GEOMETRY
“Solving Systems of Equations”
One has many decisions to make in life. One decision we will all hopefully be faced with at some point in our
life is what job to take. The decision to accept a certain job may be based on location, type of work, hours, people, and
benefits, among other things…oh yes, money is probably important somewhat too!
You are faced with that decision in this situation. You have been offered two jobs where the factors are nearly
identical. Job A pays $10.75 an hour and offers a bonus of $0.40 for each item completed. Job B pays $11.95 an hour
and offers a similar bonus of $0.25 for each item completed. So which job should you take? An equation for each job
could be created where y = hourly wage and x = # of items completed. The equations would be:
JOB A: y = 10.75 + 0.40x
JOB B: y = 11.95 + 0.25x
So what is the main factor that will decide which job to accept? It would be nice to know how many items a
worker could complete per hour. If these equations were graphed on a coordinate system, they would look as follows:
y = 10.75 + 0.40x
y = 11.95 + 0.25x
Job B appears to be the best at first but as more units are produced, Job A has the best wage. How many units
must be completed for the jobs to have the same hourly wage? EIGHT!!! How can this be computed? Solving a
system of equations allows us to answer this question and thus make an intelligent decision about which job to
accept!!!
COMPLETE ALL OF THE FOLLOWING ITEMS WITHOUT THE USE OF A CALCULATOR!!!
Systems of equations include more than one set of equations. Systems contain two variables. From a graphical
standpoint, the solution to a system is the place of intersection. This place of intersection is the ordered pair that the
graphs of the two equations have in common.
Solution (Point of
Intersection)
From a numerical standpoint, the solution to a system of equations is the number for each variable that makes each
equation true.
Given:
a + 2b = -2
4a – b = 10
If a = 2 and b = -2, then substituting these values in the first equation shows:
2 + 2(-2) = -2 TRUE
If a = 2 and b = -2, then substituting these values in the second equation shows:
4(2) – (-2) = 10 TRUE
Therefore, since both equations are true, a = 2 and b = -2 is the solution to that particular
system!
To find the solution, one must know how to graph equations and/or solve a system algebraically. These are elements
of coordinate geometry. We will focus on graphing equations later in the course; however, since numerical and
Algebraic skills are two items that we use on a regular basis, we are going to learn how to solve systems in this way.
OBJECTIVE:
to solve systems of equations by various methods.
SOLVING SYSTEMS OF EQUATIONS: METHOD ONE-SUBSTITUTION
*Given two equations with the same two variables in each, one can substitute one equation into the other if one of the
variables is already by itself. Notice that y is already by itself.
EXAMPLE:
x + 2y = 12
SOLUTION: Since y = 2x + 1, put 2x + 1 in for y in the
y = 2x + 1
equation x + 2y = 12 and solve for x.
x + 2(2x + 1) = 12
x + 4x + 2 = 12
5x + 2 = 12
5x = 10
x=2
To get y, plug 2 in for x in the equation
y = 2x + 1
y = 2(2) + 1
y=5
The solution to the system is x = 2, y = 5
*Here is another example of substitution. Notice that x is already by itself.
EXAMPLE:
x = 3y - 4
SOLUTION: Since x = 3y - 4, put 3y - 4 in for x in the
4x - 5y = -30
equation 4x - 5y = -30 and solve for y.
4(3y – 4) - 5y = -30
12y – 16 - 5y = -30
7y – 16 = -30
7y = -14
y = -2
To get x, plug -2 in for y in the equation
x = 3y - 4
x = 3(-2) - 4
x = -10
The solution to the system is x = -10, y = -2
How do we know how to substitute? Use the equation that has x or y by itself to substitute into the other equation.
But what if neither equation has x or y by itself? If an x or y exists in either equation, try to isolate
the variable.
*Here is an example of isolating a variable to solve a system by substitution. Look for variables with a coefficient of 1
and get that variable by itself. Variables with a coefficient of –1 can also be moved around and work as well.
EXAMPLE:
5x + y = 2
SOLUTION: Since a y exists in the equation 5x + y = 2,
5x – 3y = 4
isolate y so that y = 2 – 5x.
