Albert Teen

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Linear Inequalities 2

# Linear Inequalities 2

### Linear Inequalities 2

Linear inequalities may contain more than one inequality sign, or multiply variables.

Sometimes you will have to solve inequalities with two inequality symbols. It is best to think about these as two separate inequalities.

$3 < 2x + 5 < 11$

This can also be thought of as: $2x+5>3$ and $2x+5<11$

What is this inequality the same as?

$2 < 3x + 11 < 26$

In order to solve an inequality with two symbols, we need to solve each individual inequality.

Let's solve $3 < 2x + 5 < 11$

1

$3<2x+5$

2

Subtract 5 from both sides for $3<2x+5$

3

Divide both sides by 2 for $2x>-2$

4

Now tackle the right side

$2x+5<11$

5

Subtract 5 from both sides for $2x+5<11$

6

Divide both sides by 2 for $2x<6$

7

Put these two results together

Our individual inequalities give the answers $x<3$ and $x>-1$. We can combine these into a single inequality, so the final answer is $-1 < x < 3$.

1

We can show $-1 on a number line

The open circle indicates that the inequality does not include -1 or 3.

Solve this:

$2 < 3x + 11 < 26$

Now solve $1 < 4 - x < 20$

We can also have inequalities with more than one letter.

$y=2,4,6,8$

$x=1,3,5,7,9$ What is the range of $x+y$?

1

Find the smallest value of $x+y$

This involves using the smallest value of each variable.$x=1$ and y=2$$y=2\rightarrow$$y=2\rightarrow1+2=3

2

Find the largest value of $x+y$

Now we need to choose the largest values of each variable.$x=9$ and y=8$$y=8\rightarrow$$y=8\rightarrow9+8=17

3

These are the upper and lower bounds for $x+y$

Each of the other values of $x+y$ will be between the upper and lower bounds.

4

Make an inequality

The upper bound is 17, so $x+y\le17$. The lower bound is 3, so $3\le x+y$. We can arrange these two inequalities into a single inequality$3 \leq x + y \leq 17$

5

Check that $3 \leq x + y \leq 17$ matches the values

Have a look at the values given in the question. Does this inequality seem possible? It looks like it is, so we can confirm this is the final answer!