Systems of Equations
supplement 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 every
equation in the group. For example, in the following system of equations:
2 x − 5 y = −3
−x + 2 y = 3
the solution is x = 1, y = 2 because these values “work” in both equations.
However, x = −3, y = 0 is not a solution for the system even though it is a
solution for the second equation.
Three methods of solving sytems of equations will be discussed in this
supplement: the graphical method, substitution, and elimination.
The Graphical Method
The graphical method consists of graphing every equation in the system and
then using the graph to find the coordinates of the point(s) where the graphs
intersect. The point of intersection is the solution.
Example 1. Use the graphical method to solve the
following system of equations.
x − y = −2
−2 x + y = 3
Solution: Carefully graph both equations very precisely.
If you don’t graph neatly, your point of intersection
will 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.
2x − 3y = 0
x + 3y = 9
The graphical method also works if the equations are not all linear.
Example 2. Use the graphical method to solve
y = x2
−x + y = 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
Now You Try It #2. Use the graphical method to solve
x2 + y2 = 9
−x + y = 3
The graphical method is particularly helpful in understanding why a system
of equations might have no solution.
Example 3. Use the graphical method to solve
x + y = −1
2x + 2 y = 4
Solution: After carefully graphing the equations, as
shown below, it becomes apparent the two lines are
parallel. Since parallel lines do not intersect, there
is no solution to this system of equation. A system
of equations that yields parallel lines is said to be
inconsistent: the lines are parallel, the equations are
inconsistent.
Now You Try It #3. Use the graphical method to solve
2x − 3y = 6
−4 x + 6 y = 6
As easy as the graphical method is to use, it is not very helpful when the
solution is not integers. Imagine trying to use the graphical method to find the
37
16
, 241
solution if the point of intersection is ( 119
).
Graphical Method Strengths: visualizing possible solutions.
Graphical Method Weakness: imprecise.
page 3
The Substitution Method
To solve a system of equations using substitution, first solve one of the
equations for a variable, then substitute it into the other equation(s). Always
substitute into the other equation(s) and always use parentheses.
Example 4. Use substitution to solve the following sytem
of equations.
3x + 4y = 8
−2 x + 5 y = 3
Solution. Step 1. Solve one of the equations for a variable.
Let’s solve Eq. 1 for y.
3x + 4 y = 8
4 y = 8 − 3x
y = 2 − 43 x
Step 2. Substitute this into the other equation
(Eq. 2) and simplify.
−2 x + 5 y = 3
−2 x + 5(2 − 43 x) = 3
−2 x + 10 − 154 x = 3
−2 x − 154 x = −7
− 234 x = −7
x = −7( − 234 )
x=
28
23
Step 3. “Back-substitute” this value of x into the
equation from Step 1. to solve for y.
y = 2 − 43 x
y = 2 − 43 ( 28
)
23
21
y = 2 − 23
y=
25
23
The solution is x = 28
, y = 25
. You should put these
23
23
values into both original equations to check your
work.
page 4
Now You Try It #4. Use substitution to solve
3x − 4 y = 7
2x − 3y = 9
If there is no solution to the system of equations (parallel lines), the
substitution method will result in something nonsensical, such as 4 = –2. When
you 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 systems
of equations that are difficult or impossible to graph. For example, the system
2 x + y − 2 z = −2
3x − 2 y + z = 2
−2 x − 2 y + 3 z = 3
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 diagram
below illustrates this.
Using substitution allows you to ignore the three-dimensional or higher
dimensional aspects of the graphical interpretations of such equations.
Example 5. Use substitution to solve the following system
of equations.
2 x + y − 2 z = −2
3x − 2 y + z = 2
−2 x − 2 y + 3 z = 3
page 5
Solution: Step 1. Solve an equation for one of the
variables. Let’s solve Eq. 1 for y.
2 x + y − 2 z = −2, so
y = −2 − 2 x + 2 z
Step 2. Substitute this value for y into BOTH of the
OTHER equations. First into Eq. 2. When you
simplify, it becomes a simple equation with two
variables. (Use parentheses to keep the signs
straight) :
3x − 2 y + z = 2
3x − 2( −2 − 2 x + 2 z ) + z = 2
3x + (4 + 4 x − 4 z ) + z = 2
7 x + 4 − 3z = 2
7 x − 3 z = −2
Do the same substituting into Eq. 3.
