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Natural numbers Whole numbers Integers Absolute value Adding and subtracting integersHere you will learn strategies on how to add and subtract rational numbers including using visual models, the number line, as well as using standard algorithms.
Students will first learn about integers in 6 th grade math as part of their work with the number system and expand that knowledge to operations with integers in the 7 th grade.
Adding and subtracting rational numbers is when you add or subtract two or more rational numbers together.
To add or subtract two rational numbers with the same denominator, we can add or subtract the numerators and write the result over the common denominator.
To add or subtract two rational numbers with different denominators, we must find a common denominator to add or subtract the numerators.
You can also add and subtract rational numbers using visual models or a number line.
How does this apply to 7 th grade math?
Use this quiz to check your grade 2, 3, 4 and 7 students’ understanding of addition and subtraction. 15+ questions with answers covering a range of 2nd, 3rd, 4th and 7th grade addition and subtraction topics to identify areas of strength and support!
DOWNLOAD FREEUse this quiz to check your grade 2, 3, 4 and 7 students’ understanding of addition and subtraction. 15+ questions with answers covering a range of 2nd, 3rd, 4th and 7th grade addition and subtraction topics to identify areas of strength and support!
DOWNLOAD FREEIn order to add and subtract rational numbers using counters:
In order to add and subtract rational numbers using a number line:
-2 \, \cfrac{1}{2}+7= \, ?
Use what you know about integer zero pairs.
There is \cfrac{1}{2} of a positive counter and 4 whole positive counters left.
2The answer is the leftover counters.
-2 \, \cfrac{1}{2}+7=4 \cfrac{1}{2}
-4.5-(-6)= \, ?
Model the problem with counters and use zero pairs when necessary.
Model -4.5 with the counters.
There are not enough negatives to take away -6. Add 2 zero pairs in order to remove -6.
This leaves 2 positive counters and \cfrac{1}{2} of a negative counter.
There is 1 positive counter left and \cfrac{1}{2} of a positive counter left.
The answer is the leftover counters.
-4.5 - (-6) = 1.5
-11.1+(-0.3)= \, ?
To add, start at the first number and move to the second number; to subtract, start from the second number and move to the first number.
-11.1 + (-0.3) is addition. Start at -11.1 and move in the negative direction (left) 0.3 places. You land at -11.4.
Write your answer.
-11.1 + (-0.3) = -11.4
In order to add and subtract rational numbers using algorithms:
Solve 3 \, \cfrac{4}{5}-5 \, \cfrac{1}{6} \, .
Make sense of the calculation – relating to positive numbers when necessary.
3 \, \cfrac{4}{5}-5 \, \cfrac{1}{6} \, is the distance from 5 \, \cfrac{1}{6} to 3 \, \cfrac{4}{5} \, .
5 \, \cfrac{1}{6}-3 \, \cfrac{4}{5} \, is the distance from 3 \, \cfrac{4}{5} to 5 \, \cfrac{1}{6} \, , which is the exact same part of the number line, just the opposite direction.
This means you can solve 5 \, \cfrac{1}{6}-3 \, \cfrac{4}{5} \, and use the opposite as the answer for 3 \, \cfrac{4}{5}-5 \, \cfrac{1}{6} \, .
Use an algorithm.
5 \, \cfrac{1}{6}-3 \, \cfrac{4}{5}
=\cfrac{31}{6}-\cfrac{19}{5} \hspace{1.4cm} *Convert to improper fractions
=\cfrac{31 \times 5}{6 \times 5}-\cfrac{19 \times 6}{5 \times 6} \quad *Create a common denominator
=\cfrac{155}{30}-\cfrac{114}{30} \hspace{1.15cm} *Subtract the numerators
=\cfrac{41}{30} which simplifies to 1 \, \cfrac{11}{30}
Decide if the final answer is positive or negative.
There are two ways to think about this…
3 \, \cfrac{4}{5}-5 \, \cfrac{1}{6}=-1 \, \cfrac{11}{30}
Solve -125.5+(-34.56).
Make sense of the calculation – relating to positive numbers when necessary.
-125.5 + (-34.56) combines two negative amounts, so the answer will be an even larger negative amount. You can combine like you would with positive numbers, to see what the total amount is.
Use an algorithm.
Decide if the final answer is positive or negative.
Adding two negative numbers results in a more negative number. For example, if you owe \$125.50 and then you make another purchase of \$34.56, now you owe even more. Your debt has grown, and debt is negative.
