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SAT II Math I : Solving Inequalities

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Example Questions

Example Question #102 : Single Variable Algebra

Give the solution set of the inequality

$\frac{2x-5}{x-7} > 0$ $\displaystyle \frac{2x-5}{x-7} > 0$

Possible Answers:

$\left (-\infty, 2\frac{1}{2} \right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right )$

$\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$

$\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 2\frac{1}{2}, 7 \right ) \cup \left ( 7, \infty\right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 2\frac{1}{2}, 7 \right ) \cup \left ( 7, \infty\right )$

$\left ( 2\frac{1}{2}, 7 \right )$ $\displaystyle \left ( 2\frac{1}{2}, 7 \right )$

$\left ( 2\frac{1}{2}, 7 \right )\cup \left ( 7, \infty\right )$ $\displaystyle \left ( 2\frac{1}{2}, 7 \right )\cup \left ( 7, \infty\right )$

Correct answer:

$\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$

Explanation:

Two numbers of like sign have a positive quotient.

Therefore, $\frac{2x-5}{x-7} > 0$ $\displaystyle \frac{2x-5}{x-7} > 0$ has as its solution set the set of points at which 2x-5 $\displaystyle 2x-5$ and x- 7 $\displaystyle x- 7$ are both positive or both negative.

To find this set of points, we identify the zeroes of both expressions.

$\displaystyle 2x- 5 = 0$

2x= 5 $\displaystyle 2x= 5$

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

x-7 = 0 $\displaystyle x-7 = 0$

x = 7 $\displaystyle x = 7$

Since $\frac{2x-5}{x-7}$ $\displaystyle \frac{2x-5}{x-7}$ is nonzero we have to exclude $x = 2\frac{1}{2}$ $\displaystyle x = 2\frac{1}{2}$ ; x = 7 $\displaystyle x = 7$ is excluded anyway since it would bring about a denominator of zero. We choose one test point on each of the three intervals $\left (-\infty, 2\frac{1}{2} \right ) , \left ( 2\frac{1}{2}, 7 \right ) , \left ( 7, \infty\right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right ) , \left ( 2\frac{1}{2}, 7 \right ) , \left ( 7, \infty\right )$ and determine where the inequality is correct.

$\left (-\infty, 2\frac{1}{2} \right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right )$

Choose x = 0 $\displaystyle x = 0$ :

$\frac{2 \cdot 0-5}{0-7} = \frac{-5}{-7} = \frac{5}{7} > 0$ $\displaystyle \frac{2 \cdot 0-5}{0-7} = \frac{-5}{-7} = \frac{5}{7} > 0$ - True.

$\left ( 2\frac{1}{2}, 7 \right )$ $\displaystyle \left ( 2\frac{1}{2}, 7 \right )$

Choose x=3 $\displaystyle x=3$ :

$\frac{2 \cdot 3-5}{3-7} = \frac{6}{-4} =- \frac{3}{2} > 0$ $\displaystyle \frac{2 \cdot 3-5}{3-7} = \frac{6}{-4} =- \frac{3}{2} > 0$ - False.

$\left ( 7, \infty\right )$ $\displaystyle \left ( 7, \infty\right )$

Choose x = 8 $\displaystyle x = 8$ :

$\frac{2 \cdot 8-5}{8-7} = \frac{11}{1} =11> 0$ $\displaystyle \frac{2 \cdot 8-5}{8-7} = \frac{11}{1} =11> 0$ - True.

The solution set is $\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$ $\displaystyle \left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$

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Example Question #1 : Solving Inequalities

Solve for x.

$2x-7\geq 1$ $\displaystyle 2x-7\geq 1$

Possible Answers:

$x\geq -3$ $\displaystyle x\geq -3$

$x\leq 4$ $\displaystyle x\leq 4$

$x\geq 5$ $\displaystyle x\geq 5$

x> 4 $\displaystyle x> 4$

$x\geq 4$ $\displaystyle x\geq 4$

Correct answer:

$x\geq 4$ $\displaystyle x\geq 4$

Explanation:

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, add 7 to each side.

$2x-7+7\geq 1+7$ $\displaystyle 2x-7+7\geq 1+7$

$2x\geq 8$ $\displaystyle 2x\geq 8$

Now, divide both sides by 2.

$x\geq 4$ $\displaystyle x\geq 4$

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Example Question #104 : Single Variable Algebra

Solve for x.

