Add and Subtract Fractions with Different Denominators

4.5 Add and Subtract Fractions with Different Denominators

The topics covered in this section are:

  1. Find the least common denominator (LCD)
  2. Convert fractions to equivalent fractions with the LCD
  3. Add and subtract fractions with different denominators
  4. Identify and use fraction operations
  5. Use the order of operations to simplify complex fractions
  6. Evaluate variable expressions with fractions

4.5.1 Find the Least Common Denominator

In the previous section, we explained how to add and subtract fractions with a common denominator. But how can we add and subtract fractions with unlike denominators?

Let’s think about coins again. Can you add one quarter and one dime? You could say there are two coins, but that’s not very useful. To find the total value of one quarter plus one dime, you change them to the same kind of unit—cents. One quarter equals $25$ cents and one dime equals $10$ cents, so the sum is $35$ cents. See the figure below.

A quarter and a dime are shown. Below them, it reads 25 cents plus 10 cents. Below that, it reads 35 cents.
Together, a quarter and a dime are worth $35$ cents, or $\frac{35}{100}$ of a dollar.

Similarly, when we add fractions with different denominators we have to convert them to equivalent fractions with a common denominator. With the coins, when we convert to cents, the denominator is $100$. Since there are $100$ cents in one dollar, $25$ cents is $\frac{25}{100}$ and $10$ cents is $\frac{10}{100}$. So we add $\frac{25}{100} + \frac{10}{100}$ to get $\frac{35}{100}$, which is $35$ cents.

You have practiced adding and subtracting fractions with common denominators. Now let’s see what you need to do with fractions that have different denominators.

First, we will use fraction tiles to model finding the common denominator of $\frac{1}{2}$ and $\frac{1}{3}$.

We’ll start with one $\frac{1}{2}$ tile and $\frac{1}{3}$ tile. We want to find a common fraction tile that we can use to match both $\frac{1}{2}$ and $\frac{1}{3}$ exactly.

If we try the $\frac{1}{4}$ pieces, $2$ of them exactly match the $\frac{1}{2}$ piece, but they do not exactly match the $\frac{1}{3}$ piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into two pieces, each labeled 1 fourth. Underneath the second rectangle are two pieces, each labeled 1 fourth. These rectangles together are longer than the rectangle labeled as 1 third.

If we try the $\frac{1}{5}$ pieces, they do not exactly cover the $\frac{1}{2}$ piece or the $\frac{1}{3}$ piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into three pieces, each labeled 1 sixth. Underneath the second rectangle is an equally sized rectangle split vertically into 2 pieces, each labeled 1 sixth.

If we try the $\frac{1}{6}$ pieces, we see thatexactly $3$ of them cover the $\frac{1}{2}$ piece, and exactly $2$ of them cover the $\frac{1}{3}$ piece.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle are three smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 half rectangle. Below the 1 third rectangle are two smaller rectangles, each labeled 1 fifth. Together, these rectangles are longer than the 1 third rectangle.

If we were to try the $\frac{1}{12}$ pieces, they would also work.

Two rectangles are shown side by side. The first is labeled 1 half. The second is shorter and is labeled 1 third. Underneath the first rectangle is an equally sized rectangle split vertically into 6 pieces, each labeled 1 twelfth. Underneath the second rectangle is an equally sized rectangle split vertically into 4 pieces, each labeled 1 twelfth.

Even smaller tiles, such as $\frac{1}{24}$ and $\frac{1}{48}$, would also exactly cover the $\frac{1}{2}$ piece and the $\frac{1}{3}$ piece.

The denominator of the largest piece that covers both fractions is the least common denominator (LCD) of the two fractions. So, the least common denominator of $\frac{1}{2}$ and $\frac{1}{3}$ is $6$.

Notice that all of the tiles that cover $\frac{1}{2}$ and $\frac{1}{3}$ have something in common: Their denominators are common multiples of $2$ and $3$, the denominators of $\frac{1}{2}$ and $\frac{1}{3}$. The least common multiple (LCM) of the denominators is $6$, and so we say that $6$ is the least common denominator (LCD) of the fractions $\frac{1}{2}$ and $\frac{1}{3}$.

