4.5 Add and Subtract Fractions with Different Denominators

Learning objectives

Learning objectives

• Find the least common denominator (LCD)

• Convert fractions to equivalent fractions with the LCD

• Add and subtract fractions with different denominators

• Identify and use fraction operations

• Use the order of operations to simplify complex fractions

• Evaluate variable expressions with fractions

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 Figure 4.7.

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.

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

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

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

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 $\frac{1}{3}.$

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 4.63

Solution

1. Factor the denominators. Write $12=2^2\cdot3$ and $18=2\cdot3^2$, as shown by the prime-factor trees in the figures.
2. Choose the needed prime powers. The least common denominator must contain $2^2$ and $3^2$, so use $2^2\cdot3^2=36$.
3. State the result. The LCD of $12$ and $18$, and therefore the LCD for $\frac{7}{12}$ and $\frac{5}{18}$, is $36$.

Takeaway: Find the LCD by prime-factorizing both denominators and taking every prime factor with its greatest occurring power.

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.

Example 4.64

Solution

1. Factor the denominators. Write $15=3\cdot5$ and $24=2^3\cdot3$ to expose every prime factor that the common denominator must contain.
2. Choose the greatest powers. The denominator needs $3^1$ from $15$ and $2^3\cdot3^1$ from $24$, so it must include $2^3\cdot3\cdot5$.
3. Compute the LCD. Multiplying those prime powers gives $2^3\cdot3\cdot5=120$, which is divisible by both $15$ and $24$.
4. State the result. Therefore, the least common denominator for $\frac{8}{15}$ and $\frac{11}{24}$ is $120$.

Takeaway: The least common denominator is the least common multiple of the denominators, so $15$ and $24$ give an LCD of $120$.

Convert Fractions to Equivalent Fractions with the LCD

Earlier, we used fraction tiles to see that the LCD of $\frac{1}{4}$ when $\frac{1}{6}$ 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

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.

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 4.65

Solution

1. Confirm the LCD. The denominators $4$ and $6$ have prime factorizations $2^2$ and $2\cdot3$, so their LCD is $2^2\cdot3=12$.
2. Convert the first fraction. Since $12\div4=3$, multiply $\frac{1}{4}$ by $\frac{3}{3}$ to obtain $\frac{1\cdot3}{4\cdot3}=\frac{3}{12}$.
3. Convert the second fraction. Since $12\div6=2$, multiply $\frac{1}{6}$ by $\frac{2}{2}$ to obtain $\frac{1\cdot2}{6\cdot2}=\frac{2}{12}$.
4. Read the model. The figures display $4\cdot3=12$, $6\cdot2=12$, and the two converted fractions $\frac{3}{12}$ and $\frac{2}{12}$, which agree with the computation.
5. State the result. The equivalent fractions with denominator $12$ are $\frac{3}{12}$ and $\frac{2}{12}$.

Takeaway: To create equivalent fractions with a common denominator, multiply both parts of each fraction by the same factor.

Example 4.66

Solution

1. Find the LCD. The prime factorizations $15=3\cdot5$ and $24=2^3\cdot3$ give $LCD=2^3\cdot3\cdot5=120$.
2. Convert $\frac{8}{15}$. Because $120\div15=8$, multiply numerator and denominator by $8$ to get $\frac{8\cdot8}{15\cdot8}=\frac{64}{120}$.
3. Convert $\frac{11}{24}$. Because $120\div24=5$, multiply numerator and denominator by $5$ to get $\frac{11\cdot5}{24\cdot5}=\frac{55}{120}$.
4. Interpret the figures. The images show $15\cdot8=120$, $24\cdot5=120$, the equal-factor property, and the resulting fractions $\frac{64}{120}$ and $\frac{55}{120}$.
5. State the result. With the common denominator $120$, the equivalent fractions are $\frac{64}{120}$ and $\frac{55}{120}$.

Takeaway: Do not reduce converted fractions too early, because the shared denominator is needed to combine their numerators directly.

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.

