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

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 = − 4 + b Simplify + 4 + 4 Add 3 = b y-intercept: (0, 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

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

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 + 21 Add + 2x + 2x 2x + 7y = 21

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test 2. Given f (x) = x2 + 8x – 1, find f (− 7). f (− 7) = (− 7)2 + 8(− 7) – 1 Substitute f (− 7) = 49 – 56 – 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.

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

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test 3 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

Algebra I Concept Test # 9 – Two Variable Inequalities Practice Test 3 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.

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 0 + 4y = − 16 − 8x = − 16 4y = − 16 x = 2 y = − 4 x – intercept = (2, 0) y – intercept = (0, − 4)

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. 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 0 + 0 < − 16 0 < − 16 Note: Since the resulting inequality is false, shade the opposite side that contains the test point.

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.

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.