-125.5 + (-34.56) = -160.06
Solve 125.5+(-34.56).
Make sense of the calculation – relating to positive numbers when necessary.
125.5 + (-34.56) combines a positive and negative amount. Combining positive and negative creates 0 pairs, which has the same effect as subtracting two positive numbers. You can subtract like we would with positive numbers, to see what the total difference is.
Use an algorithm.
Decide if the final answer is positive or negative.
Now to decide if the answer is positive or negative, look at the original numbers being added. Since 125.5 is larger, the amount left over will also be positive.
125.5 + (-34.56) = 90.94
1) Use the model below to add -7 \, \cfrac{3}{4}+6.
There are 6 zero pairs with one negative counter and \cfrac{3}{4} of a negative counter leftover.
-7 \, \cfrac{3}{4}+6=-1 \, \cfrac{3}{4}
2) Use the number line to subtract -1.5-(- 0.9).
Solving -1.5-(-0.9) is the same as finding the distance from -0.9 to -1.5.
From -0.9 move left 0.6 and you get to -1.5. Moving left is in the negative direction.
So, -1.5-(-0.9) = -0.6
3) Use the number line to add 0.75 + (-1.25).
Start at 0.75 and move 1.25 places in the negative direction (left).
You land at -0.5.
0.75 + (-1.25) = -0.5
4) Subtract: -14 \, \cfrac{2}{3}-\left(-8 \, \cfrac{1}{2}\right)= \, ?
-14 \, \cfrac{2}{3}-\left(-8 \, \cfrac{1}{2}\right) is starting with -14 \, \cfrac{2}{3} taking away -8 \, \cfrac{1}{2}.
It can also be thought of as the distance from -8 \, \cfrac{1}{2} \, to -14 \, \cfrac{2}{3}.
Because of absolute value, this is the same as the distance from 8 \, \cfrac{1}{2} \, to 14 \, \cfrac{2}{3}.
So, you can solve for the difference with 14 \, \cfrac{2}{3}-8 \, \cfrac{1}{2}.
\begin{aligned} & 14 \, \cfrac{2}{3}-8 \frac{1}{2} \\\\ & =14 \, \cfrac{2 \times 2}{3 \times 2}-8 \frac{1 \times 3}{2 \times 3} \\\\ & =14 \, \cfrac{4}{6}-8 \frac{3}{6} \\\\ & =6 \, \cfrac{1}{6} \end{aligned}
Since the original equation started with more negatives, the result will still be negative – although there are less now after subtracting 8 \, \cfrac{1}{2} negatives. Also, moving from -8 \, \cfrac{1}{2} to -14 \, \cfrac{2}{3} is to the left or in the negative direction.
-14 \, \cfrac{2}{3}-\left(-8 \, \cfrac{1}{2}\right)=-6 \, \cfrac{1}{6}
5) Add: -13.2 + (-4.8) = \, ?
-13.2 + (-4.8) combines two negative numbers, so the sum will be more negative.
You can solve for the sum of the negatives like you would positives since there are no zero pairs canceling values out:
13.2 + 4.8 = 18.
Since the original equation is adding negatives, the answer will also be negative.
-13.2 + (-4.8) = -18
6) On a January day in New York, the morning temperature was -3.5 degrees Fahrenheit. Later that day, the temperature increased by 14.3 degrees. What is the new temperature in degrees?
-10.8 degree
17.8 degrees
-17.8 degrees
10.8 degree
-3.5 increased by 14.3 degrees is -3.5 + 14.3.
Since you can add in any order, you can also solve for 14.3 + (-3.5).
If we start with 14.3 and add -3.5 the answer will go down 3.5, so you can use 14.3-3.5 to solve.
14.3-3.5 = 10.8, so 14.3 + (-3.5) = 10.8 which means -3.5 + 14.3 = 10.8.
All positive numbers have an inverse that is negative and all negative numbers have an inverse that is positive. For example, -4.5 and 4.5 are inverses. When you add inverses together their sum is always 0 – think zero pairs. Thinking about inverses is one way to solve addition and subtraction of rational numbers.
In order to add negative fractions and mixed fractions (mixed numbers), each fraction must have the same denominator. A common strategy is to find the least common multiple (LCM) of the denominators and use this to solve (sometimes called the least common denominator or LCD). Once the denominators are common, you can add the numerators to solve.
See also: Adding fractions
Operations with rational numbers can be expanded to solve algebraic expressions and equations, including ones with polynomials and inequalities.
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