Page 779 - Algebra 1

Example
3
Solving Using Substitution
Solve each system of equations by substitution. a. y=x2+5x-1
y = 5x + 3 SOLUTION
x2 +5x-1=5x+3 -_ 5 _ x _ - _ 3 -_ 5 _ x _ - _ 3
x2 - 4 = 0
(x + 2)(x - 2) = 0
x + 2 = 0 and x - 2 = 0 x = -2 x = 2
Substitute the quadratic equation into the linear equation.
Add the expression -5x - 3 to both sides. Recognize the left side of the equation as a
difference of squares. Factor.
Solve both equations.
Math Reasoning
Verify Show that the ordered pairs (-2, -7) and (2, 13) are solutions to Example 3a and that the ordered pairs
(-2, -7) and
(-3, -9) are the solutions to Example 3b.
Determine the corresponding values of y by substituting the values of x into either equation.
1
6
y = 5x + 3
y = 5(-2) + 3 y = -10 + 3
y = -7
y = 5x + 3 y = 5(2) + 3 y = 10 + 3
y = 13
= + (-2, -7)
-16 y = 5x + 3
(2, 13)
8
y x2 5x 1
x
-2
2
4
-8
The solutions are the ordered pairs (-2, -7) and (2, 13). The solutions appear at the intersections of the two graphs.
b. y=x2+7x+3 y = 2x - 3
SOLUTION
x2 +7x+3=2x-3 -_ 2 _ x _ + _ 3 -_ 2 _ x _ + _ 3
x2 +5x+6=0
(x + 3)(x + 2) = 0
x + 3 = 0 and x + 2 = 0
x = -3 and x = -2 Determine the values of y.
Substitute the quadratic equation into the linear equation.
Add the expression -2x + 3 to both sides. Recognize the left side of the equation as a
trinomial that can be factored. Factor.
Solve both equations.
y
x
-4
-2
2
4
y=x2 +7x+3
y = 2x - 3
y = 2(-3) - 3 y = -6 - 3
y = -9
y = 2x - 3
y = 2(-2) - 3 y = -4 - 3
y = -7
y=
2x -
3
(-2, -7) -6
-8
764 Saxon Algebra 1
(-3,
The solutions are the ordered pairs (-3, -9) and (-2, -7).
-9)
y