We will check each equation by substituting in the values of the ordered triple for [latex]x,y[/latex], and [latex]z[/latex].

[latex]\begin{array}{ccccc}\begin{array}{r}\hfill x+y+z=2\\ \hfill \left(3\right)+\left(-2\right)+\left(1\right)=2\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}6x - 4y+5z=31\\ \hfill 6\left(3\right)-4\left(-2\right)+5\left(1\right)=31\\ \hfill 18+8+5=31\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{}5x+2y+2z=13\\ \hfill 5\left(3\right)+2\left(-2\right)+2\left(1\right)=13\\ \hfill \text{}15 - 4+2=13\\ \hfill \text{True}\end{array}\end{array}[/latex]

The ordered triple [latex]\left(3,-2,1\right)[/latex] is indeed a solution to the system.

The third equation states that z = −1, so we substitute this into the second equation to obtain a solution for y.

[latex]\begin{array}y-\frac{1}{2}(-1)=4\\y+\frac{1}{2}=4\\y=4-\frac{1}{2}\\y=\frac{8}{2}-\frac{1}{2}\\y=\frac{7}{2}\end{array}[/latex]

Now we have two of our solutions and we can substitute them both into the first equation to solve for x.

[latex]\begin{array}x-\frac{1}{3}\left(\frac{7}{2}\right)+\frac{1}{2}\left(-1\right)=1\\x-\frac{7}{6}-\frac{1}{2}=1\\x-\frac{7}{6}-\frac{3}{6}=1\\x-\frac{10}{6}=1\\x=1+\frac{10}{6}\\x=\frac{6}{6}+\frac{10}{6}\\x=\frac{16}{6}=\frac{8}{3}\end{array}[/latex]

Now we have our ordered triple, remember that where you place the solutions matters!

Answer

[latex](x,y,z)=\left(\frac{8}{3},\frac{7}{2},-1\right)[/latex]

Now we can substitute the value for z that we obtained into equation (2).

[latex]\begin{array}{rrr}-2y+(6)=6\\-2y=6-6\\-2y=0\\\,\,\,\,y=0\end{array}[/latex]

Be careful here not to get confused with a solution of y = 0 and an inconsistent solution.  It’s ok for variables to equal 0.

Now we can substitute z = 6 and y = 0 back into the original equation.

[latex]\begin{array}{rrr}x-y+z=5\\x-0+6=5\\x+6=5\\x=5-6\\x=-1\end{array}[/latex]

Answer

[latex](x,y,z)=(-1,0,6)[/latex]

If you multiply equation (2) by 2 and add (2) and (3) together, you can eliminate y and solve for z.

First, multiply equation (2) by 2.

[latex]\begin{array}\,\,\,\,\,\,2(-y-4z=7)\\-2y-8z=14\end{array}[/latex]

Next, add equations (2) and (3) together to eliminate y and solve for z.

[latex]\begin{array}\,\,\,\,\,\,-2y-8z=14\\\underline{+2y+8z=-12}\\0+0=2\\0=2\end{array}[/latex]

Recall that when we were solving systems with two variables, a non-true solution such as [latex]0=2[/latex] implied that there was no solution to the system.

This can happen with a system of three variables as well.

Answer

No solution, or DNE

First, we can multiply equation (1) by [latex]-2[/latex] and add it to equation (2).

[latex]\begin{array} −4x−2y+6z=0 \hfill& \text{equation }\left(1\right)\text{multiplied by }−2 \\ 4x+2y−6z=0\hfill&\left(2\right) \end{array}[/latex]

We do not need to proceed any further. The result we get is an identity, [latex]0=0[/latex], which tells us that this system has an infinite number of solutions.

As shown in the figure below, two of the planes are the same and they intersect the third plane on a line. The solution set is infinite, as all points along the intersection line will satisfy all three equations.

Answer

There are an infinite number of solutions.

To set up the system, first choose the variables. In this case the unknown values are the number of small, medium, and large photos.

S = number of small photos sold

M = number of medium photos sold

L = number of large photos sold

The total of her sales must be $300 to pay for the booth. We can find the total by multiplying the cost for each size by the number of that size sold.

10S = money received for small photos

15M = money received for medium photos

40L = money received for large photos

Total Sales: 10S + 15M + 40L = 300

The number of small photos is the same as the total of medium and large photos combined.

S = M + L

She sells twice as many medium photos as large photos.

M = 2L

To make things easier, rewrite the equations to be in the same format, with all variables on the left side of the equal sign and only a constant number on the right.

[latex]\begin{cases}10S+15M+40L=300\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,S–M–L=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M–2L=0\end{cases}[/latex]

Now solve the system.

Step 1: First choose two equations and eliminate a variable. Since one equation has no S variable, it may be helpful to use the other two equations and eliminate the S variable from them. Multiply the second equation by −10 and add.

[latex]\begin{array}-10(S–M–L=0)\\-10s+10M+10L=0\end{array}[/latex]

[latex]\begin{array}{ccc}10S+15M+40L=300\\\underline{+(-10s+10M+10L=0)}\\25M+50L=300\end{array}[/latex]

Step 2: The second equation for our two-variable system will be the remaining equation (that has no S variable). Eliminate a second variable using the equations from steps 1 and 2. While you could multiply the second equation by 25 to eliminate L, the numbers will stay nicer if you divide the first equation by 25. Don’t forget to be careful of the signs!

Divide first:

[latex]\begin{array}{ccc}\frac{25}{25}M+\frac{25}{50}L=\frac{300}{25}\\M+2L=12\end{array}[/latex]

Now eliminate L by adding M-2L=0  from this new equation.

[latex]\begin{array}M+2L=12\\\underline{+(M–2L=0)}\\2M=12\\M=\frac{12}{2}=6\end{array}[/latex]

Step 3: Use M=6 and one of the equations containing just two variables to solve for the second variable.  It’s best to use one of the original equations—in case an error was made in multiplication.

[latex]\begin{array}{ccc}M=2L\\6=2L\\3=L\end{array}[/latex]

Step 4: Use the two found values and one of the original equations to solve for the third variable.

[latex]\begin{array}{ccc}S–M–L=0\\S-6-3=0\\S-9=0\\s=9\end{array}[/latex]

Step 5:Check your answer. With application problems, it’s sometimes easier (and better) to use the original wording of the problem rather than the equations you write.

She usually sells as many small photos as medium and large photos combined.

• Medium and large photos combined =6 + 3 = 9, which is the number of small photos.

She also sells twice as many medium photos as large.

• Medium photos is 6, which is twice the number of large photos (3).

A booth at the art fair costs $300.

• Andrea receives $10(9) or $90 for the 9 small photos, $15(6) or $90 for the 6 medium photos, and $40(3) or $120 for the large photos. $90 + $90 + $120 = $300.

Answer

If Andrea sells 9 small photos, 6 medium photos, and 3 large photos, she’ll receive exactly the amount of money needed to pay for the booth.