Example 4.67

Solution

1. Find a common denominator. Since $2$ and $3$ have no common factor greater than $1$, their LCD is $2\cdot3=6$.
2. Rewrite both fractions. Multiply $\frac{1}{2}$ by $\frac{3}{3}$ to get $\frac{3}{6}$, and multiply $\frac{1}{3}$ by $\frac{2}{2}$ to get $\frac{2}{6}$.
3. Add the numerators. The sum is $\frac{3+2}{6}=\frac{5}{6}$, as depicted by the common-denominator figure.
4. Reduce and state the result. Because $5$ and $6$ share no factor other than $1$, the reduced result is $\frac{5}{6}$.

Takeaway: Convert unlike denominators to the LCD first, then add only the numerators and simplify the result.

Example 4.68

Solution

1. Find the LCD. Since $4=2^2$ is a multiple of $2$, the LCD of $2$ and $4$ is $4$.
2. Rewrite the fractions. The fraction $\frac{1}{2}$ becomes $\frac{2}{4}$ by multiplying both parts by $2$, while $-\frac{1}{4}$ already has denominator $4$.
3. Subtract the negative. The expression becomes $\frac{2}{4}-\left(-\frac{1}{4}\right)=\frac{2}{4}+\frac{1}{4}$.
4. Combine and simplify. Adding the numerators gives $\frac{3}{4}$, which is already reduced.
5. State the result. Therefore, $\frac{1}{2}-\left(-\frac{1}{4}\right)=\frac{3}{4}$.

Takeaway: Subtracting a negative fraction is equivalent to adding the positive fraction, but the denominators must still be made common first.

Example 4.69

Solution

1. Find the LCD. The prime factorizations $12=2^2\cdot3$ and $18=2\cdot3^2$ give $LCD=2^2\cdot3^2=36$.
2. Convert the first fraction. Multiply $\frac{7}{12}$ by $\frac{3}{3}$ to get $\frac{21}{36}$.
3. Convert the second fraction. Multiply $\frac{5}{18}$ by $\frac{2}{2}$ to get $\frac{10}{36}$.
4. Add the numerators. Then $\frac{21}{36}+\frac{10}{36}=\frac{31}{36}$, matching the equivalent-fraction figure.
5. Reduce and state the result. Since $31$ is prime and $36$ is not divisible by $31$, the reduced result is $\frac{31}{36}$.

Takeaway: The missing prime factors reveal the conversion factors, and a final numerator-denominator check confirms whether reduction is possible.

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 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 4.70

Solution

1. Find the LCD. The factorizations $15=3\cdot5$ and $24=2^3\cdot3$ give $LCD=2^3\cdot3\cdot5=120$.
2. Convert $\frac{7}{15}$. Multiply by $\frac{8}{8}$ to obtain $\frac{7\cdot8}{15\cdot8}=\frac{56}{120}$.
3. Convert $\frac{19}{24}$. Multiply by $\frac{5}{5}$ to obtain $\frac{19\cdot5}{24\cdot5}=\frac{95}{120}$.
4. Subtract the numerators. The difference is $\frac{56-95}{120}=-\frac{39}{120}$.
5. Reduce and state the result. Divide numerator and denominator by $3$ to get $-\frac{13}{40}$, which is in lowest terms.

Takeaway: After converting to the LCD, keep the subtraction sign attached to the signed numerator and reduce the resulting fraction.

Example 4.71

Solution

1. Find the LCD. The factorizations $30=2\cdot3\cdot5$ and $42=2\cdot3\cdot7$ give $LCD=2\cdot3\cdot5\cdot7=210$.
2. Convert the negative fraction. Multiply $-\frac{11}{30}$ by $\frac{7}{7}$ to obtain $-\frac{77}{210}$.
3. Convert the positive fraction. Multiply $\frac{23}{42}$ by $\frac{5}{5}$ to obtain $\frac{115}{210}$.
4. Add the signed numerators. The sum is $\frac{-77+115}{210}=\frac{38}{210}$.
5. Reduce and state the result. Dividing numerator and denominator by $2$ gives the reduced result $\frac{19}{105}$.

