How to add complex fractions - SSAT Upper Level Quantitative

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Simplify.

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Try to simplify the fractions rather than trying to find the least common denominator. It should look like this:

$\frac{x(x-5)}{(x-5)(x+5)}$+\frac{(x-1)(x+2)}{(2x-1)(x+2)}$ .

Remember, to break down the quadratic equation by finding the binomials, two numbers that are factors of c must add up to make b. The may not be easy to factor but when put into the quadratic formula

$\left(-b\pm $$\frac{\sqrt{b^2$-4ac}$}{2a}\right)$ , we get roots of and $\frac{1}{2}$ . Then set equations equal to so that when solving for , the answer is the root.

Now after cancellations, we get $\frac{x}{x+5}$+\frac{x-1}{2x-1}$ .

The common denominator is or or .

Remember when foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables. For the left fraction numerator you multiply by what it's lacking and same with the right fraction numerator.

The overall fraction should look like

Once simplified, the answer will be shown.

Add:

$$\frac{\frac{1}{5}$+\frac{5}{6}$}{$\frac{1}{4}$+\frac{2}{5}$}+\frac{5}{6}$=?$

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First, add the fractions found in the numerator and denominator of the complex fraction. To add two fractions, they need to have a common denominator. You can create a common denominator by multiplying the numerator and denominator of one fraction by as a fraction made up of the denominator of the other fraction. For example, if you need to create a common denominator for $\frac{1}{3}$ and $\frac{1}{2}$ , you would multiply $\frac{1}{3}$ by $\frac{2}{2}$ and $\frac{1}{2}$ by $\frac{3}{3}$ . Since this just multiplies each fraction by , it doesn't change the fractions' values, just the way in which they are represented.

$\frac{1}{5}$+\frac{5}{6}$=($\frac{1}{5}$\cdot $\frac{6}{6}$)+($\frac{5}{6}$\cdot$\frac{5}{5}$)=\frac{6}{30}$+\frac{25}{30}$=\frac{31}{30}$

$\frac{1}{4}$+\frac{2}{5}$=($\frac{1}{4}$\cdot$\frac{5}{5}$)+($\frac{2}{5}$\cdot$\frac{4}{4}$)=\frac{5}{20}$+\frac{8}{20}$=\frac{13}{20}$

Now, divide those two fractions.

$\frac{31}{30}$\div$\frac{13}{20}$=\frac{31}{30}$\times$\frac{20}{13}$=\frac{62}{39}$

Finally, add $\frac{5}{6}$ :

$\frac{62}{39}$+\frac{5}{6}$=($\frac{62}{39}$\cdot $\frac{6}{6}$)+($\frac{5}{6}$\cdot$\frac{39}{39}$)=\frac{372}{234}$+\frac{195}{234}$=\frac{567}{234}$\div 3=\frac{189}{78}$\div 3=\frac{63}{26}$

Solve.

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Find the least common denominator between the two fractions. In our case both and go into thus, this is our common denominator.

The top is multiplied by 2x for the left fraction and to get:

.

For the right fraction, the top is multiplied by 7.

To solve we need to add these two fractions together.

Final answer should read,

.

Solve.

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By inspection, the denominator of the left fraction is a perfect square $\left ( x+3\right $)^2$$ . Since the right fraction already has a , just multiply top and bottom by so the denominators of both fraction are the same.

The top should read with a bottom of $\left ( x+3\right $)^2$$ .

After simplification, the answer is revealed.

Simplify.

$\frac{1-x}{x-1}$+\frac{x-1}{1-x}$

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If you can recognize that $\frac{1-x}{x-1}$ is -1 then the answer is quick.

Otherwise just find the least common denominator which is .

The numerator of the new fraction will be . It may be tempting to think should have been the right answer, however, the question does say simplify and yes this can be simplified. If you factor the out of the numerator, it's . The bottom does look very similar to the top only the signs are flipped. This means if you factor out a on the bottom, the quadratic equation should match and cancel out leaving you with the answer of .

Solve.

$\frac{\frac{1}{2}$+\frac{1}{7}$}{$\frac{1}{5}$+\frac{1}{8}$}+\frac{1}{3}$

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First focus on the big fraction specificall,y the numerator. We need to find a common denominator in the numerator in order to add the two fractions and simplify the expression.

