whether or not the equations are linearly independent. Three equations are linearly independent if each equation cannot be produced by a linear combination of the other two. Remember that a linear combination means that one equation can be written as the sum of multiples of the others.

When solving a system of three equations and three variables, there are a few general guidelines that can be helpful:

- Start by trying to eliminate the first variable in the second row.
- Next eliminate the first and second variables in the third row. This will create zero coefficients in the lower right hand corner.
- Repeat this process for the upper right hand corner and you should end up with a very nice diagonal indicating what \(x, y\) and \(z\) equal.

Take the system of equations mentioned above:

\(x+2 y-z=0\)

\(7 x-0 y+z=14\)

\(0 x+y+z=10\)

There are a number of ways to solve this system. Common techniques involve swapping rows,

dividing and multiplying a row by a constant and adding or subtracting a multiple of one row to

another.

Step 1: Swap rows 2 and 3 . Change -0 to \(+0 .\)

\(\begin{aligned} x+2 y-z &=0 \\ 0 x+y+z &=10 \\ 7 x+0 y+z &=14 \end{aligned}\)

Step 2: Subtract 7 times row 1 from row 3 , then replace row 3 .

\(\begin{aligned} x+2 y-z &=0 \\ 0 x+y+z &=10 \\ 0 x-14 y+8 z &=14 \end{aligned}\)

Step 3: Add 14 times row 2 to row 3 and then replace row \(3 .\)

\(\begin{aligned} x+2 y-z &=0 \\ 0 x+y+z &=10 \\ 0 x+0 y+22 z &=154 \end{aligned}\)

Step 4: Divide row 3 by 22 .

\(\begin{aligned} x+2 y-z &=0 \\ 0 x+y+z &=10 \\ 0 x+0 y+z &=7 \end{aligned}\)

Step 5: Subtract row 3 from row 2 and then replace row 2.

\(x+2 y-z=0\)

\(0 x+y+0 z=3\)

\(0 x+0 y+z=7\)

Step 6: Add row 3 to row 1 and then replace row 1.

\(x+2 y+0 z=7\)

\(0 x+y+0 z=3\)

\(0 x+0 y+z=7\)

Step 7: Subtract 2 times row 2 from row 1 and then replace row 1

\(x+0 y+0 z=1\)

\(0 x+y+0 z=3\)

\(0 x+0 y+z=7\)

The solution to the system is (1,3,7) exactly as stated earlier.

## Examples

Earlier, you were asked what you are allowed to do when solving a system of three equations. When solving a system of three equations with three unknowns, you are allowed to add and subtract rows, swap rows and scale rows. These three operations should allow you to eliminate the coefficients of the variables in a systematic way.

Is the following system linearly independent or dependent? How do you know?

\(3 x+2 y+z=8\)

\(x+y+z=3\)

\(5 x+4 y+3 z=14\)

\(6 x+6 y+6 z=18\)

With four equations and three unknowns there must be at least one dependent equation. The simplest method of seeing linearly dependence is to notice that one equation is just a multiple of the other. In this case the fourth equation is just six times the second equation and so it is dependent.

Most people will not notice that the third equation is also dependent. It is common to start doing a problem and notice somewhere along the way that all the variables in a row disappear. This means that the original equations were dependent. In this case, the third equation is the first equation plus two times the second equation. This means they are dependent.

Reduce the following system to a system of two equations and two unknowns.

\(3 x+2 y+z=7\)

\(4 x+0 y+z=6\)

\(6 x-y+0 z=5\)

Strategically swapping rows so that the zero coefficients do not live on the diagonal is a clever starting move.

Step 1: Swap rows 2 and 3.

\(3 x+2 y+z=7\)

\(6 x-y+0 z=5\)

\(4 x+0 y+z=6\)

Step 2: Scale row 3 by a factor of 3. Subtract 2 times row 1 from row 2 and replace row 2.

\(\begin{aligned} 3 x+2 y+z &=7 \\ 0 x-5 y-2 z &=-9 \\ 12 x+0 y+3 z &=18 \end{aligned}\)

Step 3: Subtract 4 times row 1 from row 3 and replace row 3.

\(3 x+2 y+z=7\)

\(0 x-5 y-2 z=-9\)

\(0 x-8 y-z=-10\)

Step 4: Scale the second row by 8 and the third row by 5.

\(\begin{aligned} 3 x+2 y+z &=7 \\ 0 x-40 y-16 z &=-72 \\ 0 x-40 y-5 z &=-50 \end{aligned}\)

Step 5: Subtract row 2 from row 3 and replace row 3.

\(\begin{aligned} 3 x+2 y+z &=7 \\ 0 x-40 y-16 z &=-72 \\ 0 x+0 y+11 z &=+22 \end{aligned}\)

Step 6: Divide row 3 by 11.

\(\begin{aligned} 3 x+2 y+z &=7 \\ 0 x-40 y-16 z &=-72 \\ 0 x+0 y+z &=2 \end{aligned}\)

Now that \(z=2\), rewrite the system so it becomes a system of two equations and two unknowns.