Takeaway: Carry the negative sign into the numerator during conversion so that signed numerator addition remains consistent.

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 4.72

Solution

1. Find the LCD. Since $5$ and $8$ are relatively prime, the LCD is $5\cdot8=40$.
2. Convert the first fraction. Multiply $\frac{3}{5}$ by $\frac{8}{8}$ to obtain $\frac{24}{40}$.
3. Convert the second fraction. Multiply $\frac{x}{8}$ by $\frac{5}{5}$ to obtain $\frac{5x}{40}$.
4. Combine the numerators. Because the denominators now agree, the sum is $\frac{24+5x}{40}$.
5. Check for simplification. The terms $24$ and $5x$ are not like terms, and the entire fraction cannot be reduced, so the final result is $\frac{24+5x}{40}$.

Takeaway: A variable and a constant in the same numerator cannot be combined unless the variable has a factor that makes them like terms.

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

Example 4.73

Solution

1. Identify part a. The first expression uses addition, so write both fractions with a common denominator before combining their numerators.
2. Simplify part a. Since the LCD of $4$ and $6$ is $12$, the sum becomes $-\frac{3}{12}+\frac{2}{12}=-\frac{1}{12}$.
3. Identify part b. The second expression uses division, so multiply the first fraction by the reciprocal of the second rather than finding a common denominator.
4. Simplify part b. Rewrite the expression as $-\frac{1}{4}\cdot\frac{6}{1}=-\frac{6}{4}$.
5. Reduce and state both results. Dividing numerator and denominator by $2$ gives $-\frac{3}{2}$ for part b, while part a equals $-\frac{1}{12}$.

Takeaway: Identify the operation before simplifying: division by a fraction uses a reciprocal, giving $-\frac{1}{4}\div\frac{1}{6}=-\frac{3}{2}$.

Example 4.74

Solution

1. Identify part a. The first expression uses subtraction, so rewrite both fractions with their least common denominator, which is $30$.
2. Simplify part a. Converting gives $\frac{25x}{30}-\frac{9}{30}$, and subtracting numerators produces $\frac{25x-9}{30}$.
3. Identify part b. The second expression uses multiplication, so no common denominator is needed and each numerator is multiplied by the other numerator.
4. Simplify part b. The product is $\frac{15x}{60}$, and canceling the common factors $5$ and $3$ gives $\frac{x}{4}$.
5. State both results. Part a is $\frac{25x-9}{30}$, while part b is $\frac{x}{4}$ after full reduction.

Takeaway: Subtraction with unlike denominators requires the LCD $30$ and gives $\frac{25x-9}{30}$, while multiplication needs no common denominator and gives $\frac{x}{4}$.

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,

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.

Example 4.75

Solution

1. Simplify the numerator. The exponent applies only to $\frac{1}{2}$, so $\left(\frac{1}{2}\right)^2=\frac{1}{4}$.
2. Simplify the denominator. The exponent term becomes $3^2=9$, so the denominator is $4+9=13$.
3. Rewrite the division. The complex fraction is now $\frac{1}{4}\div13$, which becomes $\frac{1}{4}\cdot\frac{1}{13}$ when multiplying by the reciprocal.
4. State the result. Multiplying the numerators and denominators gives $\frac{1}{52}$, already in lowest terms.

Takeaway: Simplify the numerator, exponent, and denominator separately before dividing, which gives $\frac{1}{52}$.

Example 4.76

Solution

1. Simplify the numerator. The LCD of $2$ and $3$ is $6$, so $\frac{1}{2}+\frac{2}{3}=\frac{3}{6}+\frac{4}{6}=\frac{7}{6}$.
2. Simplify the denominator. The LCD of $4$ and $6$ is $12$, so $\frac{3}{4}-\frac{1}{6}=\frac{9}{12}-\frac{2}{12}=\frac{7}{12}$.
3. Divide the simplified parts. The complex fraction becomes $\frac{7}{6}\div\frac{7}{12}$, which is rewritten as $\frac{7}{6}\cdot\frac{12}{7}$.
4. Cancel and state the result. Canceling the common factors $7$ and $6$ gives $2$, so the original complex fraction simplifies to $2$.