The top should have a common denominator of .

$\frac{1}{2}$\cdot $\frac{7}{7}$+\frac{1}{7}$\cdot $\frac{2}{2}$=\frac{7}{14}$+\frac{2}{14}$

Therefore the top should become $\frac{9}{14}$ .

Now lets look at the denominator of the big fraction. We again need to find a common denominator for the two components. In this case it will be .

$\frac{1}{5}$\cdot $\frac{8}{8}$+\frac{1}{8}$\cdot $\frac{5}{5}$=\frac{8}{40}$+\frac{5}{40}$

Therefore the bottom should become $\frac{13}{40}$ .

Since we need to have a single fraction added to a $\frac{1}{3}$ , we need to multiply top and bottom by $\frac{40}{13}$ so we can add the fractions easier.

$\frac{\frac{9}{14}$}{$\frac{13}{40}$}=\frac{9}{14}$\cdot $\frac{40}{13}$=\frac{360}{182}$=\frac{180}{91}$

So far, the new equation to solve is $\frac{180}{91}$ $+\frac{1}{3}$$ .

Same thing, find the least common denominator and solve to arrive at the final answer.

$\frac{180}{91}$\cdot $\frac{3}{3}$+\frac{1}{3}$\cdot $\frac{91}{91}$=\frac{540}{273}$+\frac{91}{273}$=\frac{631}{273}$

Solve.

$1+\frac{1}{1+\frac{1}${x}}$

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Although intimidating, just focus on the inner fraction and work your way up.

Common denominator is so that should lead to

$1+\frac{1}{1\cdot \left(\frac{x}${x} \right)+\frac{1}{x}$} \rightarrow$ $1+\frac{1}{\frac{x+1}${x}}$ .

By multiplying everything by $\frac{x}{x+1}$ this should lead to $1+\frac{x}{x+1}$$ .

Common denominator is now and this should arrive at,

$1\cdot $\frac{x+1}{x+1}$+\frac{x}{x+1}$=\frac{x+1+x}{x+1}$=\frac{2x+1}{x+1}$$

which is the final answer when everything is simplified.

Simplify.

$$\frac{1}{1+\frac{1}${1+x}}+\frac{2}{2+\frac{2}${2+x}}$

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Work on each fraction and start from the bottom and work your way up.

For the left fraction, it should be

$$\frac{1}{\frac{x+2}${x+1}}$ since is the common denominator.

Then multiply by the inverse to get $\frac{x+1}{x+2}$ .

For the right, it's the same concept as the final fraction to get is $\frac{x+2}{x+3}$ .

Finally find the common denominator of these new complex fractions and solve to get the final answer.

$\frac{x+1}{x+2}$\cdot $\frac{x+3}{x+3}$+\frac{x+2}{x+3}$\cdot $\frac{x+2}{x+2}$

Simplify.

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

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By careful inspection, the fractions all reduce to $\frac{1}{x}$ .

$\frac{2}{2x}$=\frac{2\cdot 1}{2\cdot x}$=\frac{1}{x}$

$\frac{3}{3x}$=\frac{3\cdot 1}{3\cdot x}$=\frac{1}{x}$

$\frac{4}{4x}$=\frac{4\cdot 1}{4\cdot x}$=\frac{1}{x}$

$\frac{5}{5x}$=\frac{5\cdot 1}{5\cdot x}$=\frac{1}{x}$

All the fractions have values in the numerator and denominator that cancel out.

There are five of these so multiply $\frac{1}{x}$ by to get the final answer.

Answer should be $\frac{5}{x}$ .

Simplify.

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Don't try to find the least common denominator as it will take a lot of time and a definite mistake in arithmetic. Instead try to see if these quadratics can simplify.

Upon inspection, we get

$\frac{(x+2)(x+3)}{(x+2)(x+7)}$+\frac{(x+1)(x+5)}{(x+1)(x+6)}$ .

Remember, to break down the quadratic equation by finding the binomials, two numbers that are factors of c must add up to make b.

With the cancellations, we should get

$\frac{x+3}{x+7}$+\frac{x+5}{x+6}$ .

The common denominator will be

Remember when foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables. The left fraction numerator is multiplied by and the right fraction numerator is multiplied by .

Overall, the new fraction should read

.