Substitute 2 – 5x in for y in 5x – 3y =4 and
solve for x.
5x – 3(2 – 5x) = 4
5x – 6 + 15x = 4
20x – 6 = 4
20x = 10
x=½
To get y, plug ½ in for x in the equation
rewrote, y = 2 – 5x
y = 2 – 5(½)
y = -½
The solution to the system is x = ½, y = -½
Solve the following linear systems by substitution! SHOW YOUR STEPS FOR EACH!!!
1.
x – 2y = -25
y = 3x
2.
2x + 3y = 31
y=x+7
3.
3x + 5y = 25
x = 2y – 10
SOLVING SYSTEMS OF EQUATIONS: METHOD TWO-CANCELLATION
*Given two equations with the same two variables in each and the equations written in the same order, you are able to
add or subtract the equations to cancel one of the variables.
EXAMPLE:
4x + 3y = 16
SOLUTION: Since 3y and –3y exist in both equations,
2x – 3y = 8
think how can you eliminate the y terms.
Add or subtract the two equations together?
Since 3y + -3y = 0, and we are trying to
eliminate one of the variables, adding is the
best solution.
4x + 3y = 16
+ 2x – 3y = 8
6x = 24
x=4
To get y, plug 4 in for x in either equation
and solve.
(I’ll use the second equation)
2(4) – 3y = 8
8 – 3y = 8
-3y = 0
y=0
The solution to the system is x = 4, y = 0
*Here is another example where one needs to decide if adding or subtracting the equations will allow one variable to
cancel. Here the equations are written in the same order.
EXAMPLE:
-5x - 2y = 4
SOLUTION: Since –5x and –5x exist in both equations,
-5x + 3y = 19
think how can you eliminate the x terms.
Add or subtract the two equations together?
Since –5x - -5x = 0, and we are trying to
eliminate one of the variables, subtracting
is the best solution.
-5x - 2y = 4
-(-5x + 3y = 19)
-5y = -15
y=3
To get x, plug 3 in for y in either equation
and solve.
(I’ll use the first equation)
-5x – 2(3) = 4
-5x – 6 = 4
-5x = 10
x = -2
The solution to the system is x = -2, y = 3
*What if the coefficients of the variables are not the same or opposites? First you must decide which variable you
want to eliminate. Multiply one or both equations by a number(s) so that the coefficients of one of the variables are
the same or are the opposite. It is best to multiply by numbers so that you add. Here is an example and two ways to
solve it.
EXAMPLE:
6x + 2y = 2
-3x + 3y = -9
SOLUTION #1:
SOLUTION #2:
To eliminate the x terms, multiply the second
Eliminating the y terms can be done in many ways.
equation by 2 (the least common multiple of
The least common multiple of 2 and 3 is 6;
6 and 3 is 6).
Therefore, we can multiply the first equation by 3:
(2) (-3x + 3y = -9) = -6x + 6y = -18
(3) (6x + 2y = 2) = 18x + 6y = 6
Add or subtract the original equations together?
and the second equation by –2:
Since 6x + -6x = 0, adding is best.
(-2) (-3x + 3y = -9) = 6x – 6y = 18
6x + 2y = 2
Since 6y + -6y = 0, adding is best.
+ (-6x + 6y = -18)
18x + 6y = 6
8y = -16
+ (6x – 6y = 18)
y = -2
24x = 24
To get x, plug –2 in for y in any one of the three
x=1
equations and solve.
To get y, plug 1 in for x in any one of the four
(I’ll use the first equation.)
equations and solve.
6x + 2(-2) = 2
(I’ll use the first equation.)
6x – 4 = 2
6(1) + 2y = 2
6x = 6
6 + 2y = 2
x=1
2y = -4
The solution to the system is x = 1, y = -2
y = -2
The solution to the system is x = 1, y = -2
Each way gives the same solution!!!
Solve the following linear systems by cancellation! SHOW YOUR STEPS FOR EACH!!!
4.
2x + y = 4
x–y=2
5.
x + 2y = 4
x + y = -2
6.
2x – 3y = 0
3x – 2y = 5
7.
5x + 3y = 15
3x – y = 9