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SAT II Math II : Solving Inequalities

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Example Questions

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Example Question #31 : Single Variable Algebra

Solve $-7\leq2x-1<15$ .

Possible Answers:

$-3\leq x<8$

$-3\geq x >8$

$8\leq x$

-3<x

x= 8, -3

Correct answer:

$-3\leq x<8$

Explanation:

Remember, anything we do to one side of the inequality, we must also do to the other two sides. We can start by adding one to all three sides:

$-7\leq2x-1<15$

$-6\leq2x<16$

And now we divide each side by two:

$-3\leq x<8$

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Example Question #11 : Solving Inequalities

Solve $-9\leq -(x-5)<8$ .

Possible Answers:

14= x>3

$14\geq x>3$

-14<x

$0\leq x < 3$

$-14\leq x<3$

Correct answer:

$14\geq x>3$

Explanation:

The first thing we can do is distribute the negative sign in the middle term. Because we're not multiplying or dividing the entire inequality by a negative, we don't have the change the direction of the inequality signs:

$-9\leq -x+5<8$

Now we can subtract from the inequality:

$-14\leq -x<3$

And finally, we can multiply by . Note, this time we're multiplying the entire inequality by a negative, so we have to change the direction of the inequality signs:

$14\geq x>3$

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Example Question #31 : Single Variable Algebra

Solve $x+3 \leq 2(x-4)$ .

Possible Answers:

$11 \leq x$

$-11 \geq x$

$11 \geq x$

$-11 \leq x$

11 = x

Correct answer:

$11 \leq x$

Explanation:

We start by distributing the :

$x+3 \leq 2x-8$

Now we can add , and subtract an from each side of the equation:

$11 \leq x$

We didn't multiply or divide by a negative number, so we don't have to reverse the direction of the inequality sign.

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Example Question #31 : Single Variable Algebra

What is the solution set for x^2<5x+6 ?

Possible Answers:

x>-1

x>6

$-1\leq x \leq 6$

-1<x<6

Correct answer:

-1<x<6

Explanation:

Start by finding the roots of the equation by changing the inequality to an equal sign.

x^2=5x+6

x^2-5x-6=0

(x-6)(x+1)=0

Now, make a number line with the two roots:

Pick a number less than and plug it into the inequality to see if it holds.

For x=-2 ,

(-2)^2<5(-2)+6

4<-4 is clearly not true. The solution set cannot be x<-1 .

Next, pick a number between $-1\text{ and }6$ .

For x=0 ,

0^2<5(0)+6

0<6 is true so the solution set must include -1<x<6 .

Finally, pick a number greater than .

For x=7 ,

7^2<5(7)+6

49<41 is clearly not the so the solution set cannot be x>6 .

Thus, the solution set for this inequality is -1<x<6 .

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Example Question #35 : Single Variable Algebra

Give the solution set of the inequality:

$\frac{x^{2}+7x-18}{x-6} \ge 0$

Possible Answers:

$(- \infty, -2] \cup (6, 9]$

$(- \infty, -9] \cup (6, \infty)$

$\left [-9, 2 \right ] \cup (6, \infty)$

$\left [-2, 6 \right )\cup [9, \infty)$

$(- \infty, -9] \cup [2, 6)$

Correct answer:

$\left [-9, 2 \right ] \cup (6, \infty)$

Explanation:

First, find the zeroes of the numerator and the denominator. This will give the boundary points of the intervals to be tested.

$x^{2}+7x-18 = 0$

(x+9)(x-2) = 0

x= -9 ; x= 2

x-6 = 0

x= 6

Since the numerator may be equal to 0, x= -9 and x= 2 are included as solutions. However, since the denominator may not be equal to 0, x= 6 is excluded as a solution.

Now, test each of four intervals for inclusion in the solution set by substituting one test value from each:

$(-\infty, -9)$

Let's test x = -10 :

$\frac{x^{2}+7x-18}{x-6} \ge 0$

$\frac{(-10)^{2}+7(-10)-18}{(-10)-6} \ge 0$

$\frac{100-70-18}{-10-6} \ge 0$

$\frac{12}{-16} \ge 0$

$-\frac{3}{4} \ge 0$

This is false, so $(-\infty, -9)$ is excluded from the solution set.

