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Lesson 3 Solving Multi-Step Inequalities. Bell Ringer Solve the following equations. 1. z + 5 < -5 2. -8 + g > -4 3. -5x < 25 4. ¾s < -12 Did you get:

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Presentation on theme: "Lesson 3 Solving Multi-Step Inequalities. Bell Ringer Solve the following equations. 1. z + 5 < -5 2. -8 + g > -4 3. -5x < 25 4. ¾s < -12 Did you get:"— Presentation transcript:

1 Lesson 3 Solving Multi-Step Inequalities

2 Bell Ringer Solve the following equations. 1. z + 5 < -5 2. -8 + g > -4 3. -5x < 25 4. ¾s < -12 Did you get: Did you get: 1. z < -10 2. g > 4 3. x > -5 4. s < -16 If so……GREAT JOB! If you missed any, check the previous lessons.

3 Solving Multi-Step Inequalities In order to solve multi-step inequalities, you will use more than one operation. It can be solved by undoing the operation in the same way you would solve an equation with more than one operation.

4 Here’s a Real-World example. Ex. 1: Find the temperature in degrees Celsius for which chlorine is a gas. Ex. 1: Find the temperature in degrees Celsius for which chlorine is a gas. 9/5 C + 32 > -31 9/5 C + 32 > -31 First…subtract 32 from both sides. Last…multiply by the multiplicative inverse of 9/5. 9/5 C + 32 > -31 -32 -32 -32 -32 (5/9)9/5 C > -63 (5/9) (5/9)9/5 C > -63 (5/9) C > -35 C > -35 Therefore: chlorine is a gas when it’s greater than -35°C Written: {C, C > -35}

5 Definition If solving an inequality results in a statement that is always true, the solution is all real numbers. If solving an inequality results in a statement that is always true, the solution is all real numbers. Empty Set, {Ø}, { }: If solving an inequality results in a statement that is never true, the solution is the empty set. Empty Set, {Ø}, { }: If solving an inequality results in a statement that is never true, the solution is the empty set. The empty set has no solutions. The empty set has no solutions.

6 Example of an Empty Set. Solve 8(t + 2) – 3(t - 4) < 5(t – 7) + 8 Solve 8(t + 2) – 3(t - 4) < 5(t – 7) + 8 First, use the distributive property. 8t + 16 – 3t + 12 < 5t – 35 + 8 Now, combine like terms. 5t + 28 < 5t – 27 5t + 28 < 5t – 27 -5t -5t (subtract 5t from both sides) -5t -5t (subtract 5t from both sides) 28 < -27 28 < -27 Since this is a false statement, the solution set is the empty set Ø.

7 Example 3 Solve 5x + 28 > 2(2x +7) + 10 + x Solve 5x + 28 > 2(2x +7) + 10 + x First, use the distributive property and combine like terms. 5x + 28 > 4x + 14 + 10 + x 5x + 28 > 5x + 24 5x + 28 > 5x + 24 -5x -5x subtract 5x from both sides 28 > 24 28 > 24 Since this is a true statement, the solution is all real numbers.


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