After simplification, answer should be .

Simplify.

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Try to simplify the fraction rather than finding the least common denominator. By breaking the quadratic equation into its factors, it becomes:

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

Remember, to break down the quadratic equation by finding the binomials, two numbers that are factors of c must add up to make b. Also, if the values of a, b, and c in the quadratic equation can be reduced, then factor out that divisor to make the quadratic easier to factor.

Upon cancelling, we should get

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

The common denominator is $(x+3)\left ( 2x+2\right )$ or or . Remember when foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables. Then multiply the numerator of left fraction by and the numerator of right fraction by and the new fraction should look like,

.

Upon distributing the four and simplifying the fraction, answer is shown.

Simplify.

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Least common denominator is . Just multiply the left fraction numerator by and solve.

Therefore the answer is .

Add: $$\frac{\frac{2}{3}$}{$\frac{1}{8}$} + $\frac{3}{\frac{8}${3}}$

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Rewrite the fractions using a division sign.

$$\frac{\frac{2}{3}$}{$\frac{1}{8}$} + $\frac{3}{\frac{8}${3}}= \left($\frac{2}{3}$\div $\frac{1}{8}\right)+\left(3\div $\frac{8}{3}\right)$

To change the division sign to a multiplication sign, take the reciprocal of the second terms.

$\left($\frac{2}{3}$\div $\frac{1}{8}\right)+\left(3\div $\frac{8}{3}\right)=\left($\frac{2}{3}$\times 8\right)+\left(3\times $\frac{3}{8}\right)$

Simplify.

$\left($\frac{2}{3}$\times 8\right)+\left(3\times $\frac{3}{8}\right)= $\frac{16}{3}$+\frac{9}{8}$$

Find the LCD, and rewrite the fractions to add the numerators.

$\frac{16}{3}$+\frac{9}{8}$ = $\frac{128}{24}$+ $\frac{27}{24}$= $\frac{155}{24}$

Simplify.

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Try to simplify the fractions rather than trying to find the least common denominator. It should look like this:

$\frac{x(x-5)}{(x-5)(x+5)}$+\frac{(x-1)(x+2)}{(2x-1)(x+2)}$ .

Remember, to break down the quadratic equation by finding the binomials, two numbers that are factors of c must add up to make b. The may not be easy to factor but when put into the quadratic formula

$\left(-b\pm $$\frac{\sqrt{b^2$-4ac}$}{2a}\right)$ , we get roots of and $\frac{1}{2}$ . Then set equations equal to so that when solving for , the answer is the root.

Now after cancellations, we get $\frac{x}{x+5}$+\frac{x-1}{2x-1}$ .

The common denominator is or or .

Remember when foiling, you multiply the numbers/variables that first appear in each binomial, followed by multiplying the outer most numbers/variables, then multiplying the inner most numbers/variables and finally multiplying the last numbers/variables. For the left fraction numerator you multiply by what it's lacking and same with the right fraction numerator.

The overall fraction should look like

Once simplified, the answer will be shown.

Add:

$$\frac{\frac{1}{5}$+\frac{5}{6}$}{$\frac{1}{4}$+\frac{2}{5}$}+\frac{5}{6}$=?$

Tap to see back →

First, add the fractions found in the numerator and denominator of the complex fraction. To add two fractions, they need to have a common denominator. You can create a common denominator by multiplying the numerator and denominator of one fraction by as a fraction made up of the denominator of the other fraction. For example, if you need to create a common denominator for $\frac{1}{3}$ and $\frac{1}{2}$ , you would multiply $\frac{1}{3}$ by $\frac{2}{2}$ and $\frac{1}{2}$ by $\frac{3}{3}$ . Since this just multiplies each fraction by , it doesn't change the fractions' values, just the way in which they are represented.

$\frac{1}{5}$+\frac{5}{6}$=($\frac{1}{5}$\cdot $\frac{6}{6}$)+($\frac{5}{6}$\cdot$\frac{5}{5}$)=\frac{6}{30}$+\frac{25}{30}$=\frac{31}{30}$

$\frac{1}{4}$+\frac{2}{5}$=($\frac{1}{4}$\cdot$\frac{5}{5}$)+($\frac{2}{5}$\cdot$\frac{4}{4}$)=\frac{5}{20}$+\frac{8}{20}$=\frac{13}{20}$

Now, divide those two fractions.

