High School Math : Solving Inequalities

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Solving Inequalities

Solve for \(\displaystyle x\):

\(\displaystyle \frac{-7x-2}{3}>11\)

Possible Answers:

\(\displaystyle x>-5\)

\(\displaystyle x< -5\)

\(\displaystyle x>-3\)

\(\displaystyle x< 7\)

\(\displaystyle x>2\)

Correct answer:

\(\displaystyle x< -5\)

Explanation:

Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that \(\displaystyle >\) becomes \(\displaystyle < \), and vice versa.

\(\displaystyle \frac{-7x-2}{3}>11\)

\(\displaystyle -7x-2>33\)

\(\displaystyle -7x>35\)

\(\displaystyle \frac{-7x}{-7}< \frac{35}{-7}\)

\(\displaystyle x< -5\)

Example Question #2 : Solving Inequalities

Solve for \(\displaystyle x\):

\(\displaystyle x^2+4x>12\)

Possible Answers:

\(\displaystyle x< -2\ and\ x>6\)

\(\displaystyle x>2\)

\(\displaystyle x< -6\ and\ x>2\)

\(\displaystyle x< 6\)

\(\displaystyle x< -12\)

Correct answer:

\(\displaystyle x< -6\ and\ x>2\)

Explanation:

Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that \(\displaystyle >\) becomes \(\displaystyle < \), and vice versa. When we solve binomials, we must take extra caution because \(\displaystyle (-x)^2=x^2\).

So when we solve inequalities with binomials, we must create two scenarios: one where the value inside of the parentheses is positive and one where it is negative. For the negative scenario, we must flip the sign as we normally do for inequalities.

\(\displaystyle x^2+4x>12\)

\(\displaystyle x^2+4x+4>12+4\)

\(\displaystyle (x+2)^2>16\)

Now we must create our two scenarios:

\(\displaystyle x+2>{\sqrt{16}}\) and \(\displaystyle x+2< -{\sqrt{16}}\)

Notice that in the negative scenario, we flipped the sign of the inequality.

\(\displaystyle x+2>4\) and \(\displaystyle x+2< -4\)

\(\displaystyle x>2\) and \(\displaystyle x< -6\)

Example Question #3 : Solving Inequalities

Solve for \(\displaystyle x\):

\(\displaystyle x^2-10x< 24\)

Possible Answers:

\(\displaystyle x>-2\ and\ x< 12\)

\(\displaystyle x>0\ and\ x< 2\)

\(\displaystyle x>5\)

\(\displaystyle x< -12\)

\(\displaystyle x< 5\ and\ x>10\)

Correct answer:

\(\displaystyle x>-2\ and\ x< 12\)

Explanation:

Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that \(\displaystyle >\) becomes \(\displaystyle < \), and vice versa. When we solve binomials, we must take extra caution because \(\displaystyle (-x)^2=x^2\).

So when we solve inequalities with binomials, we must create two scenarios: one where the value inside of the parentheses is positive and one where it is negative. For the negative scenario, we must flip the sign as we normally do for inequalities.

\(\displaystyle x^2-10x< 24\)

\(\displaystyle x^2-10x+25< 24+25\)

\(\displaystyle (x-5)^2< 49\)

Now we must create our two scenarios:

\(\displaystyle x-5< {\sqrt{49}}\) and \(\displaystyle x-5>-{\sqrt{49}}\)

Notice that in the negative scenario, we flipped the sign of the inequality.

\(\displaystyle x-5< 7\) and \(\displaystyle x-5>-7\)

\(\displaystyle x< 12\) and \(\displaystyle x>-2\)

Example Question #241 : Algebra Ii

Solve the inequality for x:

\(\displaystyle 2x+4>64\)

Possible Answers:

\(\displaystyle x>28\)

\(\displaystyle x< 28\)

\(\displaystyle x< 30\)

\(\displaystyle x=30\)

\(\displaystyle x>30\)

Correct answer:

\(\displaystyle x>30\)

Explanation:

\(\displaystyle 2x+4>64\)

Subtract 4 from both sides:

\(\displaystyle 2x > 60\)

Divide both sides by 2:

\(\displaystyle x > 30\)

Example Question #51 : Basic Single Variable Algebra

Solve for \(\displaystyle x\).

\(\displaystyle -7x-4\geq10\)

Possible Answers:

\(\displaystyle x\geq-2\)

\(\displaystyle x\leq2\)

\(\displaystyle x\leq-2\)

\(\displaystyle x\geq2\)

Correct answer:

\(\displaystyle x\leq-2\)

Explanation:

\(\displaystyle -7x-4\geq10\)

Add 4 to both sides.

\(\displaystyle -7x\geq14\)

Divide both sides by –7. When dividing by a negative value, we must also change the direction of the inequality sign.

\(\displaystyle x\leq-2\)

Example Question #5 : Solving Inequalities

Solve for \(\displaystyle x\):

\(\displaystyle 9x-27\geq6x\)

Possible Answers:

\(\displaystyle x\leq9\)

\(\displaystyle x\geq3\)

\(\displaystyle x>9\)

\(\displaystyle x\geq9\)

\(\displaystyle x\leq3\)

Correct answer:

\(\displaystyle x\geq9\)

Explanation:

\(\displaystyle 9x-27\geq6x\)

Move like terms to the same sides:

\(\displaystyle 9x-6x\geq27\)

Combine like terms:

\(\displaystyle 3x\geq27\)

Divide both sides by 3:

\(\displaystyle x\geq9\)

Example Question #1 : Solving Inequalities

Solve for \(\displaystyle x\):

\(\displaystyle 5-3x\leq14\)

Possible Answers:

\(\displaystyle x\geq0\)

\(\displaystyle x\geq3\)

\(\displaystyle x\geq9\)

\(\displaystyle x\geq-3\)

\(\displaystyle x\leq3\)

Correct answer:

\(\displaystyle x\geq-3\)

Explanation:

Inequalities can be treated like any other equation except when multiplying and dividing by negative numbers. When multiplying or dividing by negative numbers, we just flip the sign of the inequality so that \(\displaystyle >\) becomes \(\displaystyle < \), and vice versa.

\(\displaystyle 5-3x\leq14\)

\(\displaystyle -3x\leq9\)

\(\displaystyle \frac{-3x}{-3}\geq\frac{9}{-3}\)

\(\displaystyle x\geq-3\)

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