1. 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.

2 7

14

(− 1)

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

− 7

5 –

–

14

1−7

5 +–

Simplify

− 216

Reduce

m = y2 – y1

x2 – x1

Formula

m =

m =

m = m = −

Substitute

© 2007-09 by S-Squared, Inc. All Rights Reserved.

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

1. 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) + b27

Substitute

Simplify− 1 = − 4 + b

Add + 4 + 4

3 = b

y-intercept: (0, 3)

2 7

1. Use the given points to find the following:

(− 7, 5) and (14, − 1)

c) Write the equation of the line in slope-intercept form.

Slope

Equation of liney = − x + 327

d) Find the slope of a line that is parallel to the line found in part c.

y-intercept

− Note: Parallel lines have the same slope.

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

2 7

m = −

y-intercept: (0, 3)

1. 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.

y = − x + 327

Note: Perpendicular lines have slopes that are opposite reciprocals.

7 2

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

1. 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.

y = x + 3

+ 2x + 2x

− 27

Standard Form: Ax + By = C Where A, B, and C are integers

Multiply each term by the LCD

7

7

(

(

(

7

(

7y = − 2x + 21

2x + 7y = 21

Add

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

2. Given f (x) = x2 + 8x – 1, find f (− 7).

Substitute

Simplify

f (− 7) = (− 7)2 + 8(− 7) – 1

f (− 7) = 49 – 56 – 1

f (− 7) = − 7 – 1

f (− 7) = − 8

3. Determine if (2, 5) is a solution of y – 2x < − 9

Substitute

Simplify

(5) – 2(2) < − 9

(5) – 4 < − 9

1 < − 9

Since 1 is not less than − 9, the ordered pair is not a solution.

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

4. Write the inequality that represents the graph below.

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

Slope

Equation y = x + 4− 43

y-intercept

− 4 3

m =

y-intercept: (0, 4)

Note: Write an equation and evaluate a test point to identify the

inequality. Use (0, 0).

Substitute (0) = (0) + 4− 43

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.

y > x + 4− 43

3m = 5

5. Use the inequality, for the following:

a) When graphing this inequality, is the line solid or dashed?

y ≥ x – 335

y-intercept = (0, − 3)

Solid

b) Use the y-intercept and slope to graph the line.

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

5

3

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

5. Use the inequality, for the following:

b cont.) Using a test point, shade the appropriate area.

y ≥ x – 335

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

y ≥ x – 335

Substitute

Simplify

0 ≥ (0) – 335

0 ≥ 0 – 3

0 ≥ − 3

Note: Since the resulting inequality is true, shade the side that contains the test point.

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

Note: We will use (0, 0) as our test point

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.

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

x – intercept = (2, 0) y – intercept = (0, − 4)

− 8x + 4y = − 16 − 8x + 4y = − 16

x - intercept y - intercept

− 8x + 4(0) = − 16− 8x + 0 = − 16

x = 2

− 8(0) + 4y = − 160 + 4y = − 16

y = − 4− 8x = − 16 4y = − 16

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

6. Use the inequality, − 8x + 4y < 16 for the following:

b cont.) Using a test point, shade the appropriate area.

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

− 8x + 4y < − 16

Substitute

Simplify

− 8(0) + 4(0) < − 16

0 < − 16

Note: Since the resulting inequality is false, shade the opposite side that contains the test point.

Note: We will use (0, 0) as our test point

0 + 0 < − 16

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.

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

− 2x ≥ 8

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

Divide− 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.

3 4 5 6-4 -3 -2 -1 71-7 -6 -5 2

1234567

-1-2-3-4-5-6-7

7. Use the inequality, − 2x ≥ 8 for the following:

− 2x ≥ 8

Substitute

Simplify

− 2(0) ≥ 8

Note: We will use (0, 0) as our test point

0 ≥ 8

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test

b cont.) Using a test point, shade the appropriate area.

Note: Since the resulting inequality is false, shade the opposite side that contains the test point.

Believe in yourself