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High School Math : How to solve one-step equations with integers in pre-algebra

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Example Questions

Example Question #342 : High School Math

x+15=6 $\displaystyle x+15=6$

Solve for $\displaystyle x$ .

Possible Answers:

$\displaystyle -9$

$\displaystyle -21$

$\displaystyle 21$

$\frac{2}{5}$ $\displaystyle \frac{2}{5}$

$\displaystyle 9$

Correct answer:

$\displaystyle -9$

Explanation:

To solve x+15=6 $\displaystyle x+15=6$ , we need to subtract $\displaystyle 15$ from both sides.

$\displaystyle x+15-15=6-15$

x=6-15 $\displaystyle x=6-15$

x=-9 $\displaystyle x=-9$

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Example Question #343 : High School Math

x+5=34 $\displaystyle x+5=34$

Solve for $\displaystyle x$ .

Possible Answers:

$\displaystyle 29$

$\frac{34}{5}$ $\displaystyle \frac{34}{5}$

$\displaystyle 39$

170 $\displaystyle 170$

$\displaystyle 24$

Correct answer:

$\displaystyle 29$

Explanation:

To solve x+5=34 $\displaystyle x+5=34$ , subtract $\displaystyle 5$ from both sides.

$\displaystyle x+5-5=34-5$

x=34-5 $\displaystyle x=34-5$

x=29 $\displaystyle x=29$

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Example Question #344 : High School Math

Solve for when x+9=32 $\displaystyle x+9=32$

Possible Answers:

x=23 $\displaystyle x=23$

x=21 $\displaystyle x=21$

x=41 $\displaystyle x=41$

x=39 $\displaystyle x=39$

Correct answer:

x=23 $\displaystyle x=23$

Explanation:

To solve for , subtract $\displaystyle 9$ from both sides of the equation:

$\displaystyle x+9-9=32-9$

x=32-9 $\displaystyle x=32-9$

x=23 $\displaystyle x=23$

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Example Question #345 : High School Math

Solve for when $\frac{x}{16}=4$ $\displaystyle \frac{x}{16}=4$

Possible Answers:

x=4 $\displaystyle x=4$

x=64 $\displaystyle x=64$

x=48 $\displaystyle x=48$

x=8 $\displaystyle x=8$

Correct answer:

x=64 $\displaystyle x=64$

Explanation:

To solve for , multiply both sides of the equation by $\displaystyle 16$ :

$\frac{x}{16}\cdot 16=4\cdot 16$ $\displaystyle \frac{x}{16}\cdot 16=4\cdot 16$

$x=4\cdot 16$ $\displaystyle x=4\cdot 16$

x=64 $\displaystyle x=64$

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Example Question #346 : High School Math

Solve for when x-18=72 $\displaystyle x-18=72$

Possible Answers:

x=80 $\displaystyle x=80$

x=54 $\displaystyle x=54$

x=75 $\displaystyle x=75$

x=90 $\displaystyle x=90$

Correct answer:

x=90 $\displaystyle x=90$

Explanation:

To solve for , add $\displaystyle 18$ to both sides of the equation:

$\displaystyle x-18+18=72+18$

x=72+18 $\displaystyle x=72+18$

x=90 $\displaystyle x=90$

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Example Question #347 : High School Math

Solve for when 12x=156 $\displaystyle 12x=156$

Possible Answers:

x=13 $\displaystyle x=13$

x=3124 $\displaystyle x=3124$

x=181 $\displaystyle x=181$

x=81 $\displaystyle x=81$

Correct answer:

x=13 $\displaystyle x=13$

Explanation:

To solve for , divide each side of the equation by $\displaystyle 12$ :

$\frac{12x}{12}=\frac{156}{12}$ $\displaystyle \frac{12x}{12}=\frac{156}{12}$

$x=\frac{156}{12}$ $\displaystyle x=\frac{156}{12}$

x=13 $\displaystyle x=13$

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Example Question #197 : Pre Algebra

Solve for when 5x=90 $\displaystyle 5x=90$

Possible Answers:

x=35 $\displaystyle x=35$

x=15 $\displaystyle x=15$

x=18 $\displaystyle x=18$

x=80 $\displaystyle x=80$

Correct answer:

x=18 $\displaystyle x=18$

Explanation:

To solve for , divide both sides of the equation by $\displaystyle 5$ :

$\frac{5x}{5}=\frac{90}{5}$ $\displaystyle \frac{5x}{5}=\frac{90}{5}$

x=18 $\displaystyle x=18$

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Example Question #351 : High School Math

Solve for when $\displaystyle x-918=1332$

Possible Answers:

x=2250 $\displaystyle x=2250$

x=2575 $\displaystyle x=2575$

x=1275 $\displaystyle x=1275$

x=289 $\displaystyle x=289$

Correct answer:

x=2250 $\displaystyle x=2250$

Explanation:

To solve for , add 918 $\displaystyle 918$ to both sides of the equation:

$\displaystyle x-918+918=1332+918$

x=2250 $\displaystyle x=2250$

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Example Question #201 : Pre Algebra

Solve for when $\frac{x}{7}=9$ $\displaystyle \frac{x}{7}=9$

Possible Answers:

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

x=54 $\displaystyle x=54$

x=9 $\displaystyle x=9$

x=63 $\displaystyle x=63$

Correct answer:

x=63 $\displaystyle x=63$

Explanation:

To solve for , multiply both sides of the equation by $\displaystyle 7$ :

$\frac{x}{7}\cdot 7=9\cdot 7$ $\displaystyle \frac{x}{7}\cdot 7=9\cdot 7$

$x=9\cdot 7$ $\displaystyle x=9\cdot 7$

x=63 $\displaystyle x=63$

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Example Question #101 : Algebraic Equations

Solve for when x+15=97 $\displaystyle x+15=97$

Possible Answers:

x=78 $\displaystyle x=78$

x=82 $\displaystyle x=82$

x=102 $\displaystyle x=102$

x=86 $\displaystyle x=86$

Correct answer:

x=82 $\displaystyle x=82$

Explanation:

To solve for , subtract $\displaystyle 15$ from each side of the equation:

$\displaystyle x+15-15=97-15$

x=97-15 $\displaystyle x=97-15$

x=82 $\displaystyle x=82$

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All High School Math Resources

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High School Math : How to solve one-step equations with integers in pre-algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #342 : High School Math

Example Question #343 : High School Math

Example Question #344 : High School Math

Example Question #345 : High School Math

Example Question #346 : High School Math

Example Question #347 : High School Math

Example Question #197 : Pre Algebra

Example Question #351 : High School Math

Example Question #201 : Pre Algebra

Example Question #101 : Algebraic Equations

All High School Math Resources

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