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Solving Compound Inequalities. 4. 0. 1. 2. 3. 5. –1. Previously you studied four types of simple inequalities. In this lesson you will study compound inequalities . A compound inequality consists of two inequalities connected by and or or.
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Solving Compound Inequalities 4 0 1 2 3 5 –1 Previously you studied four types of simple inequalities. In this lesson youwill study compound inequalities. A compound inequality consists of twoinequalities connected by and or or. Write an inequality that represents the set of numbers and graph the inequality. All real numbers that are greater than zero and less than or equal to 4. 0 < x£ 4 SOLUTION This inequality is also written as 0 < xandx£ 4.
Solving Compound Inequalities 4 2 1 0 –1 3 –2 –1 0 1 2 3 4 5 Previously you studied four types of simple inequalities. In this lesson youwill study compound inequalities. A compound inequality consists of twoinequalities connected by and or or. Write an inequality that represents the set of numbers and graph the inequality. All real numbers that are greater than zero and less than or equal to 4. 0 < x£ 4 SOLUTION This inequality is also written as 0 < xandx£ 4. All real numbers that are less than –1or greater than 2. SOLUTION x < –1 orx > 2
Compound Inequalities in Real Life 14,140 ft Glacier and permanent snow field region 7500 ft Alpine meadow region 6000 ft Timber region Timber region: 2000 ft 2000 £y < 6000 0 ft Write an inequality that describes the elevations of the regions of Mount Rainier. Timber region below 6000 ft SOLUTION Let y represent the approximate elevation (in feet).
Compound Inequalities in Real Life 14,140 ft Glacier and permanent snow field region 7500 ft Alpine meadow region 6000 ft Timber region Timber region: 2000 ft 2000 £y < 6000 0 ft Alpine meadow region: 6000 £y < 7500 Write an inequality that describes the elevations of the regions of Mount Rainier. Timber region below 6000 ft Alpine meadow region below 7500 ft SOLUTION Let y represent the approximate elevation (in feet).
Compound Inequalities in Real Life 14,140 ft Glacier and permanent snow field region 7500 ft Alpine meadow region 6000 ft Timber region Timber region: 2000 ft 2000 £y < 6000 0 ft Alpine meadow region: 6000 £y < 7500 Glacier and permanent snow field region: 7500 £y£ 14,410 Write an inequality that describes the elevations of the regions of Mount Rainier. Timber region below 6000 ft Alpine meadow region below 7500 ft Glacier and permanent snow field region SOLUTION Let y represent the approximate elevation (in feet).
Solving a Compound Inequality with And 7 6 3 2 1 0 4 5 The solution is all real numbers that are greater than or equal to 2 and less than or equal to 6. Solve –2 £ 3x – 8 £ 10. Graph the solution. SOLUTION Isolate the variable x between the two inequality symbols. –2 £ 3x – 8 £ 10 Write original inequality. 6 £ 3x£ 18 Add 8 to each expression. 2 £x£ 6 Divide each expression by 3.
Solving a Compound Inequality with Or 3x + 1 < 4 or 2x – 5 > 7 0 1 2 –1 4 6 5 7 3 3x < 3 or 2x > 12 x < 1 or x > 6 The solution is all real numbers that are less than 1or greater than 6. Solve 3x + 1 < 4 or 2x – 5 > 7. Graph the solution. SOLUTION A solution of this inequality is a solution of either of its simple parts. You can solve each part separately.
Reversing Both Inequality Symbols –5 –4 –3 –2 0 2 1 3 –1 To match the order of numbers on a number line, this compound is usually written as –3 < x < 0. The solution is all real numbers that are greater than –3and less than 0. Solve –2 < –2 – x < 1. Graph the solution. SOLUTION Isolate the variable x between the two inequality signs. –2 < –2 – x < 1 Write original inequality. 0 < –x < 3 Add 2 to each expression. 0 > x > –3 Multiply each expression by –1 and reverse both inequality symbols.
Modeling Real-Life Problems You have a friend Bill who lives three miles from school and another friend Mary who lives two miles from the same school. You wish to estimate the distance d that separates their homes. What is the smallest value d might have? SOLUTION DRAW A DIAGRAM A good way to begin this problem is to draw a diagram with the school at the center of a circle. Problem Solving Strategy Bill’s home is somewhere on the circle with radius 3 miles and center at the school. Mary’s home is somewhere on the circle with radius 2 miles and center at the school.
Modeling Real-Life Problems You have a friend Bill who lives three miles from school and another friend Mary who lives two miles from the same school. You wish to estimate the distance d that separates their homes. What is the smallest value d might have? SOLUTION If both homes are on the same line going toward school, the distance is 1 mile.
Modeling Real-Life Problems You have a friend Bill who lives three miles from school and another friend Mary who lives two miles from the same school. You wish to estimate the distance d that separates their homes. What is the largest value d might have? SOLUTION If both homes are on the same line but in opposite directions from school, the distance is 5 miles.
Modeling Real-Life Problems You have a friend Bill who lives three miles from school and another friend Mary who lives two miles from the same school. You wish to estimate the distance d that separates their homes. What is the largest value d might have? SOLUTION If both homes are on the same line but in opposite directions from school, the distance is 5 miles. Write an inequality that describes all the possible values that d might have. SOLUTION The values of d can be described by the inequality 1 £d£ 5.