$\frac{31}{30}$\div$\frac{13}{20}$=\frac{31}{30}$\times$\frac{20}{13}$=\frac{62}{39}$

Finally, add $\frac{5}{6}$ :

$\frac{62}{39}$+\frac{5}{6}$=($\frac{62}{39}$\cdot $\frac{6}{6}$)+($\frac{5}{6}$\cdot$\frac{39}{39}$)=\frac{372}{234}$+\frac{195}{234}$=\frac{567}{234}$\div 3=\frac{189}{78}$\div 3=\frac{63}{26}$

Solve.

Tap to see back →

Find the least common denominator between the two fractions. In our case both and go into thus, this is our common denominator.

The top is multiplied by 2x for the left fraction and to get:

.

For the right fraction, the top is multiplied by 7.

To solve we need to add these two fractions together.

Final answer should read,

.

Solve.

Tap to see back →

By inspection, the denominator of the left fraction is a perfect square $\left ( x+3\right $)^2$$ . Since the right fraction already has a , just multiply top and bottom by so the denominators of both fraction are the same.

The top should read with a bottom of $\left ( x+3\right $)^2$$ .

After simplification, the answer is revealed.

Simplify.

$\frac{1-x}{x-1}$+\frac{x-1}{1-x}$

Tap to see back →

If you can recognize that $\frac{1-x}{x-1}$ is -1 then the answer is quick.

Otherwise just find the least common denominator which is .

The numerator of the new fraction will be . It may be tempting to think should have been the right answer, however, the question does say simplify and yes this can be simplified. If you factor the out of the numerator, it's . The bottom does look very similar to the top only the signs are flipped. This means if you factor out a on the bottom, the quadratic equation should match and cancel out leaving you with the answer of .

Solve.

$\frac{\frac{1}{2}$+\frac{1}{7}$}{$\frac{1}{5}$+\frac{1}{8}$}+\frac{1}{3}$

Tap to see back →

First focus on the big fraction specificall,y the numerator. We need to find a common denominator in the numerator in order to add the two fractions and simplify the expression.

The top should have a common denominator of .

$\frac{1}{2}$\cdot $\frac{7}{7}$+\frac{1}{7}$\cdot $\frac{2}{2}$=\frac{7}{14}$+\frac{2}{14}$

Therefore the top should become $\frac{9}{14}$ .

Now lets look at the denominator of the big fraction. We again need to find a common denominator for the two components. In this case it will be .

$\frac{1}{5}$\cdot $\frac{8}{8}$+\frac{1}{8}$\cdot $\frac{5}{5}$=\frac{8}{40}$+\frac{5}{40}$

Therefore the bottom should become $\frac{13}{40}$ .

Since we need to have a single fraction added to a $\frac{1}{3}$ , we need to multiply top and bottom by $\frac{40}{13}$ so we can add the fractions easier.

$\frac{\frac{9}{14}$}{$\frac{13}{40}$}=\frac{9}{14}$\cdot $\frac{40}{13}$=\frac{360}{182}$=\frac{180}{91}$

So far, the new equation to solve is $\frac{180}{91}$ $+\frac{1}{3}$$ .

Same thing, find the least common denominator and solve to arrive at the final answer.

$\frac{180}{91}$\cdot $\frac{3}{3}$+\frac{1}{3}$\cdot $\frac{91}{91}$=\frac{540}{273}$+\frac{91}{273}$=\frac{631}{273}$

Solve.

$1+\frac{1}{1+\frac{1}${x}}$

Tap to see back →

Although intimidating, just focus on the inner fraction and work your way up.

Common denominator is so that should lead to

$1+\frac{1}{1\cdot \left(\frac{x}${x} \right)+\frac{1}{x}$} \rightarrow$ $1+\frac{1}{\frac{x+1}${x}}$ .

By multiplying everything by $\frac{x}{x+1}$ this should lead to $1+\frac{x}{x+1}$$ .

Common denominator is now and this should arrive at,

$1\cdot $\frac{x+1}{x+1}$+\frac{x}{x+1}$=\frac{x+1+x}{x+1}$=\frac{2x+1}{x+1}$$

which is the final answer when everything is simplified.