Algebra - SAT Math

Card 0 of 9157

Question

Simplify x/2 – x/5

Answer

Simplifying this expression is similar to 1/2 – 1/5. The denominators are relatively prime (have no common factors) so the least common denominator (LCD) is 2 * 5 = 10. So the problem becomes 1/2 – 1/5 = 5/10 – 2/10 = 3/10.

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Question

If $\dpi{100} \small \frac{p}{6}$ is an integer, which of the following is a possible value of $\dpi{100} \small p$ ?

Answer

$\dpi{100} \small \frac{0}{6}=0$ , which is an integer (a number with no fraction or decimal part). All the other choices reduce to non-integers.

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Question

Simplify: $\frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}$

Answer

$\frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}=\frac{x^{2}}{3y^{3}z}$

First, let's simplify $\frac{4}{12}$ . The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore $\frac{4}{12}=\frac{1}{3}$ .

To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with $\frac{1}{3}x^{2}y^{-3}z^{-1}$ or $\frac{x^{2}}{3y^{3}z}$

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Question

Which of the following fractions is not equivalent to $\frac{6}{45}$ ?

Answer

Let us simplify $\frac{6}{45}$ :

$\frac{6}{45}=\frac{3\times 2}{3\times 15}=\frac{2}{15}$

We can get alternate forms of the same fraction by multiplying the denominator and the numerator by the same number:

$\frac{2\times 2}{15\times 2}=\frac{4}{30}$

$\frac{2\times 1.5}{15\times 1.5}=\frac{3}{22.5}$

Now let's look at $\frac{12}{89}$ :

$6 \times 2 = 12$ , but $45 \times 2 = 90$ .

Therefore, $\frac{12}{89}$ is the correct answer, as it is not equivalent to $\frac{6}{45}$ .

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Question

Which of the following is not equal to 32/24?

Answer

24/32 = 1.33

16/12 =1.33

224/168 =1.33

4/3 = 1.33

96/72 = 1.33

160/96 = 1.67

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Question

Find the root of

$y=\frac{x^2-\sqrt3x+\sqrt5x-\sqrt{15}}{x-\sqrt3}$

Answer

$y=\frac{x^2-\sqrt3x+\sqrt5x-\sqrt{15}}{x-\sqrt3}=\frac{(x-\sqrt3)(x+\sqrt5)}{x-\sqrt3}=x+\sqrt5$

The root occurs where . So we substitute 0 for .

$y=x+\sqrt{5}$

$0=x+\sqrt{5}$

This means that the root is at $x=-\sqrt5$ .

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Question

Simplify the fraction below:

$\frac{5}{125}$

Answer

The correct approach to solve this problem is to first write factors for the numerator and the denominator:

The highest common factor is 5. Therefore, you can divide the numerator and denominator by 5 in order to get a simplified fraction.

Thus the numerator becomes,

$\frac{5}{5}=1$ and the denominator becomes $\frac{125}{5}=25$ .

Therefore the final answer is $\frac{1}{25}$ .

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Question

Simplify: $\frac{32}{88}$

Answer

Find the common factors of the numerator and denominator. They both share factors of 2,4, and 8. For simplicity, factor out an 8 from both terms and simplify.

$\frac{32}{88}= \frac{8\times 4}{8\times 11}= \frac{4}{11}$

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Question

Simply the following fraction:

$\frac{\frac{a^2}{b}}{\frac{a}{b}}$

Answer

Remember that when you divide a fraction by a fraction, that is the same as multiplying the fraction in the numerator by the reciprocal of the fraction in the denominator.

In other words,

$\frac{\frac{a^2}{b}}{\frac{a}{b}} = \frac{a^2}{b}\cdot\frac{b}{a}=\frac{a^2b}{ab}$

Simplifying this final fraction gives us our correct answer, .

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Question

Solve for .

$\frac{\frac{x-2}{4^2}}{\frac{x^2-4}{2}}=1$

Answer

To solve for , simplify the fraction. In order to do this, recall that dividing by a fraction is the same as multiplying by its reciprocal. Therefore, rewrite the equation as follows.

$\frac{\frac{x-2}{4^2}}{\frac{x^2-4}{2}}=1\Rightarrow \frac{x-2}{4^2}\times \frac{2}{x^2-4}=1$

Now, simplify the first fraction by calculating four squared.

$\frac{x-2}{16}\times \frac{2}{x^2-4}=1$

From here, factor the denominator of the second fraction.

$\frac{x-2}{16}\times \frac{2}{(x-2)(x+2)}=1$

Next, factor the 16.

$\frac{x-2}{2(8)}\times \frac{2}{(x-2)(x+2)}=1$

From here, cancel out like terms that are in both the numerator and denominator. In this particular case that includes (x-2) and 2.

$\frac{1}{8(x+2)}=1$

Now, distribute the eight.

$\frac{1}{8x+16}=1$

Next, multiply both sides by the denominator.

$(8x+16)\times \frac{1}{8x+16}=1\times (8x+16)$

The (8x+16) cancels out and leaves the following equation.

Now to solve for perform opposite operations to move all numerical values to one side of the equation leaving by itself on the other side of the equation.

$\1=8x+16 \-15=8x \\\frac{-15}{8}=x$

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Question

Find the solution to the following equation if x = 3:

y = (4x2 - 2)/(9 - x2)

Answer

Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.

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Question

I. x = 0

II. x = –1

III. x = 1

Answer

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Question

Answer

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Question

$\small h\left ( x\right )=\frac{28}{x+4}$

$\small \textup{For which of the following values of},x,\textup{is the above function undefined?}$

Answer

A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.

$\small x+4=0$

$\small x=-4$

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Question

Solve:

Answer

First, distribute, making sure to watch for negatives.

Combine like terms.

Subtract 7x from both sides.

Add 18 on both sides and be careful adding integers.

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Question

Solve:

Answer

First, distribute the to the terms inside the parentheses.

Add 6x to both sides.

This is false for any value of . Thus, there is no solution.

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Question

Solve $\left | 3-4x\right |<0$ .

Answer

By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.

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Question

,

In the above graphic, approximately determine the x values where the graph is neither increasing or decreasing.

Answer

We need to find where the graph's slope is approximately zero. There is a straight line between the x values of , and . The other x values have a slope. So our final answer is .

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Question

Answer

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Question

In the equation below, , , and are non-zero numbers. What is the value of in terms of and ?

$\frac{1}{m^3}-\frac{1}{k^2}=\frac{1}{p}$

Answer

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