High School Math : High School Math

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #51 : How To Solve One Step Equations With Integers In Pre Algebra

Solve for  when \(\displaystyle x-71=64\)

Possible Answers:

\(\displaystyle x=132\)

\(\displaystyle x=151\)

\(\displaystyle x=135\)

\(\displaystyle x=-7\)

Correct answer:

\(\displaystyle x=135\)

Explanation:

To solve for  we must get all of the numbers on the other side of the equation as .

To do this in a problem where  is being subtracted by a number, we must add the number to both sides of the equation.

In this case the number is \(\displaystyle 71\) so we add \(\displaystyle 71\) to each side of the equation to make it look like this 

\(\displaystyle x-71+71=64+71\)

The \(\displaystyle 71\)'s cancel on the left side and leave  by itself 

\(\displaystyle x=64+71\)

Then we perform the necessary addition to get the answer of \(\displaystyle x=135\).

Example Question #181 : Pre Algebra

Solve for  when \(\displaystyle x+93=151\).

Possible Answers:

\(\displaystyle x=58\)

\(\displaystyle x=93\)

\(\displaystyle x=51\)

\(\displaystyle x=244\)

Correct answer:

\(\displaystyle x=58\)

Explanation:

To solve for  we must get all of the numbers on the other side of the equation as .

To do this in a problem where a number is being added to , we must subtract the number from both sides of the equation.

In this case the number is \(\displaystyle 93\) so we subtract \(\displaystyle 93\) from each side of the equation to make it look like this 

\(\displaystyle x+93-93=151-93\)

The \(\displaystyle 93\)'s on the left side cancel to get 

\(\displaystyle x=151-93\)

Then we perform the necessary subtraction to get the answer of  \(\displaystyle x=58\).

Example Question #341 : High School Math

Solve for \(\displaystyle p\):

\(\displaystyle 4=7-p\)

Possible Answers:

\(\displaystyle p=0\)

\(\displaystyle p=2\)

\(\displaystyle p=3\)

\(\displaystyle p=1\)

Correct answer:

\(\displaystyle p=3\)

Explanation:

\(\displaystyle 4=7-p\)

\(\displaystyle 4+p=7-p+p\)

\(\displaystyle 4+p-4=7-4\)

\(\displaystyle p=3\)

Example Question #91 : Algebraic Equations

Solve for \(\displaystyle x\)

 

\(\displaystyle x - 3 = 12\)

Possible Answers:

There are infinite solutions

\(\displaystyle x = 15\)

There is no solution

\(\displaystyle x = 9\)

\(\displaystyle x = 36\)

Correct answer:

\(\displaystyle x = 15\)

Explanation:

When solving a one-step equation it is important to remember to isolate \(\displaystyle x\) by performing the inverse operations. Note that in our original equation, \(\displaystyle 3\) is being added to \(\displaystyle x\). Thus, we want to subtract \(\displaystyle 3\) from both sides to do the inverse operation. 

Then, \(\displaystyle x - 3 + 3 = 12 + 3\)

So, \(\displaystyle x = 15\)

 

Example Question #342 : High School Math

\(\displaystyle x+15=6\)

Solve for \(\displaystyle x\).

Possible Answers:

\(\displaystyle -9\)

\(\displaystyle -21\)

\(\displaystyle 21\)

\(\displaystyle \frac{2}{5}\)

\(\displaystyle 9\)

Correct answer:

\(\displaystyle -9\)

Explanation:

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

\(\displaystyle x+15-15=6-15\)

\(\displaystyle x=6-15\)

\(\displaystyle x=-9\)

Example Question #343 : High School Math

\(\displaystyle x+5=34\)

Solve for \(\displaystyle x\).

Possible Answers:

\(\displaystyle 29\)

\(\displaystyle \frac{34}{5}\)

\(\displaystyle 39\)

\(\displaystyle 170\)

\(\displaystyle 24\)

Correct answer:

\(\displaystyle 29\)

Explanation:

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

\(\displaystyle x+5-5=34-5\)

\(\displaystyle x=34-5\)

\(\displaystyle x=29\)

Example Question #344 : High School Math

Solve for  when \(\displaystyle x+9=32\)

Possible Answers:

\(\displaystyle x=23\)

\(\displaystyle x=21\)

\(\displaystyle x=41\)

\(\displaystyle x=39\)

Correct answer:

\(\displaystyle x=23\)

Explanation:

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

\(\displaystyle x+9-9=32-9\)

\(\displaystyle x=32-9\)

\(\displaystyle x=23\)

Example Question #345 : High School Math

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

Possible Answers:

\(\displaystyle x=4\)

\(\displaystyle x=64\)

\(\displaystyle x=48\)

\(\displaystyle x=8\)

Correct answer:

\(\displaystyle x=64\)

Explanation:

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

\(\displaystyle \frac{x}{16}\cdot 16=4\cdot 16\)

\(\displaystyle x=4\cdot 16\)

\(\displaystyle x=64\)

Example Question #346 : High School Math

Solve for  when \(\displaystyle x-18=72\)

Possible Answers:

\(\displaystyle x=80\)

\(\displaystyle x=54\)

\(\displaystyle x=75\)

\(\displaystyle x=90\)

Correct answer:

\(\displaystyle x=90\)

Explanation:

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

\(\displaystyle x-18+18=72+18\)

\(\displaystyle x=72+18\)

\(\displaystyle x=90\)

Example Question #347 : High School Math

Solve for  when \(\displaystyle 12x=156\)

Possible Answers:

\(\displaystyle x=13\)

\(\displaystyle x=3124\)

\(\displaystyle x=181\)

\(\displaystyle x=81\)

Correct answer:

\(\displaystyle x=13\)

Explanation:

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

\(\displaystyle \frac{12x}{12}=\frac{156}{12}\)

\(\displaystyle x=\frac{156}{12}\)

\(\displaystyle x=13\)

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