1. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
Use the given points to find the following:
(− 7, 5) and (14, − 1)
a) Find the slope of the line that goes through the two points.
y2 – y1
m =
Formula
x2 – x1 5 –
(− 1)
m =
Substitute
− 7 – 14 5 + 1
m =
Simplify −7 – 14 6 2
m =
m = −
Reduce
− 21 7
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2. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
Use the given points to find the following:
(− 7, 5) and (14, − 1)
b) Find the y-intercept of the line that goes through the two points.
Slope intercept form: y = mx + b
Note: Substitute one of the given points and the calculated slope from part a into the slope intercept form of a linear equation to find b (the y-intercept).
(− 1) = − (14) + b 2
Substitute 7
− 1 = − b
Simplify
+ 4
+ 4
Add
3 = b
y-intercept: (0, 3)

3. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
Use the given points to find the following:
(− 7, 5) and (14, − 1)
c) Write the equation of the line in slope-intercept form. 2
m = −
Slope 7
y-intercept: (0, 3)
y-intercept 2
y = − x + 3
Equation of line 7
d) Find the slope of a line that is parallel to the line found in part c. 2
Note: Parallel lines have
the same slope. − 7

4. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
Use the given points to find the following:
(− 7, 5) and (14, − 1)
e) Find the slope of the line perpendicular to the line in part c. 2
y = − x + 3 7
Note: Perpendicular lines have slopes that are
opposite reciprocals. 7 2

5. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
Use the given points to find the following:
(− 7, 5) and (14, − 1)
f) Write the equation of the line from part c in standard form.
Standard Form: Ax + By = C
Where A, B, and C are integers (
− 2 7 (
y = x + 3 ( ( 7
Multiply each term by the LCD 7 7
7y = − 2x
Add
+ 2x
+ 2x
2x + 7y = 21

6. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
2. Given f (x) = x x – 1, find f (− 7).
f (− 7) = (− 7) (− 7) – 1
Substitute
f (− 7) = – – 1
Simplify
f (− 7) = − 7 – 1
f (− 7) = − 8
3. Determine if (2, 5) is a solution of y – 2x < − 9
(5) – 2(2) < − 9
Substitute
(5) – 4 < − 9
Simplify
1 < − 9
Since 1 is not less than − 9, the ordered pair is not a solution.

7. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
4. Write the inequality that represents the graph below.
− 4
Slope
m = 3 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2
y-intercept
y-intercept: (0, 4)
Note: Write an equation and evaluate a test point to identify the inequality. Use (0, 0).
− 4 3
Equation
y = x + 4
− 4 3
Substitute
(0) = (0) + 4
Simplify
0 = 4
Note: Since (0, 0) is not in the shaded region, we need the inequality to be false. Also, the line is dashed.
− 4 3
y > x + 4

8. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test3
5. Use the inequality, for the following:
y ≥ x – 3 5
a) When graphing this inequality, is the line solid
or dashed?
Solid
b) Use the y-intercept and slope to graph the line. 3
m = 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2 5
y-intercept = (0, − 3) 5 3

9. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test3
5. Use the inequality, for the following:
y ≥ x – 3 5
b cont.) Using a test point, shade the appropriate area.
Note: We will use (0, 0) as our test point 3
y ≥ x – 3 5 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2 3
Substitute
0 ≥ (0) – 3 5
Simplify
0 ≥ 0 – 3
0 ≥ − 3
Note: Since the resulting inequality is true, shade the side that contains the test point.

10. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
6. Use the inequality, − 8x + 4y < − 16 for the following:
a) When graphing this inequality, is the line solid
or dashed?
Dashed
b) Use the x and y-intercept to graph the line.
x - intercept
y - intercept 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2
− 8x + 4y = − 16
− 8x + 4y = − 16
− 8x + 4(0) = − 16
− 8(0) + 4y = − 16
− 8x + 0 = − 16
y = − 16
− 8x = − 16
4y = − 16
x = 2
y = − 4
x – intercept = (2, 0)
y – intercept = (0, − 4)

11. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
6. Use the inequality, − 8x + 4y < for the following:
b cont.) Using a test point, shade the appropriate area.
Note: We will use (0, 0) as our test point
− 8x + 4y < − 16
Substitute
− 8(0) + 4(0) < − 16 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2
Simplify
< − 16
0 < − 16
Note: Since the resulting inequality is false, shade the opposite side that contains the test point.

12. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
7. Use the inequality, − 2x ≥ 8 for the following:
a) When graphing this inequality, is the line solid
or dashed?
Solid
b) Graph the line.
Divide
− 2x ≥ 8 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2
− 2
− 2
Note: Dividing by a negative switches the direction of the inequality.
x ≤ − 4
Note: This line is in the form of a
vertical line.

13. Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test
7. Use the inequality, − 2x ≥ 8 for the following:
b cont.) Using a test point, shade the appropriate area.
Believe in yourself
Note: We will use (0, 0) as our test point
− 2x ≥ 8
Substitute
− 2(0) ≥ 8 3 4 5 6 -4 -3 -2 -1 7 1 -7 -6 -5 2
Simplify
0 ≥ 8
Note: Since the resulting inequality is false, shade the opposite side that contains the test point.