# 3.1 Solving Systems of Equations Graphing

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```					             3.1
Solving Systems of Equations
Graphing & Substitution

Date: __________
Pam has \$120 and is spending \$5 every week.
Lorenzo has \$20 and is saving \$7.50 every week.
When will they have the same amount of money?
Let x = # of weeks    Let y = total money
Pam         120                       Lorenzo
y  120  5x          100
y  20  7.5x
x y                                                         x y
total money

80
0 120                                                       0 20
2 110                     60                                2 35
4 100                     40                                4 50
6 90                      20           In 8                 6 65
8 80                                   weeks                8 80
0
10 70                          0   2   4   6
weeks
8   10      10 95
System of Equations
•two or more equations involving
the same variables
•Solutions are an ordered pair
Which of the following ordered pairs are
solutions to the following system.
5x  2y  10
4 x  y  8
1) (3, 1)               2) (2, 0)
?
5(3)  2(1)  10       5(2)  2(0)  10
17  10             4(2)  (0)  8
No                 Yes

Solution must work in both equations!
Solving Systems by Graphing
y  2x  1                            y

x  y  5                                    (2,3)
x      x
y  x  5
x
plugging the point into both
original equations.
y = 2x – 1      x+y=5
3 = 2(2) – 1    2+3=5
(2,3) is the solution!
3=4–1           5=5
3=3
Solve.

x  y  1
2y  x  4
y           2y  x  4
x  y  1                             x  x
x     x                            2y  x  4
y  x 1     (-2,1)                2       2
1
x
y x 2
2

(2,1) is the solution to the system.
Solve.


2x  3y  3
x  6y  24
y             x  6y  24
2x  3y  3                            x        x
 2x      2x
6y  x  24
3y  2x  3
2                                        1
y  x 1                              x
y      x4
3                                         6

(-6, -3)

(6,3) is the solution.
Solve the sytem below by graphing.

2x  3y  6
x  4y  8
y                 x  4y  8
2x  3y  6                            x      x
 2x      2x
4y  x  8
3y  2x  6
2                                        1
y     x2                                y x2
3                                x       4
1
(4 , -1)
2

1
(4 ,1) is close to the solution.
2
3 Possible Outcomes
Independent        Inconsistent      Dependent
System            System           System

Lines Intersect    Lines Parallel   Lines Coincide
One                  No              Infinite
Solution           Solution          Solutions
Solve the system graphically.
y
 x  y  3
                                x  2y  4
                               x      x
x  2y  4                      2y  x  4
x  y  3                       x 2       2
x     x                                1
y x 3                            y  x 2
2

 1 1     One
  ,2
 2 2    Solution
Solve.                        y
 1
y  x  2
 3
2x  6y  12

2x         2x                          x
6y  2x 12
6       6
1
y x 2
3        Infinitely Many Solutions
Solve the system graphically, by substitution, and
by elimination.       y
 x  2y  3
                                    4y  2x  8
                                      4      4
4y  2x  8                          2    8
y x 
x  2y  3                         x   4    4
x      x                              1
2y  x  3                           y x 2
2     2                               2
1    3
y x 
2    2

No Solution
Substitution Method
Solve.
  2x
y                  y  2x        x y 6

  y  6
x              y  2x            x  2x  6
y  2 2              3x  6
y 4                   3 3
x 2

Solution is (2, 4)
Substitution
x  3y 12                 y  x 1
x  3x 1 12                     y  x 1
x  3x  3  12                    y  3 3 1
3  3
4
y23
4x  15                           4

x 33  4

Solution is (3 3 , 2 3 )
4     4
Solve using substitution.
x  3y

  4x  5
2y
x  3y                 2y  4x  5
2y  43y  5
x  3y                     2y 12y  5
1                   10y  5
x 3
 2                    10  10
1
x  1 1                          y
2
2

Solution is  1 1 ,  1 
2     2
Solve using substitution.
  x  5
2y
                 Solve the 1st equation for x.
  3y  7
2x

2y  x  5                  2x  3y  7
2y      2y              22y  5  3y  7
x  2y  5           4y 10  3y  7
y 10  7
x  2y  5                         10  10
x  23  5                         y  3
x  6  5                       1 y  31
x  1                                 y 3
(1,3)
Solve using substitution.
3x  4y  5       Solve the 2nd equation for y.

  y  6
2x
2x  y  6
2x       2x
3x  4y  5
1(y  2x  6 )
3x  4 2x  6  5
y  2x  6
3x  8x  24  5
24  24                y  2x  6
11x  29                    29
y 2        6
11     11                   11
29                   58 66 8
x                   y          
11     29 8     11 11 11
,
11 11
Solve using substitution.

x  6 y  1
Solve 1st equation for x.
                              x  6y 1
3x  10 y  31
6 y  6 y
x  1 6 y
3x 10 y  31
x  1  6(1)
3(1  6 y)  10 y  31
3  18 y  10 y  31          x  1 6
3  28 y  31            x7
3          3
28 y  28           (7, 1)
28  28
y  1

```
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