\(3 x+2 y+2=7\)

\(0 x-40 y-32=-72\)

Solving the second row shows that \(y=1\) which can be used to determine that \(x=1\).

When Kaitlyn went to the store with ten dollars she saw that she had some choices about what to buy. She could get one apple, one onion and one basket of blueberries for 9 dollars. She could get two apples and two onions for 10 dollars. She could also get two onions and one basket of blueberries for 10 dollars. Write and solve a system of equations with variables \(a, o\) and \(b\) representing each of the three things she can buy.

Here is the system of equations:

\(\begin{aligned} a+o+b &=9 \\ 2 a+2 o &=10 \\ 2 o+b &=10 \end{aligned}\)

Rewrite the system using \(x, y\) and \(z\) so that \(o\) and 0 do not get mixed up. Include coefficients of 0 so that each column represents one variable.

Step 1: Rewrite

\(1 x+1 y+1 z=9\)

\(2 x+2 y+0 z=10\)

\(0 x+2 y+1 z=10\)

Step 2: Subtract 2 times row 1 from row 2 and replace row 2.

\(1 x+1 y+1 z=9\)

\(0 x+0 y-2 z=-8\)

\(0 x+2 y+1 z=10\)

Step 3: Divide row 2 by -2.

\(\begin{aligned} 1 x+1 y+1 z &=9 \\ 0 x+0 y+1 z &=4 \\ 0 x+2 y+z &=10 \end{aligned}\)

At this point you can see from the second equation that \(z=4\). From the third equation, \(2 y+4=10\), so \(y=3\). Finally you can see from the first equation that \(x+3+4=9\) so \(x=2\). Apples cost 2 dollars each, onions cost 3 dollars each and blueberries cost 4 dollars each.

Show that the following system is dependent.

\(x+y+z=9\)

\(x+2 y+3 z=22\)

\(2 x+3 y+4 z=31\)

You could notice that the third equation is simply the sum of the other two. What happens when you do not notice and try to solve the system as if it were independent?

Step 1: Rewrite the system.

\(x+y+z=9\)

\(x+2 y+3 z=22\)

\(2 x+3 y+4 z=31\)

Step 2: Subtract 2 times row 1 from row 3 and replace row \(3 .\)

\(\begin{aligned} x+y+z &=9 \\ x+2 y+3 z &=22 \\ 0 x+1 y+2 z &=13 \end{aligned}\)

Step 3: Subtract row 1 from row 2 and replace row 2 .

\(\begin{aligned} x+y+z &=9 \\ 0 x+1 y+2 z &=13 \\ 0 x+1 y+2 z &=13 \end{aligned}\)

At this point when you subtract row 2 from row 3, all the coefficients in row 3 disappear. This means that you will end up with the following system of only two equations and three unknowns. Since the unknowns outnumber the equations, the system does not have a solution of one point.

\(\begin{aligned} x+y+z &=9 \\ 0 x+1 y+2 z &=13 \end{aligned}\)

1. An equation with three variables represents a plane in space. Describe all the ways that three planes could interact in space.

2. What does it mean for equations to be linearly dependent?

3. How can you tell that a system is linearly dependent?

4. If you have linearly independent equations with four unknowns, how many of these equations would you need in order to get one solution?

5. Solve the following system of equations:

\(3 x-4 y+z=-17\)

\(6 x+y-3 z=4\)

\(-x-y+5 z=16\)

6. Show that the following system is dependent:

\(\begin{aligned} 2 x-2 y+z &=5 \\ 6 x+y-3 z &=2 \\ 4 x+3 y-4 z &=-3 \end{aligned}\)

7. Solve the following system of equations:

\(\begin{array}{r} 4 x+y+z=15 \\ -2 x+3 y+4 z=38 \\ -x-y+3 z=16 \end{array}\)

8. Solve the following system of equations:

\(\begin{array}{c} 3 x-2 y+3 z=6 \\ x+3 y-3 z=-14 \\ -x+y+5 z=22 \end{array}\)

9. Solve the following system of equations:

\(3 x-y+z=-10\)

\(\begin{array}{l} 6 x-2 y+2 z=-20 \\ -x-y+4 z=12 \end{array}\)

10. Solve the following system of equations:

\(\begin{aligned} x-3 y+6 z &=-30 \\ 4 x+2 y-3 z &=18 \\ -8 x-3 y+2 z &=-22 \end{aligned}\)

11. Solve the following system of equations.

\(\begin{aligned} x+2 y+2 z+w &=5 \\ 2 x+y+2 z-0 w &=5 \\ 3 x+3 y+3 z+2 w &=12 \\ x+0 y+z+w &=1 \end{aligned}\)

A parabola goes through \((3,-9.5),(6,-32),\) and (-4,8)

12. Write a system of equations that you could use to solve to find the equation of the parabola. Hint: Use the general equation \(A x^{2}+B x+C=y\).

13. Solve the system of equations from #12.

A parabola goes through \((-2,3),(2,19),\) and \((1,6) .\)

14. Write a system of equations that you could use to solve to find the equation of the parabola. Hint: Use the general equation \(A x^{2}+B x+C=y\).

15. Solve the system of equations from #14.