MA 12C
Tech Math I Credit
Last Update: March 2008
Focus: Solving Linear Systems
Arcadia Valley Career Technology Center
Topic: Linear Equations and Systems
Show-Me Standards: MA1,MA4,MA5
MO Grade Level Expectations: : N2C8,
N2C9, A2B7
NCTM Standards: 4A, 5A
OBJECTIVE: The students will be able to solve systems of linear equations by graphing
and algebraically.
Terms:
Systems of Equations- two or more equations with the same variables.
Solving a System of Equations- finding the ordered pair that satisfies all the equations of the
system. This pair will be the point of intersection of the lines.
Methods of Solving a System:
1. Graphing
2. Algebraically
-Substitution and Elimination Methods
Examples:
EX1. Solve the system of equations by Graphing.
2x + y = 5
x–y=1
-write each equation in Slope-Intercept form.
2x + y = 5
2x – 2x + y = 5 – 2x
y = -2x + 5
m = -2 or -2/1 and b = 5
x–y=1
x–x–y=1–x
-y = 1 – x
-1y = 1 – x
-1
-1 -1
y = -1 + x
y=x–1
m = 1 or 1/1 and b = -1
-graph both lines on the same set of coordinate axes
y
^
● (0,5)
●
● (2,1)
●
● (0,-1)
>
x
-the point of intersection is (2,1)
therefore, the solution to the
system is (2,1).
-you can check your answer by substituting the pair (2,1) back into the two equations of
the system.
(2,1)
x = 2 and y = 1
2x + y = 5
2(2) + (1) = 5
5=5
(2,1)
x = 2 and y = 1
x–y=1
(2) – (1) = 1
1=1
-since the pair “works” in both equations(makes them both true), it must be the solution.
EX 2. Solve the system algebraically by using the Substitution Method.
x + 4y = 26
x – 5y = -10
-solve either equation for a single variable. Let’s take the second equation and solve for
the x variable in terms of y.
x – 5y + 5y = -10 + 5y
x = -10 +5y
-substitute -10 + 5y for x in the first equation.
(-10 + 5y) + 4y = 26
-10 + 9y = 26
-10 + 10 + 9y = 26 + 10
9y = 36
y=4
-we now know that the solution is (__,4) because y = 4 when we solved the two equations
simultaneously by substituting one equation into the other. Let’s find the x coordinate
that goes with it. We do this by substituting the 4 for y in either of the two original
equations of the system. Let’s choose the first equation.
x
x
x
x
+
+
+
=
4(4) = 26
16 = 26
16 – 16 = 26 – 16
10
-the solution to this system must be (10,4).
EX3. Solve the system algebraically by using the Substitution Method.
y + 7 = 2x
3y – x = -1
-solve the first equation for the y variable in terms of x.
y + 7 – 7 = 2x – 7
y = 2x – 7
-substitute 2x – 7 for y in the second equation.
3(2x – 7) – x = -1
6x – 21 – x = -1
6x – 21 + 21 – x = -1 + 21
6x – x = 20
5x = 20
x=4
-substitute 4 for x in either of the two original equations of the system. Let’s choose the
second equation this time.
3y – (4) = -1
3y – 4 + 4 = -1 + 4
3y = 3
y=1
-the solution to this system must be (4,1).
EX4. Solve the system algebraically by using the Elimination Method.
2x + 3y = 12
5x – 2y = 11
-the elimination method allows us to eliminate one of the variables by adding or subtracting
the two equations vertically by like terms. In this example, we are going to have to create
opposites with the coefficients of either the x or y terms so that when we add or subtract
the two equations together, a variable is eliminated. Let’s multiply the first equation by 2
and the second equation by 3.
2(2x + 3y = 12)
3(5x – 2y = 11)
4x + 6y = 24
15x – 6y = 33
-we now will add the two new equations vertically collecting the x terms , y terms, and
constant numbers.
4x + 6y = 24
+15x – 6y = 33
19x + 0y = 57
or
19x = 57
,
x=3
-substitute 3 for x in either of the two original equations of the system. Let’s choose the
first equation.
2(3) + 3y = 12
6 + 3y = 12
6 – 6 + 3y = 12 – 6
3y = 6
y=2
-the solution to this system must be (3,2).
-we chose to multiply 2 to the first equation and 3 to the second equation because it creates
an opposite pair of coefficients for the y terms in the equations. It created 6y and -6y
which, when added, eliminated the y variables.
EX. 5 Solve the system algebraically by using the Elimination Method.
x + y = -7
3x + y = -9
-multiply the first equation by -1. Leave the second equation alone.
-1(x + y = -7)
3x + y = -9
-1x + -1y = 7
3x + y = -9
-add the two equations together.
-1x + -1y = 7
+ 3x + y = -9
2x + 0y = -2
or
2x = -2
,
x = -1
-substitute -1 for x in either of the two original equations. Let’s choose the second equation.
3(-1) + y = -9
-3 + y = -9
-3 + 3 + y = -9 + 3
y = -6
-the solution to the system must be (-1,-6).
Ex. 6 Solve the system algebraically by using the Elimination Method.
4x – 3y = 3
8x + 5y = 50
-multiply the first equation by -2 and leave the second equation alone.
-2(4x – 3y = 3)
8x + 5y = 50
-8x + 6y = -6
8x + 5y = 50
-add the two equations together.
-8x + 6y = -6
+8x + 5y = 50
0x + 11y = 44
or
11y = 44
,
y=4
-substitute 4 for y in either of the two original equations. Let’s choose the first equation.
4x – 3(4) = 3
4x – 12 = 3
4x – 12 + 12 = 3 +12
4x = 15
x = 3¾
-the solution to the system must be (3¾,4).
Problems:
Solve the systems using the Graphing Method.
1. y = x + 4
y = -x + 6
2. y + 2x = 5
2y – 5x = 10
Solve the systems using the Substitution Method.
3. y – 5 = x
2x + y = 8
4. 5x – y = 17
3x + 2y = 5
Solve the systems using the Elimination Method.
5. 10x – 9y = 15
5x – 4y = 10
6. 3x – 2y = -31
5x + 6y = 23
7. 3x +5y = 6
2x – 4y = -7