LEAST COMMON DENOMINATOR

The least common denominator (LCD) of two fractions is the least common multiple (LCM) of their denominators.

To find the LCD of two fractions, we will find the LCM of their denominators. We follow the procedure we used earlier to find the LCM of two numbers. We only use the denominators of the fractions, not the numerators, when finding the LCD.

Example 1

Find the LCD for the fractions $\frac{7}{12}$ and $\frac{5}{18}$.

Solution

Factor each denominator into its primes..
List the primes of $12$ and the primes of $18$ lining them up in columns when possible..
Bring down the columns..
Multiply the factors. The product is the LCM.LCM $=36$
The LCM of $12$ and $18$ is $36$, so the LCD of $\frac{7}{12}$ and $\frac{5}{18}$ is $36$.LCD of $\frac{7}{12}$ and $\frac{5}{18}$ is $36$.

To find the LCD of two fractions, find the LCM of their denominators. Notice how the steps shown below are similar to the steps we took to find the LCM.

HOW TO: Find the least common denominator (LCD) of two fractions.

  1. Factor each denominator into its primes.
  2. List the primes, matching primes in columns when possible.
  3. Bring down the columns.
  4. Multiply the factors. The product is the LCM of the denominators.
  5. The LCM of the denominators is the LCD of the fractions.

Example 2

Find the least common denominator for the fractions $\frac{8}{15}$ and $\frac{11}{24}$.

Solution

To find the LCD, we find the LCM of the denominators.

Find the LCM of $15$ and $24$.

The top line shows 15 equals 3 times 5. The next line shows 24 equals 2 times 2 times 2 times 3. The 3s are lined up vertically. The next line shows LCM equals 2 times 2 times 2 times 3 times 5. The last line shows LCM equals 120.

The LCM of $15$ and $24$ is $120$. So, the LCD of $\frac{8}{15}$ and $\frac{11}{24}$ is $120$.

4.5.2 Convert Fractions to Equivalent Fractions with the LCD

Earlier, we used fraction tiles to see that the LCD of $\frac{1}{4}$ and $\frac{1}{6}$ is $12$. We saw that three $\frac{1}{12}$ pieces exactly covered $\frac{1}{4}$ and two $\frac{1}{12}$ pieces exactly covered $\frac{1}{6}$, so

$\large \frac{1}{4} = \frac{3}{12}$ and $\large \frac{1}{6} = \frac{2}{12}$.

On the left is a rectangle labeled 1 fourth. Below it is an identical rectangle split vertically into 3 equal pieces, each labeled 1 twelfth. On the right is a rectangle labeled 1 sixth. Below it is an identical rectangle split vertically into 2 equal pieces, each labeled 1 twelfth.

We say that $\frac{1}{4}$ and $\frac{3}{12}$ are equivalent fractions and also that $\frac{1}{6}$ and $\frac{2}{12}$ are equivalent fractions.

We can use the Equivalent Fractions Property to algebraically change a fraction to an equivalent one. Remember, two fractions are equivalent if they have the same value. The Equivalent Fractions Property is repeated below for reference.

EQUIVALENT FRACTIONS PROPERTY

If $a,b,c$ are whole numbers where $b \neq 0, c \neq 0$, then

$\large \frac{a}{b} = \frac{a \cdot c}{b \cdot c}$ and $\large \frac{a \cdot c}{b \cdot c} = \frac{a}{b}$

To add or subtract fractions with different denominators, we will first have to convert each fraction to an equivalent fraction with the LCD. Let’s see how to change $\frac{1}{4}$ and $\frac{1}{6}$ to equivalent fractions with denominator $12$ without using models.

Example 3

Convert $\frac{1}{4}$ and $\frac{1}{6}$ to equivalent fractions with denominator $12$, their LCD.

