Inequalities with Variables on Both Sides
• Some inequalities have variable terms on both sides of the inequality symbol. You can solve these inequalities like you solved equations with variables on both sides.
Example 1 Solve the inequality and graph the solutions. y ≤ 4 y + 18 – 4 y Subtract 4 y from both sides. -3 y ≤ 18 ___ -3 -3 y – 6 – 10 – 8 – 6 – 4 – 2 Since y is multiplied by -3, divide both sides by -3 to undo the multiplication. Since you divided by a negative, you must flip the inequality sign. 0 2 4 6 8 10
Example 2 Solve the inequality and graph the solutions. 4 x ≥ 7 x + 6 – 7 x Subtract 7 x from both sides. – 3 x ≥ 6 x ≤ – 2 Since x is multiplied by – 3, divide both sides by – 3 to undo the multiplication. Change ≥ to ≤. – 10 – 8 – 6 – 4 – 2 0 2 4 6 8 10
Example 3 • The Home Cleaning Company charges $312 to power-wash the siding of a house plus $12 for each window. Power Clean charges $36 per window, and the price includes power-washing the siding. How many windows must a house have to make the total cost from The Home Cleaning Company less expensive than Power Clean? • 312 + 12 w < 36 w • Subtract 12 w from both sides • • • 312 < 24 w Divide both sides by 24 312/24 < 24 w/24 13 < w or w > 13 There will have to be 14 windows in order for The Home Cleaning Company to be less expensive than Power Clean.
Example 4 • A-Plus Advertising charges a fee of $24 plus $0. 10 per flyer to print and deliver flyers. Print and More charges $0. 25 per flyer. For how many flyers is the cost at A-Plus Advertising less than the cost of Print and More? • 24 + 0. 10 f < 0. 25 f • Subtract 0. 10 f from both sides. • 24 + 0. 10 f – 0. 10 f < 0. 25 f – 0. 10 f • 24 < 0. 15 f • Divide both sides by 0. 15 • 24/0. 15 < 0. 15 f/0. 15 • 160 < f or f > 160 • 161 flyers or more will make A-Plus Advertising less than Print and More.
Example 5 – 12 – 9 – 6 – 3 0 3 Solve the inequality and graph the solutions. 2(k – 3) > 6 + 3 k – 3 Distribute 2 on the left side of the inequality. 2(k – 3) > 3 + 3 k And, combine the 6 and -3 on 2 k + 2(– 3) > 3 + 3 k the right side. Subtract 3 k from both sides. 2 k – 6 > 3 + 3 k – 3 k -k – 6 > 3 +6 +6 -k > 9 k < -9 Since 6 is subtracted from k, add 6 to both sides to undo the subtraction. Divide both sides by -1 (flip the inequality).
Example 6 Solve the inequality and graph the solutions. 5(2 – r) ≥ 3(r – 2) Distribute 5 on the left side of the inequality and distribute 3 on 5(2 – r) ≥ 3(r – 2) the right side of the inequality. 5(2) – 5(r) ≥ 3(r) + 3(– 2) Subtract 3 r from both sides. 10 – 5 r ≥ 3 r – 6 -3 r 10 − 8 r ≥ -6 -10 Subtract 10 from both sides -8 r ___ ≥ -16 ___ -8 -8 r<2 Divide both sides by -8 (flip the inequality). – 6 – 4 – 2 0 2 4
There are special cases of inequalities called identities and contradictions.
Example 7 Solve the inequality. 2 x – 7 ≤ 5 + 2 x – 7 ≤ 5 Subtract 2 x from both sides. True statement. The inequality 2 x − 7 ≤ 5 + 2 x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.
Example 8 Solve the inequality. 2(3 y – 2) – 4 ≥ 3(2 y + 7) Distribute 2 on the left side and 3 on the right side. 2(3 y) – 2(2) – 4 ≥ 3(2 y) + 3(7) 2(3 y – 2) – 4 ≥ 3(2 y + 7) 6 y – 4 ≥ 6 y + 21 6 y – 8 ≥ 6 y + 21 – 6 y Subtract 6 y from both sides. False statement. – 8 ≥ 21 No values of y make the inequality true. There are no solutions.
Try these…I Lesson Quiz: Part Solve each inequality and graph the solutions. 1. t < 5 t + 24 t > – 6 2. 5 x – 9 ≤ 4. 1 x – 81 x ≤ – 80 3. 4 b + 4(1 – b) > b – 9 4. 2 y – 2 ≥ 2(y + 7) no solution b < 13
Try these… 5. Rick bought a photo printer and supplies for $186. 90, which will allow him to print photos for $0. 29 each. A photo store charges $0. 55 to print each photo. How many photos must Rick print before his total cost is less than getting prints made at the photo store? Rick must print more than 718 photos.
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