How to find the solution to an inequality with multiplication

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

Questions 1 - 8

1

Quantitative Comparison

Column A:

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

Quantity A is greater.

Quantity B is greater.

The quantities are equal.

The relationship cannot be determined from the information provided.

Explanation

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.

2

Solve the inequality .

$x < \frac{27}{13}$

$x > \frac{13}{27}$

$x < \frac{12}{7}$

$x > -\frac{11}{17}$

$x > -\frac{13}{27}$

Explanation

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 $x < \frac{27}{13}$ .

3

We have , find the solution set for this inequality.

Explanation

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

4

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

n2 < n

|n2 - 1| > 1

(n-1)2 > n

16n2 - 1 = 0

n2 < 2n

Explanation

N_part_1

N_part_2

N_part_3

N_part_4

N_part_5

5

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

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

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

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

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

None of the other answers are correct.

The rational expression is undefined.

Explanation

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

Let us simplify the second expression. We know that:

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

So we can cancel out as follows:

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

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

6

Solve the inequality .

$x < -\frac{11}{4}$

$x > -\frac{11}{4}$

$x < -\frac{4}{11}$

$x > -\frac{4}{11}$

Explanation

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 $x < -\frac{11}{4}$ .

7

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

All of the answers choices are valid.

-4

-3

-2

-1

Explanation

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

8

Solve for x

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

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

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

$\small x \leq \frac{3}{5}$

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

Explanation

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

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

$\small 5x \geq -3$

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

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