Takeaway: Simplify the numerator and denominator of a complex fraction independently before dividing, yielding $2$.

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 4.77

Solution

1. Evaluate part a. Substitute $x=-\frac{1}{3}$ to obtain $-\frac{1}{3}+\frac{1}{3}$; the two terms are additive inverses, so their sum is $0$.
2. Set up part b. Substitute $x=-\frac{3}{4}$ to obtain $-\frac{3}{4}+\frac{1}{3}$, which requires a common denominator.
3. Find the LCD. The LCD of $4$ and $3$ is $12$, so the expression becomes $-\frac{9}{12}+\frac{4}{12}$.
4. Add and state both results. The signed numerators give $-\frac{5}{12}$, so the values are $0$ for part a and $-\frac{5}{12}$ for part b.

Takeaway: Substitute each value for $x$; the results are $0$ when $x=-\frac{1}{3}$ and $-\frac{5}{12}$ when $x=-\frac{3}{4}$.

Example 4.78

Solution

1. Substitute the value. Replace $y$ with $-\frac{2}{3}$ to obtain $-\frac{2}{3}-\frac{5}{6}$, as indicated by the substitution figure.
2. Find the LCD. The denominators $3$ and $6$ have LCD $6$, so $-\frac{2}{3}=-\frac{4}{6}$.
3. Subtract the numerators. Then $-\frac{4}{6}-\frac{5}{6}=\frac{-4-5}{6}=-\frac{9}{6}$.
4. Reduce and state the result. Divide numerator and denominator by $3$ to obtain the reduced value $-\frac{3}{2}$.

Takeaway: After substitution, preserve the negative sign during conversion and reduction to avoid changing the value of the expression.

Example 4.79

Solution

1. Substitute the values. Replace $x$ with $\frac{1}{4}$ and $y$ with $-\frac{2}{3}$ in $2x^2y$, giving $2\left(\frac{1}{4}\right)^2\left(-\frac{2}{3}\right)$ as the figures show.
2. Evaluate the square. The square applies only to $x$, so $\left(\frac{1}{4}\right)^2=\frac{1}{16}$.
3. Multiply the factors. The product is $2\cdot\frac{1}{16}\cdot\left(-\frac{2}{3}\right)=-\frac{4}{48}$.
4. Reduce the fraction. Divide numerator and denominator by their common factor $4$.
5. State the result. The reduced value is $-\frac{1}{12}$.

Takeaway: An exponent applies only to the base immediately before it, and every common factor must be removed from the final product.

Example 4.80

Solution

1. Substitute the values. Replace $p$ with $-4$, $q$ with $-2$, and $r$ with $8$ to get $\frac{-4+(-2)}{8}$, as shown in the figures.
2. Simplify the numerator. The sum $-4+(-2)=-6$, so the expression becomes $\frac{-6}{8}$.
3. Reduce the fraction. Divide numerator and denominator by their greatest common factor $2$.
4. State the result. The reduced value is $-\frac{3}{4}$.

Takeaway: Substitute every variable before simplifying a quotient and keep parentheses around a sum in the numerator.

Section 4.5 Exercises

Section 4.5 Exercises

Find the Least Common Denominator (LCD)

In the following exercises, find the least common denominator (LCD) for each set of fractions.

Convert Fractions to Equivalent Fractions with the LCD

In the following exercises, convert to equivalent fractions using the LCD.

Add and Subtract Fractions with Different Denominators

In the following exercises, add or subtract. Write the result in simplified form.

Identify and Use Fraction Operations

In the following exercises, perform the indicated operations. Write your answers in simplified form.

Use the Order of Operations to Simplify Complex Fractions

In the following exercises, simplify.

Mixed Practice

In the following exercises, simplify.

In the following exercises, evaluate the given expression. Express your answers in simplified form, using improper fractions if necessary.

Everyday Math

Writing Exercises

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well prepared for the next section? Why or why not?

