Unit 1 Day 7: Solving Inequalities with Variables on Both Sides Essential Questions: How do we solve inequalities with variables on both sides? When does an inequality have no solution or a solution of all real numbers?

you must do to the other side Solving Equation Rule Any change applied to one side of an equation must be applied to the other side in order to keep the balance. What you do to one side you must do to the other side

Solving Linear Inequalities When solving linear inequalities, treat each problem the same as when you solve a regular equation. THE ONLY DIFFERENCE: when you multiply or divide by a negative number, you must flip the inequality symbol! < changes to > > changes to <

Vocabulary No solution: when the variable in an equation or inequality is eliminated and you are left with a false statement. All real numbers: when the variable in an equation or inequality is eliminated and you are left with a true statement

Example 1: Solve the inequalities. 7x + 19 > -2x + 55 6x + 22 < -3x + 31 + 2x +2x + 3x + 3x 9x + 22 < 31 9x + 19 > 55 - 19 -19 - 22 -22 9x > 36 9x < 9 9 9 9 9 x > 4 x < 1

Example 2: Solve the inequalities. x + 2 > 3x + 1 -8x + 7 < 4x – 5 - 3x - 3x - 4x - 4x -2x + 2 > 1 -12x + 7 < - 5 - 2 - 2 - 7 - 7 -2x > -1 -12x < -12 -2 -2 -12 -12 x < 2 1 x > 1

Example 3: Solve the inequality. (-12x + 16) < 10 – 3(-x – 2) 4 1 -3x + 4 < 10 + 3x + 6 -3x + 4 < 16 + 3x - 3x - 3x -6x + 4 < 16 - 4 - 4 -6x < 12 -6 -6 x > -2

Example 4: Solve the inequality. (12x – 4) < 2(7 – 5x) 2 1 6x - 2 < 14 - 10x + 10x + 10x 16x - 2 < 14 + 2 + 2 16x < 16 16 16 x < 1

Example 5: Solve the inequalities. 12 – 2a < - 5a – 9 x – 2x + 3 > 3 – x + 5a + 5a - x + 3 > 3 - x 12 + 3a < - 9 + x + x - 12 - 12 3 > 3 3a < - 21 true statement infinite solutions 3 3 a < -7

Example 6: Solve the inequalities. 5x + 24 < 5(x - 5) 6y - (3y - 6) > 5y - 4 + 24 5x < 5x - 25 6y - 3y + 6 > 5y - 4 - 5x - 5x 3y > 5y + 6 - 4 24 < -25 - 5y - 5y - 4 -2y + 6 > false statement no solutions - 6 - 6 > -10 -2y -2 -2 y < 5

Example 7: Phone Company A charges an activation fee of 36 cents and then 3 cents per minute. Phone Company B charges 6 cents per minute with no activation fee. For what value of x is Phone Company A more expensive than Phone Company B? Phone Company A is more expensive when the number of minutes is less than 12. If you talk for more than 12 minutes, Phone Company A is a good choice. .36 + .03x > .06x - .03x - .03x .36 > .03x .03 .03 12 > x x < 12

Example 8: Justin and Tyson are beginning an exercise program to train for football season. Justin weighs 150 pounds and hopes to gain 2 pounds per week. Tyson weighs 195 pounds and hopes to lose 1 pound per week. If the plan works, for how many weeks will Justin weigh less than Tyson? Justin Tyson Justin will weigh less than Tyson up until the 15 week mark. 150 + 2x < 195 - 1x + 1x + 1x 150 + 3x < 195 - 150 - 150 3x < 45 3 3 x < 15

Summary Essential Questions: How do we solve inequalities with variables on both sides? When does an inequality have no solution or a solution of all real numbers? Take 1 minute to write 2 sentences answering the essential questions.