$3x+2\geq -7$ $\displaystyle 3x+2\geq -7$

Possible Answers:

$x\geq -1$ $\displaystyle x\geq -1$

$x\geq -3$ $\displaystyle x\geq -3$

$x\leq -3$ $\displaystyle x\leq -3$

$x\geq 3$ $\displaystyle x\geq 3$

$x\geq 1$ $\displaystyle x\geq 1$

Correct answer:

$x\geq -3$ $\displaystyle x\geq -3$

Explanation:

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, subtract 2from each side.

$3x+2-2\geq -7-2$ $\displaystyle 3x+2-2\geq -7-2$

$3x\geq -9$ $\displaystyle 3x\geq -9$

Now, divide both sides by 2.

$x\geq -3$ $\displaystyle x\geq -3$

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Example Question #105 : Single Variable Algebra

Solve the following inequality:

3x+7<9 $\displaystyle 3x+7< 9$

Possible Answers:

x<-4 $\displaystyle x< -4$

x<7 $\displaystyle x< 7$

x<3 $\displaystyle x< 3$

x< 9 $\displaystyle x< 9$

$x< \frac{2}{3}$ $\displaystyle x< \frac{2}{3}$

Correct answer:

$x< \frac{2}{3}$ $\displaystyle x< \frac{2}{3}$

Explanation:

To solve for an inequality, you solve like you would for a single variable expression and get $\displaystyle x$ by itself.

First, subtract $\displaystyle 7$ from both sides to get,

3x<2 $\displaystyle 3x< 2$ .

Then divide both sides by $\displaystyle 3$ and your final answer will be,

$x< \frac{2}{3}$ $\displaystyle x< \frac{2}{3}$ .

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Example Question #101 : Single Variable Algebra

Solve the inequality: 3-(3-2x)<10 $\displaystyle 3-(3-2x)< 10$

Possible Answers:

x<5 $\displaystyle x< 5$

$x>-\frac{1}{5}$ $\displaystyle x>-\frac{1}{5}$

$x<\frac{1}{5}$ $\displaystyle x< \frac{1}{5}$

x<-5 $\displaystyle x< -5$

x>5 $\displaystyle x>5$

Correct answer:

x<5 $\displaystyle x< 5$

Explanation:

Simplify the left side.

3-3+2x<10 $\displaystyle 3-3+2x< 10$

The inequality becomes:

2x<10 $\displaystyle 2x< 10$

Divide by two on both sides.

$\frac{2}{2}x<\frac{10}{2}$ $\displaystyle \frac{2}{2}x< \frac{10}{2}$

The answer is: x<5 $\displaystyle x< 5$

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Example Question #106 : Single Variable Algebra

Solve the inequality: $2x+6\leq 9x-3$ $\displaystyle 2x+6\leq 9x-3$

Possible Answers:

$x\geq-\frac{3}{11}$ $\displaystyle x\geq-\frac{3}{11}$

$x\leq-\frac{3}{11}$ $\displaystyle x\leq-\frac{3}{11}$

$x\leq-\frac{9}{7}$ $\displaystyle x\leq-\frac{9}{7}$

$x\geq-\frac{9}{7}$ $\displaystyle x\geq-\frac{9}{7}$

$x\geq\frac{9}{7}$ $\displaystyle x\geq\frac{9}{7}$

Correct answer:

$x\geq\frac{9}{7}$ $\displaystyle x\geq\frac{9}{7}$

Explanation:

Subtract $\displaystyle 2x$ on both sides.

$2x+6-2x\leq 9x-3-2x$ $\displaystyle 2x+6-2x\leq 9x-3-2x$

$6\leq 7x-3$ $\displaystyle 6\leq 7x-3$

Add 3 on both sides.

$6+3\leq 7x-3+3$ $\displaystyle 6+3\leq 7x-3+3$

$9\leq 7x$ $\displaystyle 9\leq 7x$

Divide by 7 on both sides.

$\frac{9}{7}\leq \frac{7x}{7}$ $\displaystyle \frac{9}{7}\leq \frac{7x}{7}$

$\frac{9}{7}\leq x$ $\displaystyle \frac{9}{7}\leq x$

The answer is: $x\geq\frac{9}{7}$ $\displaystyle x\geq\frac{9}{7}$

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SAT II Math I : Solving Inequalities

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

All SAT II Math I Resources

Example Questions

Example Question #102 : Single Variable Algebra

Example Question #1 : Solving Inequalities

Example Question #104 : Single Variable Algebra

Example Question #105 : Single Variable Algebra

Example Question #101 : Single Variable Algebra

Example Question #106 : Single Variable Algebra

All SAT II Math I Resources

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