Download Inequalities with Variables on Both Sides

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 ≤ 4y + 18
y ≤ 4y + 18
–4y –4y
Subtract 4y from both sides.
-3y ≤ 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.
4x ≥ 7x + 6
4x ≥ 7x + 6
–7x –7x
Subtract 7x from both sides.
–3x ≥ 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 + 12w < 36w
• Subtract 12w from both sides
•
•
•
•
•
312 < 24w
Divide both sides by 24
312/24 < 24w/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 + 0.10f < 0.25f
• Subtract 0.10f from both sides.
• 24 + 0.10f – 0.10f < 0.25f – 0.10f
• 24 < 0.15f
• Divide both sides by 0.15
• 24/0.15 < 0.15f/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 + 3k – 3
Distribute 2 on the left side of
the inequality.
2(k – 3) > 3 + 3k
And, combine the 6 and -3 on
2k + 2(–3) > 3 + 3k
the right side.
Subtract 3k from both sides.
2k – 6 > 3 + 3k
–3k
– 3k
-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 3r from both sides.
10 – 5r ≥ 3r – 6
-3r -3r
10 − 8r ≥ -6
-10
-10
Subtract 10 from both sides
-8r ≥ -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.
2x – 7 ≤ 5 + 2x
2x – 7 ≤ 5 + 2x
–2x
–2x
–7 ≤ 5
Subtract 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.
Example 8
Solve the inequality.
2(3y – 2) – 4 ≥ 3(2y + 7)
Distribute 2 on the left side
and 3 on the right side.
2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)
2(3y – 2) – 4 ≥ 3(2y + 7)
6y – 4 – 4 ≥ 6y + 21
6y – 8 ≥ 6y + 21
–6y
–6y
Subtract 6y from both sides.
False statement.
–8 ≥ 21
No values of y make the inequality true.
There are no solutions.
Try
these…
Lesson Quiz: Part I
Solve each inequality and graph the solutions.
1. t < 5t + 24 t > –6
2. 5x – 9 ≤ 4.1x – 81 x ≤ –80
3. 4b + 4(1 – b) > b – 9
4. 2y – 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.