Solution

Find the LCD.The LCD of $\frac{1}{4}$ and $\frac{1}{6}$ is $12$.
Find the number to multiply $4$ to get $12$.$4 \cdot 3 =12$
Find the number to multiply $6$ to get $12$.$6 \cdot 2 = 12$
Use the Equivalent Fractions Property to convert each fraction to an equivalent fraction with the LCD, multiplying both the numerator and denominator of each fraction by the same number.$\frac{3}{4} = \frac{3 \cdot 3}{4 \cdot 3}$
$\frac{1}{6} = \frac{1 \cdot 2}{6 \cdot 2}$
Simplify the numerators and denominators.$\frac{3}{12}$
$\frac{2}{12}$

We do not reduce the resulting fractions. If we did, we would get back to our original fractions and lose the common denominator.

HOW TO: Convert two fractions to equivalent fractions with their LCD as the common denominator.

  1. Find the LCD.
  2. For each fraction, determine the number needed to multiply the denominator to get the LCD.
  3. Use the Equivalent Fractions Property to multiply both the numerator and denominator by the number you found in Step 2.
  4. Simplify the numerator and denominator.

Example 4

Convert $\frac{8}{15}$ and $\frac{11}{24}$ to equivalent fractions with denominator $120$, their LCD.

Solution

The LCD is $120$. We will start Step 2.
Find the number that must multiply $15$ to get $120$.$15 \cdot 8 = 120$
Find the number that must multiply $24$ to get $120$.$24 \cdot 5 = 120$
Use the Equivalent Fractions Property.$\frac{8 \cdot 8}{15 \cdot 8}$ and $\frac{11 \cdot 5}{24 \cdot 5}$
Simplify the numerators and denominators.$\frac{64}{120}$ and $\frac{55}{120}$

4.5.3 Add and Subtract Fractions with Different Denominators

Once we have converted two fractions to equivalent forms with common denominators, we can add or subtract them by adding or subtracting the numerators.

HOW TO: Add or subtract fractions with different denominators.

  1. Find the LCD.
  2. Convert each fraction to an equivalent form with the LCD as the denominator.
  3. Add or subtract the fractions.
  4. Write the result in simplified form.

Example 5

Add: $\frac{1}{2} + \frac{1}{3}$.

Solution

$\frac{1}{2} + \frac{1}{3}$
Find the LCD of $2, 3$.
$2=2$
$3=3$
_______________
LCD $=2 \cdot 3$
LCD $=6$
Change into equivalent fractions with the LCD $6$.$\frac{1 \cdot 3}{2 \cdot 3} + \frac{1 \cdot 2}{3 \cdot 2}$
Simplify the numerators and denominators.$\frac{3}{6} + \frac{2}{6}$
Add.$\frac{5}{6}$

Remember, always check to see if the answer can be simplified. Since $5$ and $6$ have no common factors, the fraction $\frac{5}{6}$ cannot be reduced.

Example 6

Add: $\frac{1}{2} – (- \frac{1}{4})$.

Solution

$\frac{1}{2} – (- \frac{1}{4})$
Find the LCD of $2, 4$.
$2=2$
$4=2 \cdot 2$
_______________
LCD $=2 \cdot 2$
LCD $=4$
Rewrite as equivalent fractions using the LCD $4$.$\frac{1 \cdot 2}{2 \cdot 2} – (- \frac{1}{4})$
Simplify the first fraction.$\frac{2}{4} – (- \frac{1}{4})$
Subtract.$\frac{2-(-1)}{4}$
Simplify.$\frac{3}{4}$

One of the fractions already had the least common denominator, so we only had to convert the other fraction.

Example 6

Add: $\frac{7}{12} + \frac{5}{18}$.

Solution

$\frac{7}{12} + \frac{5}{18}$
Find the LCD of $12, 18$.
$12=2 \cdot 2 \cdot 3$
$18=2 \cdot 3 \cdot 3$
_______________
LCD $=2 \cdot 2 \cdot 3\cdot 3$
LCD $=36$
Rewrite as equivalent fractions with the LCD.$\frac{7 \cdot 3}{12 \cdot 3} + \frac{5 \cdot 2}{18 \cdot 2}$
Simplify the numerators and denominators.$\frac{21}{36} + \frac{10}{36}$
Add.$\frac{31}{36}$

Because $31$ is a prime number, it has no factors in common with $36$. The answer is simplified.

