Inequalities - GMAT Quantitative Reasoning

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Question

Find the domain of $y=\sqrt{x^{2}-4}$ .

Answer

We want to see what values of x satisfy the equation. $x^{2}-4$ is under a radical, so it must be positive.

$x^{2}-4\geq 0$

$x^{2}\geq 4$

$x\geq 2, x\leq -2$

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Question

How many integers $\dpi{100} \small (x)$ can complete this inequality?

$7< 2x-3 <15$

Answer

$7< 2x-3 <15$

3 is added to each side to isolate the $\dpi{100} \small x$ term:

$10< 2x <18$

Then each side is divided by 2 to find the range of $\dpi{100} \small x$ :

$5< x <9$

The only integers that are between 5 and 9 are 6, 7, and 8.

The answer is 3 integers.

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Question

Solve the inequality:

$-3x+7\geq13$

Answer

$-3x+7\geq13$

$-3x\geq6$

When multiplying or dividing by a negative number on both sides of an inequality, the direction of the inequality changes.

$x\leq-2$

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Question

Solve $5 < 3x + 10 \leq 16$ .

Answer

$5 < 3x + 10 \leq 16$

Subtract 10: $-5 < 3x \leq 6$

Divide by 3: $-5/3 < x \leq 2$

We must carefully check the endpoints. is greater than $-\frac{5}{3}$ and cannot equal $-\frac{5}{3}$ , yet CAN equal 2. Therefore $-\frac{5}{3}$ should have a parentheses around it, and 2 should have a bracket: is in

$\left ( -\frac{5}{3}, 2 \right ]$

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Question

Solve .

Answer

Subtract 3 from both sides:

Divide both sides by :

Remember: when dividing by a negative number, reverse the inequality sign!

Now we need to decide if our numbers should have parentheses or brackets. is strictly greater than , so should have a parentheses around it. Since there is no upper limit here, is in $(-2, \infty )$ .

Note: Infinity should ALWAYS have a parentheses around it, NEVER a bracket.

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Question

Solve $(x-1)^{2}(x+4)<0$ .

Answer

$(x-1)^{2}$ must be positive, except when . When , $(x-1)^{2}=0$ .

Then we know that the inequality is only satisfied when , and $x eq 1$ . Therefore , which in interval notation is $(-\infty , -4)$ .

Note: Infinity must always have parentheses, not brackets. has a parentheses around it instead of a bracket because is less than , not less than or equal to .

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Question

Solve .

Answer

The roots we need to look at are $x=0, x=1,\ and\ x=-1.$

:

Try

, so

does not satisfy the inequality.

:

Try $x=-\frac{1}{2}$

$-\frac{1}{2}\cdot (-\frac{1}{2}-1)(-\frac{1}{2}+1)=\frac{3}{8}>0$

so does satisfy the inequality.

:

Try $x=\frac{1}{2}$

$\frac{1}{2}\left (\frac{1}{2}-1 \right )\left ( \frac{1}{2}+1 \right )=-\frac{3}{8}<0$

so does not satisfy the inequality.

:

Try .

so satisfies the inequality.

Therefore the answer is and .

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Question

Find the solution set for :

Answer

Subtract 7:

Divide by -1. Don't forget to switch the direction of the inequality signs since we're dividing by a negative number:

$-3 \cdot (-1) > -x \cdot (-1) > 8\cdot (-1)$

Simplify:

or in interval form, .

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Question

Which of the following is equivalent to ?

Answer

To solve this problem we need to isolate our variable .

We do this by subtracting from both sides and subtracting from both sides as follows:

Now by dividing by 3 we get our solution.

or

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Question

How many integers satisfy the following inequality:

Answer

$\frac{9}{4}<m<\frac{9}{2}$

There are two integers between 2.25 and 4.5, which are 3 and 4.

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Question

What is the lowest value the integer can take?

Answer

$\frac{17}{3}<n<\frac{32}{3}$

The lowest value n can take is 6.

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Question

What value of will make the following expression negative:

$0.4(50-x)\times10+0.6x\times5$

Answer

Our first step is to simplify the expression. We need to remember our order of operations or PEMDAS.

First distribute the 0.4 to the binomial.

$0.4(50-x)\times10+0.6x\times5 < 0$

$(20-0.4x)\times 10 +0.6x \times 5<0$

Now distribute the 10 to the binomial.

$200-4x+0.6x \times 5<0$

Now multiply 0.6 by 5

Remember to flip the sign of the inequation when multiplying or dividing by a negative number.

220 will make the expression negative.

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Question

Solve for .

$3x+12\leq 48$

Answer

To solve this problem all we need to do is solve for :

$3x+12\leq48$

$3x\leq36$

$x\leq12$

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Question

What value of satisfies both of the following inequalities?

