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

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.

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

3x+7 geq -2x+4

3x geq -2x-3

5x geq -3

xgeq -$\frac{3}{5}$

$(x-3)circ$\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}$

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 which is the same as .

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 .

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 which is the same as .

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.

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

3x+7 geq -2x+4

3x geq -2x-3

5x geq -3

xgeq -$\frac{3}{5}$

$(x-3)circ$\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}$

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 .

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 which is the same as .

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.