When we use the Equivalent Fractions Property, there is a quick way to find the number you need to multiply by to get the LCD. Write the factors of the denominators and the LCD just as you did to find the LCD. The “missing” factors of each denominator are the numbers you need.

The first line says 12 equals 2 times 2 times 3. There is a blank space next to the 3. The next line says 18 equals 2 times 3 times 3. There is a blank space between the 2 and the first 3. There are red lines drawn from the blank spaces. This is labeled as missing factors. There is a horizontal line. Below the line, it says LCD equals 2 times 2 times 3 times 3. Below this, it says LCD equals 36.

The LCD, $36$, has $2$ factors of $2$ and $2$ factors of $3$.

Twelve has two factors of $2$, but only one of $3$—so it is ‘missing‘ one $3$. We multiplied the numerator and denominator of $\frac{7}{12}$ by $3$ to get an equivalent fraction with denominator $36$.

Eighteen is missing one factor of $2$—so you multiply the numerator and denominator $\frac{5}{18}$ by $2$ to get an equivalent fraction with denominator $36$. We will apply this method as we subtract the fractions in the next example.

Example 7

Subtract: $\frac{7}{15} – \frac{19}{24}$.

Solution

$\frac{7}{15} – \frac{19}{24}$
Find the LCD.
$15=3 \cdot 5$
$24=2 \cdot 2 \cdot 2 \cdot 3$
____________________________
LCD $=2 \cdot 2 \cdot 2\cdot 3 \cdot 5$
LCD $=120$
$15$ is ‘missing’ three factors of $2$
$24$ is ‘missing’ a factor of $5$
Rewrite as equivalent fractions with the LCD.$\frac{7 \cdot 8}{15 \cdot 8} – \frac{19 \cdot 5}{24 \cdot 5}$
Simplify each numerator and denominator.$\frac{56}{120} – \frac{95}{120}$
Subtract.$- \frac{39}{120}$
Rewrite showing the common factor of $3$.$- \frac{13 \cdot 3}{40 \cdot 3}$
Remove the common factor to simplify.$- \frac{13}{40}$

Example 7

Add: $- \frac{11}{30} + \frac{23}{42}$.

Solution

$- \frac{11}{30} + \frac{23}{42}$
Find the LCD.
$30=2 \cdot 3 \cdot 5$
$42=2 \cdot 3 \cdot 7$
______________________
LCD $=2 \cdot 3 \cdot 5\cdot 7$
LCD $=210$
Rewrite as equivalent fractions with the LCD.$- \frac{11 \cdot 7}{30 \cdot 7} + \frac{23 \cdot 5}{42 \cdot 5}$
Simplify each numerator and denominator.$- \frac{77}{210} + \frac{115}{210}$
Add.$\frac{38}{210}$
Rewrite showing the common factor of $2$.$\frac{19 \cdot 2}{105 \cdot 2}$
Remove the common factor to simplify.$\frac{19}{105}$

In the next example, one of the fractions has a variable in its numerator. We follow the same steps as when both numerators are numbers.

Example 8

Add: $\frac{3}{5} + \frac{x}{8}$.

Solution

$\frac{3}{5} + \frac{x}{8}$
Find the LCD.
$5=5$
$8=2 \cdot 2 \cdot 2$
______________________
LCD $=2 \cdot 2 \cdot 2 \cdot 5$
LCD $=40$
Rewrite as equivalent fractions with the LCD.$\frac{3 \cdot 8}{5 \cdot 8} + \frac{x \cdot 5}{8 \cdot 5}$
Simplify the numerators and denominators.$\frac{24}{40} + \frac{5x}{40}$
Add.$\frac{24 + 5x}{40}$

We cannot add $24$ and $5x$ since they are not like terms, so we cannot simplify the expression any further.

4.5.4 Identify and Use Fraction Operations

By now in this chapter, you have practiced multiplying, dividing, adding, and subtracting fractions. The following table summarizes these four fraction operations. Remember: You need a common denominator to add or subtract fractions, but not to multiply or divide fractions

SUMMARY OF FRACTION OPERATIONS

Fraction multiplication: Multiply the numerators and multiply the denominators.

$\large \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

Fraction division: Multiply the first fraction by the reciprocal of the second.

