How to add complex fractions - SSAT Upper Level Quantitative

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Question

Add:

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

Answer

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 $\frac{5}{6}$ :

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

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Question

Solve.

$\frac{6}{7x}+\frac{3}{2x^{2}}$

Answer

Find the least common denominator between the two fractions. In our case both and go into $14x^{2}$ thus, this is our common denominator.

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

$14x^{2} \rightarrow \frac{6}{7x}\cdot \frac{2x}{2x}=\frac{12x}{14x^2}$ .

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

$\frac{3}{2x^2}\cdot \frac{7}{7}=\frac{21}{14x^2}$

To solve we need to add these two fractions together.

Final answer should read,

$\frac{12x}{14x^2}+\frac{21}{14x^2}=\frac{12x+21}{14x^2}$ .

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Question

Solve.

$\frac{6}{x^{2}+6x+9}+\frac{x-2}{x+3}$

Answer

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.

$\frac{6}{x^2+6x+9}+\frac{(x-2)(x+3)}{(x+3)(x+3)}$

$\frac{6}{x^2+6x+9}+\frac{x^2+3x-2x-6}{x^2+6x+9}$

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

After simplification, the answer is revealed.

$\frac{6+x^2+x-6}{x^2+6x+9}=\frac{x^2+x}{x^2+6x+9}$

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Question

Simplify.

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

Answer

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 $\frac{2x^{2}-4x+2}{-x^2+2x-1}$ 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 .

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Question

Solve.

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

Answer

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.

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

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Question

Solve.

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

Answer

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.

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Question

Simplify.

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

Answer

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.

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

$\frac{x^2+4x+3}{x^2+5x+6}+\frac{x^2+4x+4}{x^2+5x+6}$

$\frac{2x^2+8x+7}{x^2+5x+6}$

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Question

Simplify.

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

Answer

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 $\frac{5}{x}$ .

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Question

Simplify.

$\frac{x^2+5x+6}{x^2+9x+14}+\frac{x^2+6x+5}{x^2+7x+6}$

Answer

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

$\frac{x^2+6x+3x+18+x^2+7x+5x+35}{x^2+13x+42}$ .

After simplification, answer should be $\frac{2x^2+21x+53}{x^2+13x+42}$ .

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Question

Simplify.

$\frac{2x^2+2x-4}{x^2+2x-3}+\frac{x^2-3x-10}{2x^2-8x-10}$

Answer

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,

$\frac{4(x^2+x+2x+2)+x^2+2x+3x+6}{2x^2+8x+6}$ .

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

$\frac{5x^2+17x+14}{2x^2+8x+6}$

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Question

Simplify.

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

Answer

Least common denominator is $x^{3}$ . Just multiply the left fraction numerator by and solve.

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

Therefore the answer is $\frac{2x+1}{x^3}$ .

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Question

Simplify.

$\frac{x^2-5x}{x^2-25}+\frac{x^2+x-2}{2x^2+3x-2}$

Answer

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

$\frac{2x^2-x+x^2-x+5x-5}{2x^2+9x-5}$

Once simplified, the answer will be shown.

$\frac{3x^2+3x-5}{2x^2+9x-5}$

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Question

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

Answer

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.

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

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Question

Simplify.

$\frac{x^2-5x}{x^2-25}+\frac{x^2+x-2}{2x^2+3x-2}$

Answer

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

$\frac{2x^2-x+x^2-x+5x-5}{2x^2+9x-5}$

Once simplified, the answer will be shown.

$\frac{3x^2+3x-5}{2x^2+9x-5}$

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Question

Add:

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

Answer

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 $\frac{5}{6}$ :

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

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Question

Solve.

$\frac{6}{7x}+\frac{3}{2x^{2}}$

Answer

Find the least common denominator between the two fractions. In our case both and go into $14x^{2}$ thus, this is our common denominator.

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

$14x^{2} \rightarrow \frac{6}{7x}\cdot \frac{2x}{2x}=\frac{12x}{14x^2}$ .

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

$\frac{3}{2x^2}\cdot \frac{7}{7}=\frac{21}{14x^2}$

To solve we need to add these two fractions together.

Final answer should read,

$\frac{12x}{14x^2}+\frac{21}{14x^2}=\frac{12x+21}{14x^2}$ .

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Question

Solve.

$\frac{6}{x^{2}+6x+9}+\frac{x-2}{x+3}$

Answer

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.

$\frac{6}{x^2+6x+9}+\frac{(x-2)(x+3)}{(x+3)(x+3)}$

$\frac{6}{x^2+6x+9}+\frac{x^2+3x-2x-6}{x^2+6x+9}$

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

After simplification, the answer is revealed.

$\frac{6+x^2+x-6}{x^2+6x+9}=\frac{x^2+x}{x^2+6x+9}$

Compare your answer with the correct one above

Question

Simplify.

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

Answer

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 $\frac{2x^{2}-4x+2}{-x^2+2x-1}$ 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 .

Compare your answer with the correct one above

Question

Solve.

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

Answer

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.

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

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Question

Solve.

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

Answer

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.

Compare your answer with the correct one above

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