1. Change to equivalent fractions with the LCD:

$\frac{3}{4}$ and $\frac{5}{6},$ LCD $=12$

2. Change to equivalent fractions with the LCD:

$-\frac{7}{12}$ and $\frac{11}{15},$ LCD $=60$

3. Change to equivalent fractions with the LCD:

$\frac{13}{24}$ and $\frac{17}{32},$ LCD $96$

4. Change to equivalent fractions with the LCD:

$\frac{9}{28}$ and $\frac{27}{32},$ LCD $224$

5. Add: $\frac{1}{4}+\frac{1}{3}.$

6. Add: $\frac{1}{2}+\frac{1}{5}.$

1. Simplify: $\frac{1}{2}-(-\frac{1}{8}).$

2. Simplify: $\frac{1}{3}-(-\frac{1}{6}).$

3. Add: $\frac{7}{12}+\frac{11}{15}.$

4. Add: $\frac{13}{15}+\frac{17}{20}.$

5. Subtract: $\frac{13}{24}-\frac{17}{32}.$

6. Subtract: $\frac{21}{32}-\frac{9}{28}.$

1. Add: $-\frac{13}{42}+\frac{17}{35}.$

2. Add: $-\frac{19}{24}+\frac{17}{32}.$

3. Add: $\frac{y}{6}+\frac{7}{9}.$

4. Add: $\frac{x}{6}+\frac{7}{15}.$

5. Simplify each expression:

1. ⓐ $-\frac{3}{4}-\frac{1}{6}$

2. ⓑ $-\frac{3}{4}\cdot\frac{1}{6}$

6. Simplify each expression:

1. ⓐ $\frac{5}{6}\mathrm{÷}(-\frac{1}{4})$

2. ⓑ $\frac{5}{6}-(-\frac{1}{4})$

1. Simplify:

1. ⓐ $\frac{(27a-32)}{36}$

2. ⓑ $\frac{2a}{3}$

2. Simplify:

1. ⓐ $\frac{(24k+25)}{30}$

2. ⓑ $\frac{24k}{5}$

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

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

5. Evaluate: $x+\frac{3}{4}$ when

1. ⓐ $x=-\frac{7}{4}$

2. ⓑ $x=-\frac{5}{4}$

6. Evaluate: $y+\frac{1}{2}$ when

1. ⓐ $y=\frac{2}{3}$

2. ⓑ $y=-\frac{3}{4}$

1. Evaluate: $y-\frac{1}{2}$ when $y=-\frac{1}{4}.$

2. Evaluate: $x-\frac{3}{8}$ when $x=-\frac{5}{2}.$

3. Evaluate. $3a{b}^{2}$ when $a=-\frac{2}{3}$ and $b=-\frac{1}{2}.$

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

5. Evaluate: $\frac{a+b}{c}$ when $a=-8,b=-7,$ and $\,c=6.$

6. Evaluate: $\frac{x+y}{z}$ when $x=9,y=-18,$ and $z=-6.$

1. $\frac{1}{4}$ and $\frac{1}{5},$ LCD $=20$

2. $\frac{7}{12}$ and $\frac{5}{8},$ LCD $=24$

3. $\frac{11}{16}$ and $-\frac{5}{12},$ LCD $=48$

4. $\frac{1}{3},\frac{3}{4},$ and $\frac{3}{5},$ LCD $=60$

5. $\frac{1}{4}+\frac{1}{5}$

6. $\frac{1}{3}+\frac{1}{8}$

1. $\frac{1}{4}-(-\frac{1}{8})$

2. $\frac{1}{2}-(-\frac{1}{6})$

3. $\frac{3}{4}+\frac{2}{5}$

4. $\frac{5}{12}+\frac{3}{8}$

