# 1.3 Add and subtract integers  (Page 3/10)

 Page 3 / 10

Fill in $<,>,\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}=$ for each of the following pairs of numbers:

$|-5|___-|-5|$ $8___-|-8|$ $-9___-|-9|$ $\text{−}\left(-16\right)___-|-16|$

## Solution

$\begin{array}{cccccc}& & & \hfill \phantom{\rule{1.6em}{0ex}}|-5|& ___\hfill & -|-5|\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{1.6em}{0ex}}5& ___\hfill & -5\hfill \\ \text{Order.}\hfill & & & \hfill \phantom{\rule{1.6em}{0ex}}5& >\hfill & -5\hfill \\ & & & \hfill \phantom{\rule{1.6em}{0ex}}|-5|& >\hfill & -|-5|\hfill \end{array}$

$\begin{array}{cccccc}& & & \hfill \phantom{\rule{2.3em}{0ex}}8& ___\hfill & -|-8|\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{2.3em}{0ex}}8& ___\hfill & -8\hfill \\ \text{Order.}\hfill & & & \hfill \phantom{\rule{2.3em}{0ex}}8& >\hfill & -8\hfill \\ & & & \hfill \phantom{\rule{2.3em}{0ex}}8& >\hfill & -|-8|\hfill \end{array}$

$\begin{array}{cccccc}& & & \hfill \phantom{\rule{1.6em}{0ex}}9& ___\hfill & -|-9|\hfill \\ \text{Simplify.}\hfill & & & \hfill \phantom{\rule{1.6em}{0ex}}-9& ___\hfill & -9\hfill \\ \text{Order.}\hfill & & & \hfill \phantom{\rule{1.6em}{0ex}}-9& =\hfill & -9\hfill \\ & & & \hfill \phantom{\rule{1.6em}{0ex}}-9& =\hfill & -|-9|\hfill \end{array}$

$\begin{array}{cccccc}& & & \hfill \text{−}\left(-16\right)& ___\hfill & -|-16|\hfill \\ \text{Simplify.}\hfill & & & \hfill 16& ___\hfill & -16\hfill \\ \text{Order.}\hfill & & & \hfill 16& >\hfill & -16\hfill \\ & & & \hfill \text{−}\left(-16\right)& >\hfill & -|-16|\hfill \end{array}$

Fill in<,>, or $=$ for each of the following pairs of numbers: $|-9|___-|-9|$ $2___-|-2|$ $-8___|-8|$
$\text{−}\left(-9\right)___-|-9|.$

> > < >

Fill in<,>, or $=$ for each of the following pairs of numbers: $7___-|-7|$ $\text{−}\left(-10\right)___-|-10|$
$|-4|___-|-4|$ $-1___|-1|.$

> > > <

We now add absolute value bars to our list of grouping symbols. When we use the order of operations, first we simplify inside the absolute value bars as much as possible, then we take the absolute value    of the resulting number.

## Grouping symbols

$\begin{array}{cccccccc}\text{Parentheses}\hfill & \left(\right)\hfill & & & & & \text{Braces}\hfill & \left\{\right\}\hfill \\ \text{Brackets}\hfill & \left[\right]\hfill & & & & & \text{Absolute value}\hfill & \phantom{\rule{0.15em}{0ex}}|\phantom{\rule{0.2em}{0ex}}|\hfill \end{array}$

In the next example, we simplify the expressions inside absolute value bars first, just like we do with parentheses.

Simplify: $24-|19-3\left(6-2\right)|.$

## Solution

$\begin{array}{cccccc}& & & & & \hfill 24-|19-3\left(6-2\right)|\hfill \\ \text{Work inside parentheses first: subtract}\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}6.\hfill & & & & & \hfill 24-|19-3\left(4\right)|\hfill \\ \text{Multiply}\phantom{\rule{0.2em}{0ex}}3\left(4\right).\hfill & & & & & \hfill 24-|19-12|\hfill \\ \text{Subtract inside the absolute value bars.}\hfill & & & & & \hfill 24-|7|\hfill \\ \text{Take the absolute value.}\hfill & & & & & \hfill 24-7\hfill \\ \text{Subtract.}\hfill & & & & & \hfill 17\hfill \end{array}$

Simplify: $19-|11-4\left(3-1\right)|.$

16

Simplify: $9-|8-4\left(7-5\right)|.$

9

Evaluate: $|x|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}x=-35$ $|\text{−}y|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}y=-20$ $\text{−}|u|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}u=12$ $\text{−}|p|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}p=-14.$

## Solution

$|x|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}x=-35$

 $|x|$ Take the absolute value. 35

$|\text{−}y|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}y=-20$
 $|-y|$ Simplify. $|20|$ Take the absolute value. 20

$\text{−}|u|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}u=12$
 $-|u|$ Take the absolute value. $-12$

$\text{−}|p|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}p=-14$
 $-|p|$ Take the absolute value. $-14$

Evaluate: $|x|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}x=-17$ $|\text{−}y|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}y=-39$ $\text{−}|m|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}m=22$ $\text{−}|p|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}p=-11.$

$17$ $39$ $-22$ $-11$

Evaluate: $|y|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}y=-23$ $|\text{−}y|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}y=-21$ $\text{−}|n|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}n=37$ $\text{−}|q|\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}q=-49.$

$23$ $21$ $-37$ $-49$

Most students are comfortable with the addition and subtraction facts for positive numbers. But doing addition or subtraction with both positive and negative numbers may be more challenging.

We will use two color counters to model addition and subtraction of negatives so that you can visualize the procedures instead of memorizing the rules.

We let one color (blue) represent positive. The other color (red) will represent the negatives. If we have one positive counter and one negative counter, the value of the pair is zero. They form a neutral pair. The value of this neutral pair is zero.

We will use the counters to show how to add the four addition facts using the numbers $5,-5$ and $3,-3.$

$\begin{array}{cccccccccc}\hfill 5+3\hfill & & & \hfill -5+\left(-3\right)\hfill & & & \hfill -5+3\hfill & & & \hfill 5+\left(-3\right)\hfill \end{array}$

To add $5+3,$ we realize that $5+3$ means the sum of 5 and 3.

 We start with 5 positives. And then we add 3 positives. We now have 8 positives. The sum of 5 and 3 is 8.

Now we will add $-5+\left(-3\right).$ Watch for similarities to the last example $5+3=8.$

To add $-5+\left(-3\right),$ we realize this means the sum of $-5\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}-3.$

 We start with 5 negatives. And then we add 3 negatives. We now have 8 negatives. The sum of −5 and −3 is −8.

In what ways were these first two examples similar?

• The first example adds 5 positives and 3 positives—both positives.
• The second example adds 5 negatives and 3 negatives—both negatives.

4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
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Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
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@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
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@Geogie my bad that was meant for u
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