You want to make a fruit salad for a summer picnic. Three pounds of strawberries plus five pounds of grapes plus one pound of melon cost $20. Three pounds of strawberries plus two pounds of grapes plus two pounds of melon cost $21. Four pounds of strawberries plus three pounds of grapes plus three pounds of melon cost $30. How much does each fruit cost?
Guidance
An equation in three variables, such as
A system of three equations in three variables consists of three planes in space. These planes could intersect with each other or not as shown in the diagrams below.
 In the first diagram, the three planes intersect at a single point and thus a unique solution exists and can be found.
 The second diagram illustrates one way that three planes can exist and there is no solution to the system. It is also possible to have three parallel planes or two that are parallel and a third that intersects them. In any of these cases, there is no point that is in all three planes.
 The third diagram shows three planes intersecting in a line. Every point on this line is a solution to the system and thus there are infinite solutions.
To solve a system of three equations in three variables, we will be using the linear combination method. This time we will take two equations at a time to eliminate one variable and using the resulting equations in two variables to eliminate a second variable and solve for the third. This is just an extension of the linear combination procedure used to solve systems with two equations in two variables.
Example A
Determine whether the point, (6, 2, 5), is a solution to the system:
Solution: In order for the point to be a solution to the system, it must satisfy each of the three equations.
First equation:
Second equation:
Third equation:
The point, (6, 2, 5), satisfies all three equations. Therefore, it is a solution to the system.
Example B
Solve the system using linear combinations:
Solution:
We can start by taking two equations at a time and eliminating the same variable. We can take the first two equations and eliminate
Result from equations 1 and 2:
Result from equations 2 and 3:
Now we have reduced our system to two equations in two variables. We can eliminate
Now use this value to find
Finally, we can go back to one of the original three equations and use our
Therefore the solution is (5, 2, 3).
Don’t forget to check your answer by substituting the point into each equation.
Equation 1:
Equation 2:
Equation 3:
Example C
Solve the system using linear combinations:
Solution: We can start by combining equations 1 and 2 together by multiplying the first equation by 5.
Since the result is a false equation, there is no solution to the system. If you end up with 0 = 0, then there will be infinitely many solutions.
Intro Problem Revisit The system of linear equations represented by this situation is:
The easiest way to solve this system is to first solve
Now we can substitute this value for m into the other two equations. When we do so we get a new system of linear equations:
Simplifying both equations results in:
If we multiply the first of these equations by –2, we get the new system of equations:
Now we can add these two equations to eliminate the s variable. When we do so, we get g = 2.
Next, we can substitute this value of g into either of these two equations to get the value of s :
Finally we substitute these values for g and s into one of our original equations.
Therefore, strawberries cost $3 per pound, grapes cost $2 per pound, and melon costs $4 per pound.
Guided Practice
1. Is the point, (3, 2, 1), a solution to the system:
2. Solve the following system using linear combinations:
3. Solve the following system using linear combinations:
Answers
1. Check to see if the point satisfies all three equations.
Equation 1:
Equation 2:
Equation 3:
Since the third equation is not satisfied by the point, the point is not a solution to the system.
2. Combine the first and second equations to eliminate
Result from equations 1 and 2:
Result from equations 1 and 3:
Now we have reduced our system to two equations in two variables. We can eliminate
Now find
Finally, we can go back to one of the original three equations and use our
Therefore the solution is (1, 0, 4).
Don’t forget to check your answer by substituting the point into each equation.
Equation 1:
Equation 2:
Equation 3:
3. Combine equations 1 and two to eliminate
Result from equations 1 and 2:
Result from equations 2 and 3:
Now we have reduced our system to two equations in two variables. Now we can combine these two equations and attempt to eliminate another variable.
Since the result is a false equation, there is no solution to the system.
Explore More
 Is the point, (2, 3, 5), the solution to the system:
 Is the point, (1, 3, 8), the solution to the system:
 Is the point, (0, 3, 5), the solution to the system:
 Is the point, (1, 1, 1), the solution to the system:
Solve the following systems in three variables using linear combinations.
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3x−2y+z4x+y−3z9x−2y+2z=0=−9=20

 .


11x+15y+5z3x+4y+z7x+13y+3z=1=−2=3

 .


2x+y+7z3x−2y−z4x−y+3z=5=−1=5

 .


x+3y−4z2x+5y−3z−x−3y+z=−3=3=−3

 .


3x+2y−5z3x+2y+5z6x+4y−10z=−8=−8=−16

 .


x+2y−z2x+4y+z3x−y+8z=−1=10=6

 .


x+y+z2x−y−z4x+y+z=−3=6=0

 .


4x+y+3z8x+2y+6z3x−3y−z=8=15=5

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2x+3y−zx−2y+3z−x+y−2z=−1=−4=3

 .


x−3y+4z−x+2y−5z2x+5y−3z=14=−13=−5

 .


x+y+zx+y−z2x+2y+z=3=3=6