5. $\frac{7}{16}-\frac{5}{12}$

6. $\frac{5}{8}-\frac{7}{12}$

1. $\frac{5}{6}-\frac{3}{4}$

2. $-\frac{9}{20}+\frac{17}{30}$

3. $-\frac{23}{30}+\frac{5}{48}$

4. $-\frac{33}{49}-\frac{18}{35}$

5. $-\frac{3}{4}-(-\frac{4}{5})$

6. $-\frac{7}{20}-(-\frac{5}{8})$

1. $1+\frac{5}{6}$

2. $1-\frac{3}{10}$

3. $\frac{y}{2}+\frac{2}{3}$

4. $\frac{x}{5}-\frac{1}{4}$

5. 1. ⓐ $\frac{2}{3}+\frac{1}{6}$

2. ⓑ $\frac{2}{3}\mathrm{÷}\frac{1}{6}$

6. 1. ⓐ $-\frac{4}{5}-\frac{1}{8}$

2. ⓑ $-\frac{4}{5}\cdot\frac{1}{8}$

1. 1. ⓐ $\frac{3a}{8}\mathrm{÷}\frac{7}{12}$

2. ⓑ $\frac{3a}{8}-\frac{7}{12}$

2. 1. ⓐ $\frac{4}{15}\cdot(-\frac{5q}{9})$

2. ⓑ $\frac{4}{15}+(-\frac{5q}{9})$

3. $-\frac{5}{12}\mathrm{÷}(-\frac{5}{9})$

4. $-\frac{1}{8}+\frac{7}{12}$

5. $\frac{5}{9}-\frac{1}{6}$

6. $\frac{7}{12}\cdot(-\frac{8}{35})$

1. $-\frac{3}{8}-\frac{x}{11}$

2. $\frac{10y}{13}\cdot\frac{8}{15y}$

3. $\frac{{(\frac{1}{3})}^{2}}{5+{2}^{2}}$

4. $\frac{{(\frac{3}{4})}^{2}}{{(\frac{5}{8})}^{2}}$

5. $\frac{5}{\frac{1}{4}+\frac{1}{3}}$

6. $\frac{\frac{3}{4}+\frac{1}{2}}{\frac{5}{6}-\frac{2}{3}}$

1. $\frac{\frac{3}{4}-\frac{3}{5}}{\frac{1}{4}+\frac{2}{5}}$

2. $\frac{1}{3}+\frac{2}{5}\cdot\frac{3}{4}$

3. $1-\frac{5}{6}\mathrm{÷}\frac{1}{12}$

4. $\frac{2}{3}+\frac{1}{4}+\frac{3}{5}$

5. $\frac{2}{5}+\frac{5}{8}-\frac{3}{4}$

6. $8(\frac{15}{16}-\frac{5}{6})$

1. $(\frac{3}{4}+\frac{1}{6})\mathrm{÷}(\frac{5}{8}-\frac{1}{3})$

2. $x+\frac{2}{3}$ when

1. ⓐ $x=-\frac{1}{6}$

2. ⓑ $x=-\frac{5}{3}$

3. $x+(-\frac{11}{12})$ when

1. ⓐ $x=\frac{11}{12}$

2. ⓑ $x=\frac{3}{4}$

4. $x-\frac{2}{5}$ when

1. ⓐ $x=\frac{3}{5}$

2. ⓑ $x=-\frac{3}{5}$

5. $x-\frac{1}{3}$ when

1. ⓐ $x=\frac{2}{3}$

2. ⓑ $x=-\frac{2}{3}$

6. $\frac{5}{12}-w$ when

1. ⓐ $w=\frac{1}{4}$

2. ⓑ $w=-\frac{1}{4}$

1. $5{m}^{2}n$ when $m=-\frac{2}{5}$ and $n=\frac{1}{3}$

2. $8{u}^{2}{v}^{3}$ when $\,u=-\frac{3}{4}$ and $v=-\frac{1}{2}$

3. $\frac{m+n}{p}$ when $m=-6,n=-2,p=4$

4. Baking Vanessa is baking chocolate chip cookies and oatmeal cookies. She needs $1\frac{1}{4}$ cups of sugar for the chocolate chip cookies, and $1\frac{1}{8}$ cups for the oatmeal cookies How much sugar does she need altogether?

5. Explain how to find the LCD of two fractions.