How to find the solution to an inequality with multiplication - GRE Quantitative Reasoning

Card 0 of 8

Question

Quantitative Comparison

Column A:

Column B: $\frac{5}{x}$

Answer

For quantitative comparison questions involving a shared variable between quantities, the best approach is to test a positive integer, a negative integer, and a fraction. Half of our work is eliminated, however, because the question stipulates that x > 0. We only need to check a positive integer and a positive fraction between 0 and 1. Plugging in 2, we see that quantity A is greater than quantity B. Checking 1/2, however, we find that quantity B is greater than quantity A. Thus the relationship cannot be determined.

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Question

If –1 < n < 1, all of the following could be true EXCEPT:

Answer

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Question

(√(8) / -x ) < 2. Which of the following values could be x?

Answer

The equation simplifies to x > -1.41. -1 is the answer.

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Question

Solve for x

$\small 3x+7 \geq -2x+4$

Answer

$\small 3x+7 \geq -2x+4$

$\small 3x \geq -2x-3$

$\small 5x \geq -3$

$\small x\geq -\frac{3}{5}$

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Question

Fill in the circle with either $<$ , $>$ , or $=$ symbols:

$(x-3)\circ\frac{x^2-9}{x+3}$ for $x\geq 3$ .

Answer

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

Let us simplify the second expression. We know that:

$(x^2-9)=(x+3)(x-3)$

So we can cancel out as follows:

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

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

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Question

We have , find the solution set for this inequality.

Answer

$x^2-4<0 \Rightarrow x^2<4 \Rightarrow -2<x<2$

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Question

Solve the inequality .

Answer

Start by simplifying the expression by distributing through the parentheses to .

Subtract from both sides to get .

Next subtract 9 from both sides to get . Then divide by 4 to get $-\frac{11}{4} > x$ which is the same as $x < -\frac{11}{4}$ .

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Question

Solve the inequality .

Answer

Start by simplifying each side of the inequality by distributing through the parentheses.

This gives us .

Add 6 to both sides to get .

Add to both sides to get .

Divide both sides by 13 to get $x < \frac{27}{13}$ .

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