Systems of Equationssupplement for UND’s Math 107 precalculus course

by Dave Morstad

Introduction

A system of equations is any group of equations.

A solution to a system of equations is a solution that works in everyequation in the group. For example, in the following system of equations:

2 52 3

x y

x y

− = −8− + =

the solution is x y= =1 2, because these values “work” in both equations. However, x y= − =3 0, is not a solution for the system even though it is asolution for the second equation.

Three methods of solving sytems of equations will be discussed in thissupplement: the graphical method, substitution, and elimination.

The Graphical Method

The graphical method consists of graphing every equation in the system andthen using the graph to find the coordinates of the point(s) where the graphsintersect. The point of intersection is the solution.

Example 1. Use the graphical method to solve thefollowing system of equations.

x y

x y

− = −− + =

22 3

Solution: Carefully graph both equations very precisely. If you don’t graph neatly, your point of intersectionwill be way off.

A graph is on the next page. The solution is x = –1, y = 1. Substitute these values into both equations to check the solution.

page 1

Solution to Example 1.

Now You Try It #1. Use the graphical method to solve.

2 3 03 9

x y

x y

− =+ =

The graphical method also works if the equations are not all linear.

Example 2. Use the graphical method to solve

y x

x y

=− + =

2

2

Solution: Carefully graph both equations very precisely. The graph below reveals two solutions: x = –1, y = 1 and x = 2, y = 4. Check both solutions.

page 2

page 3

Now You Try It #2. Use the graphical method to solve

x y

x y

2 2 93

+ =− + =

The graphical method is particularly helpful in understanding why a systemof equations might have no solution.

Example 3. Use the graphical method to solvex y

x y

+ = −+ =

12 2 4

Solution: After carefully graphing the equations, asshown below, it becomes apparent the two lines areparallel. Since parallel lines do not intersect, thereis no solution to this system of equation. A systemof equations that yields parallel lines is said to beinconsistent: the lines are parallel, the equations are inconsistent.

Now You Try It #3. Use the graphical method to solve

2 3 64 6 6x y

x y

− =− + =

As easy as the graphical method is to use, it is not very helpful when thesolution is not integers. Imagine trying to use the graphical method to find thesolution if the point of intersection is ( )37

11916241, .

Graphical Method Strengths: visualizing possible solutions.Graphical Method Weakness: imprecise.

The Substitution Method

To solve a system of equations using substitution, first solve one of theequations for a variable, then substitute it into the other equation(s). Alwayssubstitute into the other equation(s) and always use parentheses.

Example 4. Use substitution to solve the following sytemof equations.

3 4 82 5 3

x y

x y

+ =− + =

Solution. Step 1. Solve one of the equations for a variable. Let’s solve Eq. 1 for y.

3 4 84 8 3

2 34

x y

y x

y x

+ == −

= −

Step 2. Substitute this into the other equation (Eq. 2) and simplify.

− + =

− + − =

− + − =

− − = −

−

2 5 32 5 2 32 10 32 7

34

154

154

234

x y

x x

x x

x x

( )

x

x

x

= −

= − −

=

77 4

232823

( )

Step 3. “Back-substitute” this value of x into theequation from Step 1. to solve for y.

y x

y

y

y

= −

= −

= −

=

222

3434

2823

2123

2523

( )

The solution is x y= =2823

2523, . You should put these

values into both original equations to check yourwork.

page 4

Now You Try It #4. Use substitution to solve

3 4 72 3 9

x y

x y

− =− =

If there is no solution to the system of equations (parallel lines), thesubstitution method will result in something nonsensical, such as 4 = –2. Whenyou are using the substitution method and end up with an equation that is false,remember that it just means there is no solution.

One of the strengths of the substitution method is that it works for systemsof equations that are difficult or impossible to graph. For example, the system

2 2 23 2 22 2 3 3

x y z

x y z

x y z

+ − = −− + =

− − + =is not three lines, but rather is three planes. Each equation is a different plane.Solving requires finding the point where the three planes intersect. The diagrambelow illustrates this.

Using substitution allows you to ignore the three-dimensional or higherdimensional aspects of the graphical interpretations of such equations.

Example 5. Use substitution to solve the following system of equations.

2 2 23 2 22 2 3 3

x y z

x y z

x y z

+ − = −− + =

− − + =

page 5

Solution: Step 1. Solve an equation for one of thevariables. Let’s solve Eq. 1 for y.

