Algebra 1 Focus 3.1
Strategic Standard Build
1) What are the solutions for the quadratic
equation x2 – 5x = 84?
3) What is the domain of the following
relation? (4,2), (-3,7), (1,7), (-2,-3)
CA 14.0*
CA 17.0
2) The lengths of the sides of a triangle
are x, x + 3, and 12 millimeters. If the
perimeter of the triangle is 55
millimeters, what is the value of x?
y
4) Write an equation in
slope-intercept form
that represents the
following graph.
x
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)
y  2x  4

3x  4y  1
3x – 4(2x – 4) = 1
3x – 8x + 16 = 1
-5x + 16 = 1
-16 -16
-5x = -15
-5
-5
x=3
6x  2y  12

y  3x  6
II)
y = 2(3) – 4
y=6–4
y=2
6x – 2(3x – 6) = 12
6x – 6x + 12 = 12
12 = 12
True
Infinite solutions
(3,2)
You Do: Solve each system of equations.
III)
y  2x  5

2x  2y  4
2x + 2(-2x – 5) = -4
2x – 4x – 10 = -4
-2x – 10 = -4
+10 +10
-2x = 6
-2
-2
x = -3
IV)
y = -2(-3) – 5
y=6–5
y=1
(-3,1)
4x – 2(x – 2) = 14
4x – 2x + 4 = 14
2x + 4 = 14
-4
-4
2x = 10
2
2
x=5
You Do: Solve each system of equations.
x  2y  12
VI)
VII)

y  2x  3
x – 2(2x + 3) = -12
x – 4x – 6 = -12
-3x – 6 = -12
+6
+6
-3x = -6
-3
-3
x=2
y = 2(2) + 3
y=4+3
y=7
(2,7)
4x  2y  14

y  x  2
y = (5) – 2
y=5–2
y=3
V)
12x + 4(-3x – 2) = -4
12x – 12x – 8 = -4
-8 = -4
False
No Solution
(5,3)
y  2x  5

4x  2y  10
4x – 2(2x – 5) = 10
4x – 4x + 10 = 10
10 = 10
True
Infinite Solutions
y  3x  2

12x  4y  4
VIII)
2x  y  3

y  5x  11
2x – (-5x – 11) = -3
2x + 5x + 11 = -3
7x + 11 = -3
-11 -11
7x = -14
7
7
x = -2
y = -5(-2) – 11
y = 10 – 11
y = -1
(-2,-1)
Student Work
1)
x2 – 5x = 84
-84 -84
x2 – 5x – 84 = 0
-84
-12
Step for finding solutions to
Quadratic Equations
3)
1) Place in Standard Form
ax2 + bx + c = 0
(4,2), (-3,7), (1,7), (-2,-3)
Domain
{-3, -2, 1, 4}
2) Completely factor quadratic
+7
3) Use Principal of Zero Products
to create separate equations
-5
4) Solve each equation
(x – 12)(x + 7) = 0
x – 12 = 0 or
+12 +12
x = 12 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+7=0
-7 -7
x = -7
x = -7,12
2) Perimeter = S1 + S2 + S3
55 = (x) + (x + 3) + (12)
55 = x + x + 3 + 12
55 = 2x + 15
-15
-15
40 = 2x
The perimeter of a triangle is
2
2
equal to the sum of the sides.
20 = x
Perimeter = S1 + S2 + S3
Substitute values or
expressions into equation for
the Perimeter and 3 sides,
then solve for x.
Lesson Notes:
Slope-Intercept Form of
Linear Equations
y = mx + b
y
4)
Need to find value of the
slope (m) and y-intercept (b)
x
rise = -3
y–intercept = b = -1
run = 1
Slope = m =
rise
3
=
= -3
1
run
y = mx + b
y = -3x – 1
Focus 3.1 Response Form
Strategic Standards Build: Multiple Choices
1)
2)
A)
x = 12,7
B)
3)
A)
{-3, 2, 7}
x = -12,7
B)
{-3, -2, 1, 4}
C)
x = 12,-7
C)
{-3, 7}
D)
x = -12,-7
D)
{-3, -2, 1, 2, 4, 7}
A)
15
A)
y
B)
20
C)
35
D)
25
4)
B)
C)
D)
1
x 1
3
y  3x  1
1
x 1
3
y  3x  1
y
Additional Practice
5) What are the solutions for the quadratic
equation x2 – 3x = 4?
A)
x = 2,-2
B)
x = -4,1
C)
x=2
D)
x2 – 3x = 4
-4 -4
x2 – 3x – 4 = 0
-4
x = -1,4
1
-3
-4
7) What is the range of the following relation?
(-3,6), (-2,1), (0,4), (2,1)
A)
{-3, -2, 0, 1, 2, 4, 6}
B)
{-3, -2, 0, 2}
C)
{1, 4, 6}
D)
{-3, -2, 0, 1, 1, 2, 4, 6}
(x – 4)(x + 1) = 0
x–4=0
+4 +4
x=4
or
or
(-3,6), (-2,1), (0,4), (2,1)
x+1=0
-1 -1
x = -1
Range {1, 4, 6}
x = -1,4
CA 14.0*
6) The lengths of the sides of a triangle are
x, 2x – 1, and 20 yards. If the perimeter
of the triangle is 43 yards, what is the
value of x?
A)
15
B)
12
C)
31
D)
8
Perimeter = S1 + S2 + S3
43 = (x) + (2x – 1) + (20)
43 = x + 2x – 1 + 20
43 = 3x + 19
-19
-19
24 = 3x
3
3
8=x
CA 17.0
8) Which equation represents the line shown by
the graph below?
y
A)
1
y=- x–2
2
B)
y = -2x – 2
C)
y=
D)
y = 2x – 2
1
x–2
2
Slope = m =
run = 2
rise = 1
rise
1
=
run
2
x
y-intercept
b = -2
y = mx + b
y=
CA 5.0*
1
x–2
2
CA 6.0*