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Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Warm UpSolve each equation. 1. 2x = 7x + 15
2.
5. Solve and graph 5(2 – b) > 52.
3. 2(3z + 1) = –2(z + 3)
4. 3(p – 1) = 3p + 2
x = –3
b < –3
–5 –3 –2 –1–4 0–6
3y – 21 = 4 – 2y y = 5
z = –1
no solution
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Solve inequalities that contain variable terms on both sides.
Objective
Holt Algebra 1
3-5 Solving 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.
Use the properties of inequality to “collect” all the variable terms on one side and all the constant terms on the other side.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Directions: Solve the inequality and graph the solutions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 1
y ≤ 4y + 18
y ≤ 4y + 18–y –y
0 ≤ 3y + 18–18 – 18
–18 ≤ 3y
–6 ≤ y (or y –6)
To collect the variable terms on one side, subtract y from both sides.
Since 18 is added to 3y, subtract 18 from both sides to undo the addition.
Since y is multiplied by 3, divide both sides by 3 to undo the multiplication.
–10 –8 –6 –4 –2 0 2 4 6 8 10
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
4m – 3 < 2m + 6To collect the variable terms on one
side, subtract 2m from both sides.–2m – 2m
2m – 3 < + 6
Since 3 is subtracted from 2m, add 3 to both sides to undo the subtraction
+ 3 + 3
2m < 9 Since m is multiplied by 2, divide both sides by 2 to undo the multiplication.
4 5 6
Example 2
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 3
4x ≥ 7x + 6
4x ≥ 7x + 6–7x –7x
–3x ≥ 6
x ≤ –2
To collect the variable terms on one side, subtract 7x from both sides.
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
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 4
5t + 1 < –2t – 6
5t + 1 < –2t – 6+2t +2t
7t + 1 < –6– 1 < –1
7t < –77t < –77 7t < –1
–5 –4 –3 –2 –1 0 1 2 3 4 5
To collect the variable terms on one side, add 2t to both sides.
Since 1 is added to 7t, subtract 1 from both sides to undo the addition.
Since t is multiplied by 7, divide both sides by 7 to undo the multiplication.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 5: Business Application
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?
Let w be the number of windows.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 5 Continued
312 + 12 • w < 36 • w
312 + 12w < 36w– 12w –12w
312 < 24w
13 < w
The Home Cleaning Company is less expensive for houses with more than 13 windows.
To collect the variable terms, subtract 12w from both sides.
Since w is multiplied by 24, divide both sides by 24 to undo the multiplication.
HomeCleaningCompany
siding charge
plus$12 per window
# of windows
is lessthan
PowerClean
cost per window
# ofwindows.times
times
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 6
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?
Let f represent the number of flyers printed.
24 + 0.10 • f < 0.25 • f
plus
$0.10per
flyer
is lessthan
# of flyers.
A-Plus
Advertising
fee of $24
Print and
More’s cost
per flyer
# of flyers
times times
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 6 Continued
24 + 0.10f < 0.25f
–0.10f –0.10f
24 < 0.15f
160 < f
To collect the variable terms, subtract 0.10f from both sides.
Since f is multiplied by 0.15, divide both sides by 0.15 to undo the multiplication.
More than 160 flyers must be delivered to make A-Plus Advertising the lower cost company.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
You may need to simplify one or both sides of an inequality before solving it. Look for like terms to combine and places to use the Distributive Property.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 7
2(k – 3) > 6 + 3k – 3
2(k – 3) > 3 + 3k Distribute 2 on the left side of the inequality.
2k + 2(–3) > 3 + 3k
2k – 6 > 3 + 3k–2k – 2k
–6 > 3 + k
To collect the variable terms, subtract 2k from both sides.
–3 –3
–9 > k
Since 3 is added to k, subtract 3 from both sides to undo the addition.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 7 Continued
–9 > k
–12 –9 –6 –3 0 3
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 8
0.9y ≥ 0.4y – 0.5
0.9y ≥ 0.4y – 0.5–0.4y –0.4y
0.5y ≥ – 0.5
0.5y ≥ –0.5 0.5 0.5
y ≥ –1
To collect the variable terms, subtract 0.4y from both sides.
Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.
–5 –4 –3 –2 –1 0 1 2 3 4 5
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 9
5(2 – r) ≥ 3(r – 2)
5(2 – r) ≥ 3(r – 2)
5(2) – 5(r) ≥ 3(r) + 3(–2)
10 – 5r ≥ 3r – 6+6 +6
16 − 5r ≥ 3r+ 5r +5r
16 ≥ 8r
Distribute 5 on the left side of the inequality and distribute 3 on the right side of the inequality.
Since 6 is subtracted from 3r, add 6 to both sides to undo the subtraction.
Since 5r is subtracted from 16 add 5r to both sides to undo the subtraction.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 9 Continued
–6 –2 0 2–4 4
16 ≥ 8rSince r is multiplied by 8, divide
both sides by 8 to undo the multiplication.
2 ≥ r
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 10
0.5x – 0.3 + 1.9x < 0.3x + 6
2.4x – 0.3 < 0.3x + 6+ 0.3 + 0.3 2.4x < 0.3x + 6.3
–0.3x –0.3x
2.1x < 6.3
Since 0.3 is subtracted from 2.4x, add 0.3 to both sides.
Since 0.3x is added to 6.3, subtract 0.3x from both sides.
x < 3
Since x is multiplied by 2.1, divide both sides by 2.1.
Simplify.2.4x – 0.3 < 0.3x + 6
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 10 Continued
x < 3
–5 –4 –3 –2 –1 0 1 2 3 4 5
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
There are special cases of inequalities called identities and contradictions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Example 11
2x – 7 ≤ 5 + 2x
2x – 7 ≤ 5 + 2x–2x –2x
–7 ≤ 5Subtract 2x from both sides.
True statement.
The inequality 2x − 7 ≤ 5 + 2x is an identity. All values of x make the inequality true. Therefore, all real numbers are solutions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
2(3y – 2) – 4 ≥ 3(2y + 7)
2(3y – 2) – 4 ≥ 3(2y + 7)
2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)
6y – 4 – 4 ≥ 6y + 21
6y – 8 ≥ 6y + 21
Distribute 2 on the left side and 3 on the right side.
Example 12
–6y –6y
–8 ≥ 21
Subtract 6y from both sides.
False statement.No values of y make the inequality true. There are no solutions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
4(y – 1) ≥ 4y + 2
4(y – 1) ≥ 4y + 2
4(y) + 4(–1) ≥ 4y + 2
4y – 4 ≥ 4y + 2
Distribute 4 on the left side.
Example 13
Solve the inequality.
–4y –4y
–4 ≥ 2
Subtract 4y from both sides.
False statement.
No values of y make the inequality true. There are no solutions.
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Solve the inequality.
x – 2 < x + 1
x – 2 < x + 1 –x –x
–2 < 1Subtract x from both sides.True statement.
All values of x make the inequality true. All real numbers are solutions.
Example 14
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Lesson Summary: Part I
Solve each inequality and graph the solutions.
1. t < 5t + 24 t > –6
2. 5x – 9 ≤ 4.1x – 81 x ≤ –80
b < 133. 4b + 4(1 – b) > b – 9
Holt Algebra 1
3-5 Solving Inequalities with Variables on Both Sides
Lesson Summary: Part II
4. 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.