SAT Math - How to evaluate a fraction

Solve the following: $\frac{1}{3}-\frac{2}{9}$

$\frac{1}{9}$

$-\frac{1}{9}$

$-\frac{1}{3}$

$-\frac{5}{9}$

$\frac{7}{27}$

Explanation

In order to subtract the fractions, the denominator must be the same. The common denominator is 9. Rewrite the fractions.

$\frac{1}{3}-\frac{2}{9}=\frac{3}{9}-\frac{2}{9}= \frac{1}{9}$

Evaluate the following equation when and round your answer to the nearest hundredth.

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

Explanation

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

1. Plug in wherever there is an in the above equation.

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

2. Perform the above operations.

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

$\frac{125-50-15+5}{25+3}$

$\frac{65}{28}=2.32$

Evaluate: $\frac{x}{3}+\frac{11}{6}=6$

$\frac{25}{2}$

$\frac{11}{25}$

$\frac{6}{11}$

$\frac{2}{9}$

$\frac{51}{6}$

Explanation

Find the least common denominator, or LCD of $\frac{x}{3}+\frac{11}{6}=6$ is six.

Rewrite the equation with the correct denominator.

$\frac{2x}{6}+\frac{11}{6}=\frac{36}{6}$

Multiply by six on both sides of the equation and solve for .

$x=\frac{25}{2}$

Solve: $\frac{8}{3} + \frac{x}{2}=4$

$\frac{8}{3}$

$\frac{4}{3}$

$\frac{6}{5}$

$\frac{3}{16}$

Explanation

In order to solve $\frac{8}{3} + \frac{x}{2}=4$ , identify the least common denominator, or LCD. Multiply the uncommon denominators, and the LCD is 6.

Rewrite the equation.

$\frac{16}{6} + \frac{3x}{6}=\frac{24}{6}$

Multiply by six on both sides of the equation to cancel the denominators.

$x=\frac{8}{3}$

Solve Actmath_7_113_q10_1

–1

no solution

infinitely many solutions

Explanation

The common denominator of the left side is x(x–1). Multiplying the top and bottom of 1/x by (x–1) yields

Actmath_7_113_q10_2

Actmath_7_113_q10_3

Actmath_7_113_q10_4

Actmath_7_113_q10_5

Since this statement is true, there are infinitely many solutions.

If , , and , find the value of $\frac{2x+y}{z}$ .

$\frac{17}{5}$

$\frac{13}{5}$

$\frac{9}{2}$

$\frac{16}{7}$

$\frac{19}{5}$

Explanation

In order to solve $\frac{2x+y}{z}$ , we must first find the values of , , and using the initial equations provided. Starting with :

$x=\frac{55}{5}=11$

Then:

Finally:

With the values of , , and in hand, we can solve the final equation:

$\frac{2x+y}{z}$

$=\frac{2(11)+12}{10}$

$=\frac{22+12}{10}$

$=\frac{34}{10}$

$=\frac{17}{5}$

Simplify:

$\frac{\frac{81}{5}+\frac{7}{25}}{\frac{1}{9}-\frac{2}{7}}$

$-\frac{25956}{275}$

$\frac{8414}{41}$

$-\frac{551}{12}$

$-\frac{4532}{1575}$

$\frac{8417}{14}$

Explanation

Begin by simplifying the numerator and the denominator.

Numerator

$\frac{81}{5}+\frac{7}{25}$ has a common denominator of . Therefore, we have:

$\frac{405}{25}+\frac{7}{25} = \frac{412}{25}$

Denominator

$\frac{1}{9}-\frac{2}{7}$ has a common denominator of . Therefore, we have:

$\frac{7}{63}-\frac{18}{63}=-\frac{11}{63}$

Now, reconstructing our fraction, we have:

$\frac{\frac{412}{25}}{-\frac{11}{63}}$

To make this division work, you multiply the numerator by the reciprocal of the denominator:

$\frac{\frac{412}{25}}{-\frac{11}{63}} = \frac{412}{25} \cdot -\frac{63}{11} =-\frac{25956}{275}$

For this question, the following trigonometric identities apply:

$\csc x=\frac{1}{\sin x}$ ,

$\sec x=\frac{1}{\cos x}$

$\cot x=\frac{1}{\tan x}$

Simplify: $\sin ^{2}x\csc x\cot x$

$\cos x$

$\sin x\cos x$

$\sin x$

$\tan x$

$\sec x$

Explanation

To begin a problem like this, you must first convert everything to $\sin x$ and $\cos x$ alone. This way, you can begin to cancel and combine to its most simplified form.

Since $\csc x=\frac{1}{\sin x}$ and $\cot x=\frac{\cos x}{\sin x}$ , we insert those identities into the equation as follows.

$\sin ^{2}x\cdot \frac{1}{\sin x}\cdot \frac{\cos x}{\sin x}$

From here we combine the numerator and denominators of each fraction together to easily see what we can combine and cancel.

$\frac{\sin^{2}x\cdot \cos x}{\sin x\cdot \sin x}$

Since there is a $\sin ^{2}x$ in the numerator and the denominator, we can cancel them as they divide to equal 1. All we have left is $\cos x$ , the answer.