$9-3m\leq5$

$2m+7\leq13-m$

Answer

Solve for in both inequalities:

(1)

$9-3m\leq5$

Subtract 9 from each side of the inequality:

$-3m\leq-4$

Then, divide by . Remember to switch the direction of the sign from "less than or equal to" to "greater than or equal to."

$m\geq\frac{4}{3}$

(2)

$2m+7\leq13-m$

Add to both sides of the inequality:

$3m+7\leq13$

Subtract 7 from each side of the inequality:

$3m\leq6$

Divide each side of the inequality by 3:

$m\leq2$

From solving both inequalities, we find such that:

$\frac{4}{3}\leq m\leq 2$

$\frac{4}{3}\approx 1.33$

Only $\frac{3}{2}$ is in that interval $(\frac{3}{2}=1.5)$ .

$\frac{5}{4}$ and $\frac{2}{3}$ are less than 1.33, so they are not in the interval.

$\frac{7}{3}$ and $\frac{8}{3}$ are more than 2, so they are not in the interval.

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Question

Which of the following could be a value of , given the following inequality?

$8-7x\leq11x+4$

Answer

The inequality that is presented in the problem is:

$8-7x\leq11x+4$

Start by moving your variables to one side of the inequality and all other numbers to the other side:

$-7x-11x\leq4-8$

$-18x\leq-4$

Divide both sides of the equation by . Remember to flip the direction of the inequality's sign since you are dividing by a negative number!

$x\geq\frac{-4}{-18}$

Reduce:

$x\geq\frac{2}{9}$

The only answer choice with a value greater than $\frac{2}{9}$ is $\frac{4}{9}$ .

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Question

If and are two integers, which of the following inequalities would be true?

Answer

First let's solve each of the inequalities:

Don't forget to flip the direction of the sign when dividing by a negative number:

is the correct answer. is an integer greater than 3 and is greater than 9. Therefore, the sum of and is greater than 12.

is not true. and are two positive integers as is greater than 3 and is greater than 9. The sum of two positive integers cannot be a negative number.

$\frac{m}{n}<0$ is not true. and are two positive integers as is greater than 3 and is greater than 9. The division of two positive numbers is positive and therefore cannot be less than 0.

is not true. is greater than 3 and is greater than 9. The product of and cannot be less than 3.

is not true. and are positive. Therefore, the product of and is negative and cannot be greater than 0.

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Question

Give the solution set of the inequality

$x^{2} + 4x > 5x + 6$

Answer

To solve a quadratic inequality, move all expressions to the left first:

$x^{2} + 4x > 5x + 6$

$x^{2} + 4x- 5x - 6 > 5x + 6 - 5x - 6$

$x^{2} -x - 6 > 0$

The boundary points of the solution set will be the points at which:

: that is, , or

: that is, .

None of these values will be included in the solution set, since equality is not allowed by the inequality symbol.

Test the intervals

$(-\infty, -2) , (-2, 3), (3, \infty)$

by choosing a value in each interval and testing the truth of the inequality.

$(-\infty, -2)$ : Test

$x^{2} + 4x > 5x + 6$

$(-3)^{2} + 4 (-3) > 5(-3) + 6$

True; include the interval $(-\infty, -2)$ .

: Test

$x^{2} + 4x > 5x + 6$

$(0)^{2} + 4 (0) > 5(0) + 6$

False; exclude the interval .

$(3, \infty)$ : Test

$x^{2} + 4x > 5x + 6$

$(4)^{2} + 4 (4) > 5(4) + 6$

True; include the interval $(3, \infty)$ .

The solution set is $(-\infty, -2) \cup (3, \infty)$ .

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Question

If an integer satisfies both of the above inequalities, which of the following is true about ?

Answer

First, we solve both inequalities:

If satisfies both inequalities, then is greater than 5 AND is greater than 11. Therefore is greater than 11.

is the correct answer.

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Question

Solve the following inequality:

Answer

We start by simplifying our inequality like any other equation:

Now we must remember that when we divide by a negative, the inequality is flipped, so we obtain:

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Question

Solve the following inequality:

$\small 4x-7\geq13x+11$

Answer

Solving inequalities is very similar to solving equations, but we need to remember an important rule:

If we multiply or divide by a negative number, we must switch the direction of the inequality. So a "greater than" sign will become a "less than" sign and vice versa.

We are given

$4x-7\geq13x+11$

Start by moving the and the over:

$4x-7({\color{Blue} -13x})({\color{Red} +7})\geq13x+11({\color{Red} +7})({\color{Blue} -13x})$

Simplify to get the following:

$-9x\geq18$

Then, we will divide both sides of the equation by . Remember to switch the direction of the inequality sign!

$\frac{-9x}{-9}\geq\frac{18}{-9}$

$\small x\leq\frac{18}{-9}$

So,

$\small \small x\leq-2$

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