Uploaded by Eagle Ridge Randy Cox

# Solving Inequalities ```Solving Inequalities
Just as we can solve equations to find out the value of a variable, we can do the same with
inequalities, with one significant exception (more about that in a minute).
If we start with an inequality like 3𝑥 − 7 &gt; 7, we can use the same tools we are familiar with
from soliving equations, such as adding or subtracting a number from each side of the
inquality or dividing or multiplying each side of the inequality by the same number.
3𝑥 − 7 &gt; 5
+7 + 7
3𝑥
&gt; 12
−
−
3
3
𝑥
&gt; 4
Original inequality
Simplify
Divide each
side by 3 (Division property of inequalities)
Solved for x
Likewise,
1
3+ 𝑥 ≥ 0
2
−3
−3
1
𝑥 ≥ −3
2
&times;2 &times;2
𝑥 ≥ −6
The only trick is when you multiply or divide by a negative number. If you happen to do that,
the direction of the inequality must flip. Less than becomes greater than and vice versa.
On the left, as soon as we divided by -2, we had to switch the &gt; sign to &lt;. On the right, as soon
as we multiplied by -2, we had to switch the ≥ sign to ≤.
On the next page, we will test some values to make sure that flipping signs when multiplying
or dividing by a negative is correct.
Our first example was 2𝑥 + 8 &gt; 4𝑥 + 2 and our solution was 𝑥 &lt; 3. Let’s test several values
near 3 to see if the left side of the inequality does end up greater than the right whenever
𝑥 &lt; 3.
We will also test the second inquality 1.5𝑥 ≥ 2𝑥 − 5 to make sure that it is true for values
≤ 10.
𝑥
1
2
3
4
5
6
2𝑥 + 8
2(1)+8 = 10
2(2)+8 = 12
2(3)+8 = 14
2(4)+8 = 16
2(5)+8 = 18
2(6)+8 = 20
?&gt;
&gt;
&gt;
=
&lt;
&lt;
&lt;
4𝑥 + 2
4(1)+2 = 6
4(2)+2 = 10
4(3)+2 = 14
4(4)+2 = 18
4(5)+2 = 22
4(6)+2 = 26
𝑥
7
8
9
10
11
12
1.5𝑥
1.5(7) = 10.5
1.5(8) = 12
1.5(9) = 13.5
1.5(10) = 15
1.5(11) = 16.5
1.5(12) = 18
?≥
&gt;
&gt;
&gt;
=
&lt;
&lt;
2𝑥 − 5
2(7)-5 = 9
2(8)-5 = 11
2(9)-5 = 13
2(10)-5 = 15
2(11)-5 = 17
2(12)-5 = 19
Our solutions, 𝑥 &lt; 3 for the first inequality is valid for values of x &lt; 3, so flipping the sign
works! Likewise, in the second problem (1.5𝑥 ≥ 2𝑥 − 5), flipping the inequality sign to get a
solution of 𝑥 ≤ 10 works because when we input values for x that are equal to or less than 10,
we get a true result for the original inequality.
Exercises
For each inquality, solve for x and then plot the result on the number line.
3𝑥 + 2 ≤ 2𝑥 + 6
6𝑥 ≥ 36
5𝑥 + 7 &gt; 𝑥 + 23
5𝑥 + 3 ≥ 2𝑥 − 6
1
𝑥+1&lt;2
2
4𝑥 + 6 − 2𝑥 ≥ 0
2𝑥 + 2 &lt; 3𝑥 − 5
3𝑥 + 2 &gt; 2𝑥 − 5
1
− 𝑥 ≥ −2
3
−5𝑥 + 27 &lt; 5𝑥 − 23
3
6+ 𝑥 ≥𝑥+4
4
17𝑥 + 6 ≤ 13𝑥 − 2
|𝑥| ≥ 4
3𝑥 + 2 ≤ 2𝑥 + 6
−2𝑥
− 2𝑥
𝑥+2≤6
−2 − 2
𝑥≤4
6𝑥 ≥ 36
&divide;6
&divide;6
𝑥≥6
5𝑥 + 7 &gt; 𝑥 + 23
−7
−7
5𝑥
&gt; 𝑥 + 16
−𝑥
−𝑥
4𝑥 &gt; 16
&divide;4
&divide;4
𝑥&gt;4
5𝑥 + 3 ≥ 2𝑥 − 6
−3
−3
5𝑥
≥ 2𝑥 − 9
−2𝑥
− 2𝑥
3𝑥
≥
−9
&divide;3
&divide;3
𝑥 ≥ −3
1
𝑥+1&lt;2
2
−1 − 1
1
𝑥
&lt;1
2
&times;2
&times;2
𝑥&lt;2
4𝑥 + 6 − 2𝑥 ≥ 0
−6
−6
4𝑥
− 2𝑥 ≥ −6
2𝑥 ≥ −6
&divide;2 &divide;2
𝑥 ≥ −3
2𝑥 + 2 &lt; 3𝑥 − 5
2𝑥 + 2 − 2 &lt; 3𝑥 − 5 − 2
2𝑥 &lt; 3𝑥 − 7
2𝑥 − 3𝑥 &lt; 3𝑥 − 3𝑥 − 7
−𝑥 &lt; −7
−𝑥 &times; −1 &gt; −7 &times; −1
𝑥&gt;7
3𝑥 + 2 &gt; 2𝑥 − 5
3𝑥 + 2 − 2 &gt; 2𝑥 − 5 − 2
3𝑥 &gt; 2𝑥 − 7
3𝑥 − 2𝑥 &gt; −7
𝑥 &gt; −7
1
− 𝑥 ≥ −2
3
−5𝑥 + 27 &lt; 5𝑥 − 23
3
6+ 𝑥 ≥𝑥+4
4
−5𝑥 + 27 + 23 &lt; 5𝑥 − 23 + 23
−5𝑥 + 50 &lt; 5𝑥
−5𝑥 + 5𝑥 + 50 &lt; 5𝑥 + 5𝑥
50 &lt; 10𝑥
50 10𝑥
&lt;
10
10
5&lt;𝑥
𝑥&gt;5
|𝑥| ≥ 4 𝑥 ≥ 4 𝑜𝑟 𝑥 ≤ 4
3
6−6+ 𝑥 ≥ 𝑥+4−6
4
3
𝑥−𝑥 ≥𝑥−𝑥−2
4
1
− 𝑥 ≥ −2
4
1
−1 &times; − 𝑥 ≤ −1 &times; −2 𝑥 ≤ 2
4
1
− 𝑥 &times; −3 ≤ −2 &times; −3
3
𝑥≤6
17𝑥 + 6 ≤ 13𝑥 − 2
17𝑥 − 13𝑥 ≤ −2 − 6
4𝑥 ≤ −8
4
8
𝑥≤−
4
4
𝑥 ≤ −2
```