Algebra C A Extra Examples Lesson 13-3 by tsu23398

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```									Algebra Concepts and Applications                                                                 Chapter 13

Lesson 13-3

Examples                 Use substitution to solve each system of equations.
1   y = -3x
x + 2y = 10
The first equation tells you that y is equal to -3x. So, substitute -3x for y in the
second equation. Then solve for x.
x + 2y = 10
x + 2(-3x) = 10      Replace y with -3x.
x - 6x = 10
-5x = 10
-5x 10
-5
= -5       Divide each side by -5.
x = -2

Now substitute -2 for x in either equation and solve for y.
Choose the equation that is easier for you to solve.
y = -3x
y = -3(-2)   Replace x with -2.
y=6

The solution of this system of
equations is (-2, 6). You can see
from the graph that the solution
is correct. You can also check by
substituting (-2, 6) into each of
the original equations.

2   x - y = -4
x = 3y + 5
Substitute 3y + 5 for x in the first equation. Then solve for y.
x - y = -4
3y + 5 - y = -4         Replace x with 3y + 5.
2y + 5 = -4
2y + 5 - 5 = -4 - 5      Subtract 5 from each side.
2y = -9
2y -9
2
=2            Divide each side by 2.
9
y = -2
Algebra Concepts and Applications                                                              Chapter 13

9
Now substitute -2 for y into either equation and solve for x.
x = 3y + 5
9                              9
x = 3-2 + 5           Replace y with -2.
27
x=-2 +5
17
x=-2
17 9
The solution of this system of equations is - 2 , -2. Check by replacing (x, y)
17 9
with - 2 , -2 in each equation.

Example              3   -x + y = 7
3x - y = -5
Solve the first equation for y since the coefficient of y is 1.
-x + y = 7  y = x + 7

Next, find the value of x by                   Now substitute 1 for x in either
substituting x + 7 for y in the second         equation and solve for y.
equation.                                          -x + y = 7
3x - y = -5                                  -1 + y = 7       Replace x with 1.
3x - (x + 7) = -5                              -1 + y + 1 = 7 + 1
3x - x - 7 = -5                                       y =8
2x - 7 = -5
2x - 7 + 7 = -5 + 7
2x = 2
2x 2
2
=2
x =1
The solution is (1, 8).

Examples                 Use substitution to solve each system of equations.
4   y = -5x + 1
5x + y = -2
Find the value of x by substituting -5x + 1 for y in the second equation.
5x + y = -2
5x + (-5x + 1) = -2    Replace y with -5x + 1.
1 = -2

The statement 1 = -2 is false. This means
that there are no ordered pairs that are
solutions to both equations. Compare the
slope-intercept forms of the equations,
y = -5x + 1 and y = -5x - 2. Notice that the
graphs of these equations have the same
slope but different y-intercepts. Thus, the
lines are parallel, and the system has no solution.
Algebra Concepts and Applications                                                             Chapter 13

5   y = -4x + 1
8x + 2y = 2
8x + 2y    =2
8x + 2(-4x + 1)    =2    Replace y with -4x + 1.
8x - 8x + 2   =2    Distributive Property
2   =2

The statement 2 = 2 is true. This means that an ordered pair for a point on either
line is a solution to both equations. The system has infinitely many solutions.

Example              6   The Garden Center sells bags of wild flower seed that contain part seed and
Gardening Link           part sand to aid in planting. One seed mix contains 35% seed and a second
type contains 55% seed. How much of each type of mix should be added to
make 500 pounds of a 45% seed mix?
Explore Let a = the number of pounds of the 35% seed.
Let b = the number of pounds of the 55% seed.

35% Seed        55% Seed        45% Seed
Total Pounds                   a               b              500
Pounds of Seed               0.35a           0.55b         0.45(500)

Plan       Write two equations to represent the information.
a + b = 500                     total pounds
0.35a + 0.55b = 0.45(500)       total pounds of seed

Solve      Use substitution to solve this system. Since a + b = 500, a = 500 - b.
0.35a + 0.55b = 0.45(500)
0.35(500 - b) + 0.55b = 0.45(500)       Replace a with 500 - b.
175 - 0.35b + 0.55b = 225             Distributive Property
175 + 0.2b - 175 = 225 - 175        Subtract 175 from each side.
0.2b = 50
0.2b 50
0.2
= 0.2            Divide each side by 0.2.
b = 250

Now substitute 250 for b in either equation and solve for a.
a + b = 500
a + 250 = 500            Replace b with 250.
a + 250 - 250 = 500 - 250      Subtract 250 from each side.
a = 250
So, 250 pounds of 35% seed mix and 250 pounds of 55% seed mix
should be used.

Examine Check by substituting (250, 250) into the original equations. The
solution is correct.

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