$\large \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$

Fraction addition: Add the numerators and place the sum over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.

$\large \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$

Fraction subtraction: Subtract the numerators and place the difference over the common denominator. If the fractions have different denominators, first convert them to equivalent forms with the LCD.

$\large \frac{a}{c} – \frac{b}{c} = \frac{a-b}{c}$

Example 9

Simplify:

  1. $- \frac{1}{4} + \frac{1}{6}$
  2. $- \frac{1}{4} \div \frac{1}{6}$
Solution

First we ask ourselves, “What is the operation?”

Part 1. The operation is addition.

Do the fractions have a common denominator? No.

$- \frac{1}{4} + \frac{1}{6}$
Find the LCD.
$4=2 \cdot 2$
$6=2 \cdot 3$
_________________
LCD$=2 \cdot 2 \cdot 3$
LCD$=12$
Rewrite each fraction as an equivalent fraction with the LCD.$- \frac{1 \cdot 3}{4 \cdot 3} + \frac{1 \cdot 2}{6 \cdot 2}$
Simplify the numerators and denominators.$- \frac{3}{12} + \frac{2}{12}$
Add the numerators and place the sum over the common denominator.$- \frac{1}{12}$
Check to see if the answer can be simplified. It cannot.

Part 2. The operation is division. We do not need a common denominator.

$- \frac{1}{4} \div \frac{1}{6}$
To divide fractions, multiply the first fraction by the reciprocal of the second.$- \frac{1}{4} \cdot \frac{6}{1}$
Multiply.$- \frac{6}{4}$
Simplify.$- \frac{3}{2}$

Example 9

  1. $\frac{5x}{6} – \frac{3}{10}$
  2. $\frac{5x}{6} \cdot \frac{3}{10}$
Solution

Part 1. The operation is subtraction. The fractions do not have a common denominator.

$\frac{5x}{6} – \frac{3}{10}$
Rewrite each fraction as an equivalent fraction with the LCD, $30$$\frac{5x \cdot 5}{6 \cdot 5} – \frac{3 \cdot 3}{10 \cdot 3}$
$\frac{25x}{30} – \frac{9}{30}$
Subtract the numerators and place the difference over the common denominator.$\frac{25x-9}{30}$

Part 2. The operation is multiplication; no need for a common denominator.

$\frac{5x}{6} \cdot \frac{3}{10}$
To multiply fractions, multiply the numerators and multiply the denominators.$\frac{5x \cdot 3}{6 \cdot 10}$
Rewrite, showing common factors.$\frac{\cancel{5} \cdot x \cancel{3}}{2 \cdot \cancel{3} \cdot 2 \cdot \cancel{5}}$
Remove common factors to simplify.$\frac{x}{4}$

4.5.5 Use the Order of Operations to Simplify Complex Fractions

In Multiply and Divide Mixed numbers and Complex Fractions, we saw that a complex fraction is a fraction in which the numerator or denominator contains a fraction. We simplified complex fractions by rewriting them as division problems. For example,

$\large \frac{\frac{3}{4}}{\frac{5}{8}} = \frac{3}{4} \div \frac{5}{8}$

Now we will look at complex fractions in which the numerator or denominator can be simplified. To follow the order of operations, we simplify the numerator and denominator separately first. Then we divide the numerator by the denominator.

HOW TO: Simplify complex fractions.

  1. Simplify the numerator.
  2. Simplify the denominator.
  3. Divide the numerator by the denominator.
  4. Simplify if possible.

Example 10

Simplify: $\frac{( \frac{1}{2} )^{2}}{ 4+3^{2} }$.

Solution

$\frac{( \frac{1}{2} )^{2}}{ 4+3^{2} }$
Simplify the numerator.$\frac{\frac{1}{4}}{ 4+3^{2} }$
Simplify the term with the exponent in the denominator.$\frac{\frac{1}{4}}{ 4+9}$
Add the terms in the denominator.$\frac{\frac{1}{4}}{13}$
Divide the numerator by the denominator.$\frac{1}{4} \div 13$
Rewrite as multiplication by the reciprocal.$\frac{1}{4} \cdot \frac{1}{13}$
Multiply.$\frac{1}{52}$

Example 11

Simplify: $\frac{\frac{1}{2} + \frac{2}{3}}{\frac{3}{4} – \frac{1}{6}}$.

