Algebra 1 Focus 3.2
Strategic Standard Build
1) What are the solutions for the quadratic
equation x2 + 11x = 42?
3) What is the range of the following
relation? (-2,4), (1,-2), (3,-5), (4,1)
CA 14.0*
CA 17.0
2) The lengths of the sides of a triangle
are b, b + 3, and 2b – 1. If the
perimeter of the triangle is 50 inches,
what is the value of b?
y
4) Write an equation in
slope-intercept form
that represents the
following graph.
xx
See Page 2
under Student Work
CA 5.0*
CA 6.0*
Lesson Objective: You will be able to solve systems of equations.
Systems of Equations
CA 9.0*
CST TEST PREP
We Do: Solve each system of equations.
I)
⎧2x − 6y = 28
⎨
-1(⎩2x + 3y) = (−8)●-1
2x – 6y = 28
-2x – 3y = 8
0 – 9y = 36
-9y = 36
-9
-9
y = -4
-3(⎧x − 2y) = (6)●-3
II)
2x – 6(-4) = 28
2x + 24 = 28
-24 -18
2x = 4
2
2
x=2
⎨
⎩3x − 6y = 6
-3x + 6y = -18
3x – 6y = 6
0 = -12
False
No Solution
(2,-4)
You Do: Solve each system of equations.
III)
⎧−6x − 4y = 8
⎨
2(⎩3x + 2y) = (−4)●2
-6x – 4y = 8
6x + 4y = -8
0=0
True
Infinite Solutions
IV)
4(⎧4x − y)= (−13)●4
V)
⎨
⎩2x + 4y = −2
16x – 4y = -52
2x + 4y = -2
18x + 0 = -54
18x = -54
18
18
x = -3
2(-3) + 4y
-6 + 4y
+6
4y
4
x
= -2
= -2
+6
=4
4
=1
⎧8x − 6y = 14
⎨
4(⎩−2x + 5y) =(7 )●4
8x – 6y = 14
-8x + 20y = 28
0 + 14y = 42
14y = 42
14
14
y=3
(-3,1)
You Do: Solve each system of equations.
-3(⎧3x − 2y) = (4 )●-3
VI)
VII)
⎨
⎩4x − 6y = 22
-9x + 6y = -12
4x – 6y = 22
-5x + 0 = 10
-5x = 10
-5
-5
x = -2
3(-2) – 2y = 4
-6 – 2y = 4
+6
+6
-2y = 10
-2
-2
y = -5
(-2,-5)
⎧6x − 9y = 12
⎨
-3(⎩2x − 3y) = (4 )●-3
6x – 9y = 12
-6x + 9y = -12
0=0
True
Infinite Solutions
-2x + 5(3) = 7
-2x + 15 = 7
-15 -15
-2x = -8
-2
-2
x=4
(4,3)
VIII)
⎧8x + 5y = 2
⎨
-2(⎩4x − 3y) =(34 )●-2
8x + 5y =
-8x + 6y =
0 + 11y =
11y =
11
y=
2
-68
-66
-66
11
-6
4x – 3(-6) = 34
4x + 18 = 34
-18 -18
4x = 16
4
4
x=4
(4,-6)
Student Work
Step for finding solutions to
Quadratic Equations
1)
x2 + 11x = 42
-42 -42
x2 + 11x – 42 = 0
1) Place in Standard Form
ax2 + bx + c = 0
-42
+14
3)
(-2,4), (1,-2), (3,-5), (4,1)
Range
{-5, -2, 1, 4}
2) Completely factor quadratic
-3
3) Use Principal of Zero Products
to create separate equations
11
4) Solve each equation
(x + 14)(x – 3) = 0
x + 14 = 0
or
-14 -14
x = -14 or
The domain of a relation or function
is the set of x-values (inputs)
The range of a relation or function is
the set of y-values (outputs)
x–3=0
+3 +3
x=3
x = -14,3
2) Perimeter = S1 + S2 + S3
50 = (b) + (b + 3) + (2b – 1)
50 = b + b + 3 + 2b – 1
50 = 4b + 2
-2
-2
48 = 4b
The perimeter of a triangle is
4
4
equal to the sum of the sides.
12 = b
y
4)
Slope-Intercept Form of
Linear Equations
y = mx + b
run = 2
Need to find value of the
slope (m) and y-intercept (b)
rise = 3
x
y–intercept = b = 2
Perimeter = S1 + S2 + S3
Substitute values or
expressions into equation for
the Perimeter and 3 sides,
then solve for x.
Lesson Notes:
rise
3
Slope = m =
=
run
2
y = mx + b
y=
3
x+2
2
Focus 3.2 Response Form
Strategic Standards Build: Multiple Choices
1)
2)
A)
x = 6,-7
A)
{-2, 1, 3, 4}
B)
x = -14,3
B)
{-5, -2, 1, 3, 4}
C)
x = 14,-3
C)
{-5, -2, 1, 4}
D)
x = -6,7
D)
{1, 2, 3, 4, 5}
A)
9
B)
12
C)
17
D)
23
3)
4)
A)
B)
C)
D)
3
x+2
2
3
y = − x+2
2
2
y = x+2
3
2
y = − x+2
3
y=
Additional Practice
5) What are the solutions for the quadratic
equation x2 – 6x = 40? 2
A)
x = 4,-10
B)
x = -4,-10
x – 6x = 40
-40 -40
x2 – 6x – 40 = 0
-40
4
-10
C)
x = 4,10
D)
x = -4,10 (x + 4)(x – 10) = 0
-6
x+4=0
-4 -4
x = -4
7) What is the domain of the following relation?
(1,-5), (2,-3), (3,-1), (4,1)
A)
{1, 2, 3, 5}
B)
{-1, 1, 3, 5}
C)
{1, 2, 3, 4}
D)
{1, -1, -3, -5}
(1,-5), (2,-3), (3,-1), (4,1)
or
x – 10 = 0
+10 +10
or
x = 10
Domain {1, 2, 3, 4}
x = -4,10
CA 14.0*
6) The lengths of the sides of a triangle are x,
x + 4, and 2x – 5 meters. If the perimeter
of the triangle is 91 meters, what is the
value of x?
Perimeter = S1 + S2 + S3
A)
20.5
B)
23
C)
21
D)
25
91
91
91
+1
92
4
= (x) + (x + 4) + (2x – 5)
= x + x + 4 + 2x – 5
= 4x – 1
+1
= 4x
4
23 = x
CA 17.0
8) Which equation represents the line shown by
the graph below?
A)
y=−
B)
y=
C)
y=−
D)
y=
3
x +1
2
2
x +1
3
CA 5.0*
y-intercept
b=1
rise = -2
run = 3
x
2
x +1
3
y = mx + b
y =−
y
3
x +1
2
Slope = m =
−2
rise
=
run
3
2
x+1
3
CA 6.0*