2 2 22 2 2

x y z

y x z

+ − = −= − − +

, so

Step 2. Substitute this value for y into BOTH of the OTHER equations. First into Eq. 2. When yousimplify, it becomes a simple equation with twovariables. (Use parentheses to keep the signsstraight) :

3 2 23 2 2 2 2 23 4 4 4 27 4 3

x y z

x x z z

x x z z

x z

− + =− − − + + =+ + − + =+ −

( )( )

=− = −

27 3 2x z

Do the same substituting into Eq. 3.− − + =− − − − + + =− + + − + =

2 2 3 32 2 2 2 2 3 32 4 4 4 3 3

2

x y z

x x z z

x x z z

( )( )

x z

x z

+ − =− = −

4 32 1

Step 3. Now use these two resulting equations,

7 3 22 1

x z

x z

− = −− = −

to solve for x and z. Using substitution all overagain on these two equation should give you x = 1, z = 3.

Step 4. Now go back up to Step 1 andback-substitute these values of x and z into theequation for y.

y x z

y

y

= − − += − − +=

2 2 22 2 1 2 3

2( ) ( )

The solution is x y z= = =1 2 3, , . Use the original three equations to check this solution.

page 6

Now You Try It #5. Use substitution to solve

3 2 2 22 3 53 2 3 8

x y z

x y z

x y z

− + =+ + =+ + =

The Elimination Method

Elimination consists of adding equations together to eliminate variables. Sometimes you have to multiply equations by a number before you add them. The goal is to end up with one equation that has just one variable. Then you canuse back-substitution to solve for the other variable(s).

When using elimination, eliminate one variable at a time. It is alsoimportant to write down “instructions” that indicate how you are manipulating the equations going from step to step.

Example 6. Use elimination to solve 5 3 74 5 3x y

x y

+ = −+ = −

Solution Method1. Let’s get rid of the y variable. Multiplying the first equation by 5 and the secondequation by (–3) will make the y terms cancel.

5 3 74 5 3x y

x y

+ = −+ = −

5 1

3 2×

− × →( . )

( . )eq

eq 25 15 3512 15 9x y

x y

+ = −− − =

Now add the two new equations:25 15 35

12 15 9x y

x y

+ = −+ − − =( )

13 0 26x y+ = −

So x = –2. Use back-substitution to find that y = 1. This works, but Method 2 below is a little quickerand it generalizes to larger systems of equationsmuch more easily. Method 2 is what you’ll want touse most of the time.

page 7

Solution Method 2. In this method, you leave oneequation the same and replace the other equation(s). Add 5 times the first equation to –3 times thesecond equation. Put the result in for the secondequation.

5 3 74 5 3x y

x y

+ = −+ = −

( )( . ) ( )( . ) ( . )5 1 3 2 2eq eq eq+ − ⇒ → 5 3 713 0 26

x y

x y

+ = −+ = −

Next multipy the second equation so x has acoefficient of 1. Put it in for the second equation.

( )113 2 2× ⇒

→( . ) ( . )eq eq 5 3 7

0 2x y

x y

+ = −+ = −

So x = –2. Once again, use back-substituion to findthat y = 1. The solution is x = –2, y = 1.

Now You Try It #6. Use elimination to solve

3 6 37 5 11

x y

x y

− =− = −

If there is no solution to the system of equations, the elimination methodwill yield a false statement, just like substitution does. The real power of theelimination method becomes apparent in larger systems of equations.

Make sure you eliminate one variable at a time and be very systematic.Also, once you’ve got zeros in front of the variable you’re eliminating, avoiddoing anything later that will remove those zeros.

Example 7. Use elimination to solve2 2 23 2 2

6 4 4

x y z

x y z

x y z

+ − = −− + =

− − + =

Solution. First let’s knock out the y in the second and third equations.

2 2 23 2 2

6 4 4

x y z

x y z

x y z

+ − = −− + =

− − + =

( )( . ) ( . ) ( . )( . ) ( . ) ( . )2 1 2 2

1 3 3eq eq eq

eq eq eq+ ⇒

+ ⇒ → 2 2 27 0 3 2

4 0 2 2

x y z

x y z

x y z

+ − = −+ − = −

− + + =

page 8

Now we can multiply eq. 3 by 12 to give the z term a

coefficient of 1. Coefficients of 1 are nice to workwith. Trying to make any other coefficient 1 wouldintroduce fractions, which is all right, but they canbe messy to work with.

2 2 27 0 3 2

4 0 2 2

x y z

x y z

x y z

+ − = −+ − = −

− + + = ( )12 3 3( . ) ( . )eq eq⇒

→ 2 2 27 0 3 2

2 0 1

x y z

x y z

x y z

+ − = −+ − = −

− + + =

Next add 3 times eq. 3 to eq. 2 and put it in for eq. 2.

( )( . ) ( . ) ( . )3 3 2 2eq eq eq+ ⇒ → 2 2 2

0 0 12 0 1

x y z

x y z

x y z

+ − = −+ + =

− + + =

Equation 2 reveals x = 1. Back-substitute into eq. 3to find z = 3. Back-substitute both x and z into eq. 1 to get y = 2. The solution is x = 1, y = 2, z = 3.