Solution

$\frac{\frac{1}{2} + \frac{2}{3}}{\frac{3}{4} – \frac{1}{6}}$
Rewrite numerator with the LCD of $6$ and denominator with LCD of $12$.$\frac{\frac{3}{6} + \frac{4}{6}}{\frac{9}{124} – \frac{2}{126}}$
Add in the numerator. Subtract in the denominator.$\frac{\frac{7}{6}}{\frac{7}{12}}$
Divide the numerator by the denominator.$\frac{7}{6} \div \frac{7}{12}$
Rewrite as multiplication by the reciprocal.$\frac{7}{6} \cdot \frac{12}{7}$
Rewrite, showing common factors.$\frac{\cancel{7} \cdot \cancel{6} \cdot 2}{\cancel{6} \cdot \cancel{7} \cdot 1}$
Simplify.$2$

4.5.6 Evaluate Variable Expressions with Fractions

We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.

Example 12

Evaluate $x+ \frac{1}{3}$ when

  1. $x=- \frac{1}{3}$
  2. $x=- \frac{3}{4}$.
Solution

Part 1. To evaluate $x+ \frac{1}{3}$ when $x=- \frac{1}{3}$, substitute $- \frac{1}{3}$ for $x$ in the expression.

$x+ \frac{1}{3}$
Substitute $- \frac{1}{3}$ for $x$.$- \frac{1}{3} + \frac{1}{3}$
Simplify.$0$

Part 2. To evaluate $x+ \frac{1}{3}$ when $x=- \frac{3}{4}$, we substitute $- \frac{3}{4}$ for $x$ in the expression.

$x+ \frac{1}{3}$
Substitute $- \frac{3}{4}$ for $x$.$- \frac{3}{4} + \frac{1}{3}$
Rewrite as equivalent fractions with the LCD, $12$.$- \frac{3 \cdot 3}{4 \cdot 3} + \frac{1 \cdot 4}{3 \cdot 4}$
Simplify the numerators and denominators.$- \frac{9}{12} + \frac{4}{12}$
Add.$- \frac{5}{12}$

Example 13

Evaluate $y- \frac{5}{6}$ when $y=- \frac{2}{3}$.

Solution

We substitute $- \frac{2}{3}$ for $y$ in the expression

$y -\frac{5}{6}$
Substitute $- \frac{2}{3}$ for $y$.$- \frac{2}{3} – \frac{5}{6}$
Rewrite as equivalent fractions with the LCD, $6$.$- \frac{4}{6} – \frac{5}{6}$
Subtract.$- \frac{9}{6}$
Simplify.$- \frac{3}{2}$

Example 14

Evaluate $2x^{2} y$ when $x= \frac{1}{4}$ and $y=- \frac{2}{3}$.

Solution

Substitute the values into the expression. In $2x^{2} y$, the exponent applies only to $x$.

$2x^{2} y$
Substitute $\frac{1}{4}$ for $x$ and $- \frac{2}{3}$ for $y$.$2( \frac{1}{4} )^{2}{(- \frac{2}{3} )}$
Simplify exponents first.$2( \frac{1}{16} ) {(- \frac{2}{3} )}$
Multiply. The product will be negative.$- \frac{2}{1} \cdot \frac{1}{16} \cdot \frac{2}{3}$
Simplify.$- \frac{4}{48}$
Remove the common factors.$- \frac{1 \cdot \cancel{4}}{\cancel{4} \cdot 12}$
Simplify.$- \frac{1}{12}$

Example 15

Evaluate $\frac{p+q}{r}$ when $p=-4$, $q=-2$, and $r=8$.

Solution

We substitute the values into the expression and simplify.

$\frac{p+q}{r}$
Substitute $-4$ for $p$, $-2$ for $q$, and $8$ for $r$.$\frac{-4+(-2)}{8}$
Add the numerator first.$- \frac{6}{8}$
Simplify.$- \frac{3}{4}$
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