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SAT Math : Equations / Inequalities

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Example Questions

Example Question #31 : Inequalities

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

Possible Answers:

n² < 2n

|n² - 1| > 1

n² < n

16n² - 1 = 0

(n-1)² > n

Correct answer:

|n² - 1| > 1

Explanation:

N_part_1

N_part_2

N_part_3

N_part_4

N_part_5

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Example Question #15 : Inequalities

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

Possible Answers:

All of the answers choices are valid.

-3

-1

-4

-2

Correct answer:

-1

Explanation:

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

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Example Question #16 : Inequalities

Solve for x

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

Possible Answers:

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

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

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

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

Correct answer:

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

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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Example Question #17 : Inequalities

We have x^2-4<0 , find the solution set for this inequality.

Possible Answers:

x>-2

-2<x<2

x=0

x<2

x>2, x<-2

Correct answer:

-2<x<2

Explanation:

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

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Example Question #32 : Inequalities

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

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

Possible Answers:

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

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

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

None of the other answers are correct.

The rational expression is undefined.

Correct answer:

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

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:

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

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

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Example Question #33 : Inequalities

What value must take in order for the following expression to be greater than zero?

$\frac{3}{k}-\frac{5}{7}$

Possible Answers:

Correct answer:

Explanation:

is such that:

$\frac{3}{k}-\frac{5}{7}>0$

Add $\frac{5}{7}$ to each side of the inequality:

$\frac{3}{k}>\frac{5}{7}$

Multiply each side of the inequality by :

$\frac{21}{k}>5$

Multiply each side of the inequality by :

21>5k

Divide each side of the inequality by :

$k< \frac{21}{5}$

You can now change the fraction on the right side of the inequality to decimal form.

$k<\frac{21}{5}=\frac{20}{5}+\frac{1}{5}= 4.2$

The correct answer is , since k has to be less than 4.2 for the expression to be greater than zero.

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Example Question #155 : Algebra

Give the solution set of this inequality:

$|-4x+ 17| \ge 67$

Possible Answers:

$(-\infty, -21 ] \cup [12.5, \infty )$

The set of all real numbers

[-12.5, 21]

[-21, 12.5]

$(-\infty, -12.5 ] \cup [21, \infty )$

Correct answer:

$(-\infty, -12.5 ] \cup [21, \infty )$

Explanation:

The absolute value inequality

$|-4x+ 17| \ge 67$

can be rewritten as the compound inequality

$-4x+ 17 \le - 67$ or $-4x+ 17 \ge 67$

Solve each inequality separately, using the properties of inequality to isolate the variable on the left side:

$-4x+ 17 \le - 67$

Subtract 17 from both sides:

$-4x+ 17 - 17 \le - 67 - 17$

$-4x \le - 84$

Divide both sides by , switching the inequality symbol since you are dividing by a negative number:

$-4x \div (-4) \ge - 84 \div (-4)$

$x \ge 21$ ,

which in interval notation is $[21, \infty )$

The same steps are performed with the other inequality:

$-4x+ 17 \ge 67$

$-4x+ 17- 17 \ge 67 - 17$

$-4x \ge 50$

$-4x \div (-4) \le 50 \div (-4)$

$x \le -12.5$

which in interval notation is $(-\infty, -12.5 ]$ .

The correct response is the union of these two sets, which is

$(-\infty, -12.5 ] \cup [21, \infty )$ .

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Example Question #34 : Inequalities

Find the maximum value of f(x,y) , from the system of inequalities.

$y\geq 2$

$4x+5\geq y+3$

$1\leq x\leq5$

f(x,y)=3x+3y

Possible Answers:

Correct answer:

Explanation:

First step is to rewrite $4x+5\geq y+3$

$4x+5\geq y+3$

$4x+2\geq y$

Next step is to find the vertices of the bounded region. We do this by plugging in the x bounds into the equation. Don't forgot to set up the other x and y bounds, which are given pretty much.

The vertices are

(1,2)

(5,2)

(1, 7)

(5, 22)

Now we plug each coordinate into f(x,y) , and what the maximum value is.

(1,2)=3(1)+3(2)=9

(5,2)=3(5)+3(1)=21

(1, 7)=3(1)+3(7)=25

(5, 22)=3(5)+3(22)=81

So the maximum value is

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Example Question #35 : Inequalities

Each of the following is equivalent to

xy/z * (5(x + y)) EXCEPT:

Possible Answers:

5x² + y²/z

xy(5x + 5y)/z

5x²y + 5xy²/z

xy(5y + 5x)/z

Correct answer:

5x² + y²/z

Explanation:

Choice a is equivalent because we can say that technically we are multiplying two fractions together: (xy)/z and (5(x + y))/1. We multiply the numerators together and the denominators together and end up with xy (5x + 5y)/z. xy (5y + 5x)/z is also equivalent because it is only simplifying what is inside the parentheses and switching the order- the commutative property tells us this is still the same expression. 5x²y + 5xy²/z is equivalent as it is just a simplified version when the numerators are multiplied out. Choice 5x² + y²/z is not equivalent because it does not account for all the variables that were in the given expression and it does not use FOIL correctly.

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Example Question #36 : Inequalities

Let S be the set of numbers that contains all of values of x such that 2x + 4 < 8. Let T contain all of the values of x such that -2x +3 < 8. What is the sum of all of the integer values that belong to the intersection of S and T?

Possible Answers:

-7

-3

-2

Correct answer:

-2

Explanation:

First, we need to find all of the values that are in the set S, and then we need to find the values in T. Once we do this, we must find the numbers in the intersection of S and T, which means we must find the values contained in BOTH sets S and T.

S contains all of the values of x such that 2x + 4 < 8. We need to solve this inequality.

2x + 4 < 8

Subtract 4 from both sides.

2x < 4

Divide by 2.

x < 2

Thus, S contains all of the values of x that are less than (but not equal to) 2.

Now, we need to do the same thing to find the values contained in T.

-2x + 3 < 8

Subtract 3 from both sides.

-2x < 5

Divide both sides by -2. Remember, when multiplying or dividing an inequality by a negative number, we must switch the sign.

x > -5/2

Therefore, T contains all of the values of x that are greater than -5/2, or -2.5.

Next, we must find the values that are contained in both S and T. In order to be in both sets, these numbers must be less than 2, but also greater than -2.5. Thus, the intersection of S and T consists of all numbers between -2.5 and 2.

The question asks us to find the sum of the integers in the intersection of S and T. This means we must find all of the integers between -2.5 and 2.

The integers between -2.5 and 2 are the following: -2, -1, 0, and 1. We cannot include 2, because the values in S are LESS than but not equal to 2.

Lastly, we add up the values -2, -1, 0, and 1. The sum of these is -2.

The answer is -2.

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SAT Math : Equations / Inequalities

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #31 : Inequalities

Example Question #15 : Inequalities

Example Question #16 : Inequalities

Example Question #17 : Inequalities

Example Question #32 : Inequalities

Example Question #33 : Inequalities

Example Question #155 : Algebra

Example Question #34 : Inequalities

Example Question #35 : Inequalities

Example Question #36 : Inequalities

All SAT Math Resources

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