(-9, 2)

Let's test x= 0 :

$\frac{x^{2}+7x-18}{x-6} \ge 0$

$\frac{0^{2}+7 (0)-18}{0-6} \ge 0$

$\frac{-18}{-6} \ge 0$

$3 \ge 0$

This is true, so (-9, 2) is included in the solution set.

(2, 6)

Let's test $x = 3\;$ :

$\frac{x^{2}+7x-18}{x-6} \ge 0$

$\frac{3^{2}+7(3)-18}{3-6} \ge 0$

$\frac{9+21-18}{3-6} \ge 0$

$\frac{12}{-3} \ge 0$

$-4 \ge 0$

This is false, so (2, 6) is excluded from the solution set.

$(6, \infty)$

Let's test x = 7 :

$\frac{7^{2}+7 (7)-18}{7-6} \ge 0$

$\frac{49+49-18}{7-6} \ge 0$

$\frac{ 80}{1} \ge 0$

$80 \ge 1$

This is true, so $(6, \infty)$ is included in the solution set.

The solution set is therefore $\left [-9, 2 \right ] \cup (6, \infty)$ .

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Example Question #11 : Solving Inequalities

Solve the inequality: 5x-6 <19

Possible Answers:

x<5

$x\geq5$

$x\leq5$

$x\leq\frac{13}{5}$

$x<\frac{13}{5}$

Correct answer:

x<5

Explanation:

Add 6 on both sides.

5x-6 +6<19+6

5x<25

Divide by five on both sides.

$\frac{5x}{5}<\frac{25}{5}$

The answer is: x<5

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Example Question #32 : Single Variable Algebra

Solve the inequality: $3x-6\leq27$

Possible Answers:

$x\geq 11$

x<11

$x\leq 7$

$x\leq 11$

$x\geq 7$

Correct answer:

$x\leq 11$

Explanation:

Add six on both sides.

$3x-6+6\leq27+6$

$3x\leq 33$

Divide by three on both sides.

$\frac{3x}{3}\leq \frac{33}{3}$

The answer is: $x\leq 11$

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Example Question #11 : Solving Inequalities

Solve the inequality: -3x+9>-15

Possible Answers:

x<8

x>2

x<2

x<-8

x<-2

Correct answer:

x<8

Explanation:

Subtract nine from both sides.

-3x+9-9>-15-9

-3x>-24

Divide by negative 3 on both sides. We will need to switch the sign.

$\frac{-3x}{-3}>\frac{-24}{-3}$

The answer is: x<8

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Example Question #12 : Solving Inequalities

Solve the inequality: $3x-7 \leq -14$

Possible Answers:

$x\leq \frac{7}{3}$

$x\leq7$

$x\geq -\frac{7}{3}$

$x\leq -7$

$x\leq -\frac{7}{3}$

Correct answer:

$x\leq -\frac{7}{3}$

Explanation:

Add seven on both sides.

$3x-7 +7\leq -14+7$

$3x\leq -7$

Divide by three on both sides.

$\frac{3x}{3}\leq \frac{-7}{3}$

The answer is: $x\leq -\frac{7}{3}$

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Example Question #43 : Single Variable Algebra

Solve: $-7 ( x + 9 ) \geq 37 + 3x$

Possible Answers:

$x \geq -10$

$x \leq-10$

$x \geq-7$

$-10 \leq x \leq 10$

$x \leq -10, x\geq10$

Correct answer:

$x \leq-10$

Explanation:

First, we distribute the through the equation:

$-7x - 63 \geq 37 + 3x$

Now, we collect and combine terms:

$10x \leq -100$

$x \leq-10$

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SAT II Math II : Solving Inequalities

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #31 : Single Variable Algebra

Example Question #11 : Solving Inequalities

Example Question #31 : Single Variable Algebra

Example Question #31 : Single Variable Algebra

Example Question #35 : Single Variable Algebra

Example Question #11 : Solving Inequalities

Example Question #32 : Single Variable Algebra

Example Question #11 : Solving Inequalities

Example Question #12 : Solving Inequalities

Example Question #43 : Single Variable Algebra

All SAT II Math II Resources

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