−2 x − 2 y + 3 z = 3
−2 x − 2( −2 − 2 x + 2 z ) + 3z = 3
−2 x + ( 4 + 4 x − 4 z ) + 3 z = 3
2x + 4 − z = 3
2 x − z = −1
Step 3. Now use these two resulting equations,
7 x − 3 z = −2
2 x − z = −1
to solve for x and z. Using substitution all over
again on these two equation should give you x = 1,
z = 3.
Step 4. Now go back up to Step 1 and
back-substitute these values of x and z into the
equation for y.
y = −2 − 2 x + 2 z
y = −2 − 2(1) + 2(3)
y=2
The solution is x = 1, y = 2, z = 3. Use the original
three equations to check this solution.
page 6
Now You Try It #5. Use substitution to solve
3x − 2 y + 2 z = 2
2x + 3y + z = 5
3x + 2 y + 3z = 8
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 can
use back-substitution to solve for the other variable(s).
When using elimination, eliminate one variable at a time. It is also
important to write down “instructions” that indicate how you are manipulating the
equations going from step to step.
Example 6. Use elimination to solve
5 x + 3 y = −7
4 x + 5 y = −3
Solution Method1. Let’s get rid of the y variable.
Multiplying the first equation by 5 and the second
equation by (–3) will make the y terms cancel.
5⋅( eq. 1)
5 x + 3 y = −7 −
3⋅( eq. 2 )
→ 25 x + 15 y = −35
4 x + 5 y = −3
−12 x − 15 y = 9
Now add the two new equations:
25 x + 15 y = −35
+ ( −12 x − 15 y = 9)
13 x + 0 y = −26
So x = –2. Use back-substitution to find that y = 1.
This works, but Method 2 below is a little quicker
and it generalizes to larger systems of equations
much more easily. Method 2 is what you’ll want to
use most of the time.
page 7
Solution Method 2. In this method, you leave one
equation the same and replace the other equation(s).
Add 5 times the first equation to –3 times the
second equation. Put the result in for the second
equation.
5x + 3 y = −7 (
5 )( eq. 1) + ( −3 )( eq. 2 ) ⇒ ( eq. 2 )
→ 5 x + 3 y = −7
4 x + 5 y = −3
13x + 0 y = −26
Next multipy the second equation so x has a
coefficient of 1. Put it in for the second equation.
( 131 )⋅( eq. 2 ) ⇒ ( eq. 2 )→ 5 x + 3 y = −7
 
x + 0 y = −2
So x = –2. Once again, use back-substituion to find
that y = 1. The solution is x = –2, y = 1.
Now You Try It #6. Use elimination to solve
3x − 6 y = 3
7 x − 5 y = −11
If there is no solution to the system of equations, the elimination method
will yield a false statement, just like substitution does. The real power of the
elimination 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, avoid
doing anything later that will remove those zeros.
Example 7. Use elimination to solve
2 x + y − 2 z = −2
3x − 2 y + z = 2
−6 x − y + 4 z = 4
Solution. First let’s knock out the y in the second and third
equations.
2 x + y − 2 z = −2
2 x + y − 2 z = −2
( 2 )( eq. 1) + ( eq. 2 ) ⇒ ( eq. 2 )
( eq. 1)+ ( eq. 3 ) ⇒ ( eq. 3 )
3x − 2 y + z = 2  
→ 7 x + 0 y − 3 z = −2
−6 x − y + 4 z = 4
−4 x + 0 y + 2 z = 2
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 work
with. Trying to make any other coefficient 1 would
introduce fractions, which is all right, but they can
be messy to work with.
2 x + y − 2 z = −2
2 x + y − 2 z = −2
1
(
eq
.
3
)
⇒
(
eq
.
3
)
(
)
2
7 x + 0 y − 3z = −2  
→ 7 x + 0 y − 3z = −2
−4 x + 0 y + 2 z = 2
−2 x + 0 y + z = 1
Next add 3 times eq. 3 to eq. 2 and put it in for
eq. 2.
2 x + y − 2 z = −2
( 3 )( eq. 3 )+ ( eq. 2 ) ⇒ ( eq. 2 )
  
→
x + 0 y + 0z = 1
−2 x + 0 y + z = 1
Equation 2 reveals x = 1. Back-substitute into eq. 3
to 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 x − 2 y + 5 z = −3
2 x + 6 y − 4 z = 20
4 x − 3 y + 5 z = −5
This technique of elimination is used extensively in other areas of
mathematics, including matrices and linear programming. Systematically and
neatly re-writing all the equations in each step is very important in preventing
errors and loosing big points on exams. It also is very important in the way this
technique is generalized to other problems.
page 9
Exercises
17.