Now You Try It #7. Use elimination to solve

3 2 5 32 6 4 204 3 5 5

x y z

x y z

x y z

− + = −+ − =− + = −

This technique of elimination is used extensively in other areas ofmathematics, including matrices and linear programming. Systematically andneatly re-writing all the equations in each step is very important in preventingerrors and loosing big points on exams. It also is very important in the way this technique is generalized to other problems.

page 9

Exercises

In exercises 1-12, solve the systems ofequations graphically.

1. 3 22

x y

x y

− =+ =

2. x y

x y

− =+ =

3 04

3. y x

x y

=+ =

2

2 04. y x

y

==

2

1

5. x y

x y

+ =+ = −

2 23 6 6

6. 2 16 3 6

x y

x y

− = −− + =

7. 3 3 62 2 8

x y

x y

+ = −− + =

8. 2 3 114 1

x y

x y

− = −+ = −

9. x y

x

2 2 164+ ==

10. x y

x y

2 2 8+ ==

11. y x

x y y

=

+ + =

2

2 2 2 0

12. y x

x y y

=

+ − + =

2

2 2 4 3 0

In exercises 13-24, solve the systems ofequations using substitution.. Don’t forget touse parentheses.

13. 3 52 4 9

x y

x y

− =+ =

14. 4 2 75 1

x y

x y

+ =+ =

15. a b

a b

2 2 8+ ==

16. a b

a b

2 2 202 0

+ =− =

17. 3 16 2 0

x y

x y

− =− + =

18. 6 9 12 3 1

x y

x y

− =− + =

19. 2 5 83 4 5

c d

c d

+ =− =

20. 3 4 75 6 12c d

c d

+ =− =

21. x x y

x y

2 4 33

− + = −− =

22. x x y

x y

2 2 22

+ + =− = −

23. 3 24 72

s t

s s t

+ =

− + =24. s t

s t

2

2 2

3

1

− = −

+ =

In exercises 25-42, solve the systems ofequations using elimination. Be sure to writeout your work neatly with directions betweeneach step.

25. 2 3 14 2 10

x y

x y

+ =− =

26. − + =− = −

3 4 46 5 5

x y

x y

27. 5 3 26 7 16

x y

x y

− = −− + =

28. − + =− = −

4 9 47 5 7

x y

x y

29. 3 2 56 4 1

x y

x y

− =− + =

30. 7 3 021 9 4

x y

x y

− =− =

31. 2 3 48 12 16

x y

x y

− =− + = −

32. 5 2 315 6 9

x y

x y

− =− =

33. 3 11

3 2 10

2

2

x y

x y

− =

− + = −34. 2 3 3

2 4 4

2

2

x y

x y

+ =

− = −

35.x y z

x y z

x y z

− + =− + − = −

− + =

2 3 72 3 83 2 7

page 10

36.2 3 53 3 2 105 4 3 5

x y z

x y z

x y z

+ − = −− + =+ − = −

37.3 6 9 35 7 10 4

11 4 6 2

x y z

x y z

x y z

+ − =− + = −+ − =

38.7 3 5 86 8 2 302 5 9 14

x y z

x y z

x y z

− + =+ − =+ − =

39.2 3 12 3 24 5 2 4

x y z

x y z

x y z

− + =− + − =− − + = −

40.4 2 3 33 3 4 32 5 2

x y z

x y z

x y z

+ − = −− + = −− + =

41.

x y z w

x y z w

x y z w

x y z w

+ − + =− + + − =

+ − − = −− + +

2 2 3 112 3 3 2 5

2 5 143 2 3 = 14

42.

2 4 2 223 5 3 2 242 2 2 2 164 3 2

x y z w

x y z w

x y z w

x y

− + + =+ − + =

− − + − = −− + z w− =3 2

page 11

Solutions to Now You Try It.

NYTI #1. NYTI #2.

NYTI #3. NYTI #4. x = –15, y = –13

NYTI #5. x = 0, y = 1, z = 2

NYTI #6. x = –3, y = –2

NYTI #7. x = 1, y = 3, z = 0

page 12

Solutions to odd numbered exercises.

1. 3.

5. 7.

9. 11.

page 13

page 14

13. x y= =2914

1714

, 15. a b and a b= = = − = −2 2 2 2, ,

17. no solution 19. c d= =5723

1423

,

21. x y and x y= = − = =0 3 3 0, , 23. no solution

25. x y= = −2 1, 27. x y= =2 4,

29. no solution 31. if x any number then y x= = −, 23

43

33. x y and x y= = = − =2 1 2 1, , 35. x y z= = =1 0 2, ,

37. x y z= = =0 2 1, , 39. no solution

41. x y z w= = = =1 2 3 4, , ,