In exercises 1-12, solve the systems of
equations graphically.
1.
3x − y = 2
x+ y =2
3.
y = x2
2x + y = 0
3x − y = 1
−6 x + 2 y = 0
19. 2 c + 5d = 8
3c − 4 d = 5
18.
6x −9y =1
−2 x + 3 y = 1
20. 3c + 4d = 7
5c − 6d = 12
2.
x − 3y = 0
x+ y= 4
4.
y = x2
y =1
21.
5.
x + 2y = 2
3 x + 6 y = −6
6.
2 x − y = −1
−6 x + 3 y = 6
23.
7.
3 x + 3 y = −6
−2 x + 2 y = 8
8.
2 x − 3 y = −11
4 x + y = −1
In exercises 25-42, solve the systems of
equations using elimination. Be sure to write
out your work neatly with directions between
each step.
9.
x 2 + y 2 = 16
x=4
10.
x2 + y2 = 8
x= y
11.
x 2 − 4 x + y = −3 22.
x− y=3
x2 + 2x + 2 = y
x − y = −2
3s + t = 2
s 2 − t = −3
s 2 − 4s + 7 = t
25. 2 x + 3 y = 1
4 x − 2 y = 10
y = x2
24.
s2 + t2 =1
26.
−3 x + 4 y = 4
6 x − 5 y = −5
27.
5 x − 3 y = −2
−6 x + 7 y = 16
28.
−4 x + 9 y = 4
7 x − 5 y = −7
29.
3x − 2 y = 5
−6 x + 4 y = 1
30.
7 x − 3y = 0
21 x − 9 y = 4
In exercises 13-24, solve the systems of
equations using substitution.. Don’t forget to
use parentheses.
31.
2 x − 3y = 4
32. 5 x − 2 y = 3
−8 x + 12 y = −16
15 x − 6 y = 9
13. 3x − y = 5
2x + 4 y = 9
14.
4x + 2 y = 7
5x + y = 1
33.
a2 + b2 = 8
a=b
16.
a 2 + b 2 = 20
2a − b = 0
12.
15.
x2 + y2 + 2 y = 0
y = x2
x2 + y2 − 4 y + 3 = 0
35.
3 x 2 − y = 11
2
34. 2 x + 3 y = 3
−3 x 2 + 2 y = −10
2 x 2 − 4 y = −4
x − 2 y + 3z = 7
−2 x + y − 3 z = −8
3x − y + 2z = 7
page 10
2 x + y − 3 z = −5
36. 3x − 3 y + 2 z = 10
5 x + 4 y − 3 z = −5
3 x + 6 y − 9z = 3
37. 5 x − 7 y + 10 z = −4
11 x + 4 y − 6 z = 2
7 x − 3 y + 5z = 8
38. 6 x + 8 y − 2 z = 30
2 x + 5 y − 9 z = 14
39.
2 x − 3y + z = 1
−2 x + 3 y − z = 2
−4 x − 5 y + 2 z = −4
4 x + 2 y − 3 z = −3
40. 3x − 3 y + 4 z = −3
2 x − y + 5z = 2
41.
x + 2 y − 2 z + 3w = 11
−2 x + 3 y + 3 z − 2w = 5
x + 2 y − 5 z − w = −14
3x − 2 y + z + 3w = 14
42.
2 x − y + 4 z + 2w = 22
3 x + 5 y − 3 z + 2w = 24
−2 x − 2 y + 2 z − 2w = −16
4 x − 3 y + 2 z − 3w = 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
29
17
, y=
14
14
13.
x=
15.
a = 2, b = 2 and
17.
no solution
21.
x = 0, y = −3 and
19.
c=
23.
no solution
25.
x = 2, y = −1
27.
x = 2, y = 4
29.
no solution
31.
if x = any number , then y = 23 x − 43
33.
x = 2, y = 1 and
35.
x = 1, y = 0, z = 2
37.
x = 0, y = 2, z = 1
39.
no solution
41.
x = 1, y = 2, z = 3, w = 4
x = 3, y = 0
x = −2, y = 1
a = −2, b = −2
57
14
,d=
23
23
page 14