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High School Math : Pre-Algebra

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Example Questions

Example Question #2 : How To Solve Two Step Equations With Integers In Pre Algebra

Solve the equation for .

$\small 7p+3=10p$

Possible Answers:

$\small p=-\frac{13}{10}$

$\small p=-1$

$\small p=\frac{13}{10}$

$\small p=1$

Correct answer:

$\small p=1$

Explanation:

$\small 7p+3=10p$

Subtract from both sides to get the variables on the same side, and simplify.

$\small 7p+3-7p=10p-7p$

$\small 3=3p$

Divide both sides by .

$\small \frac{3}{3}=\frac{3p}{3}$

$\small 1=p$

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Example Question #3 : How To Solve Two Step Equations With Integers In Pre Algebra

Solve for :

4x-2=14

Possible Answers:

Correct answer:

Explanation:

We have two steps to this problem. Remember our end result is to get isolated on one side of the equation.

First we add to each side, cancelling out the on the left side. This brings the equation to 4x =16 .

Remember to get rid of a number we perform the opposite operation. The is multiplied with the , therefore to cancel it out we will divide by .

divided by equals . What we do to one side we must do to the other, therefore we divide by also. This becomes x=4 .

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Example Question #4 : How To Solve Two Step Equations With Integers In Pre Algebra

Simplify:

$(2x^{2}+5) (3x^{3}+ 3)$

Possible Answers:

$6x^{5} +15$

6x^6 +15

$6x^{5}+15x^{3}+6x^{2}+15$

$6x^6 +6x^{2} + 15x^{3} +15$

$5x^{6} + 8$

Correct answer:

$6x^{5}+15x^{3}+6x^{2}+15$

Explanation:

The requires the FOIL method (first, outside, inside, last).

Multiplying the first two monomials $2x^2 \cdot 3x^3$ equals $6x^{5}$ .

The outside two are $2x^{2} \cdot 3$ which equals $6x^{2}$ .

The two inside monomials are $5 \cdot 3x^3$ which equals 15x^3 .

The last two are $5\cdot 3$ which is .

All together this becomes $6x^{5}+15x^{3}+6x^{2}+15$ .

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Example Question #6 : How To Solve Two Step Equations With Integers In Pre Algebra

Simplify:

$2x^{5} + 5x^5$

Possible Answers:

$10x^{5}$

7x^5

$7x^{25}$

7x^1^0

Correct answer:

7x^5

Explanation:

According to the laws of exponents, if you add you do not add the exponents. Therefore 5 +2 =7 , and the variable term remains $x^{5}$ . The answer is 7x^5 .

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Example Question #421 : High School Math

Solve for $\small s$ .

26-s=13

Possible Answers:

s=2

s=13

s=-39

s=39

Correct answer:

s=13

Explanation:

26-s=13

Add $\small s$ to both sides.

26-s+s=13+s

26=13+s

Subtract 13 from both sides.

26-13=13+s-13

13=s

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Example Question #2 : How To Solve Two Step Equations With Integers In Pre Algebra

Solve for $\small n$ .

6n+14=38

Possible Answers:

n=2

n=9

n=144

n=4

Correct answer:

n=4

Explanation:

6n+14=38

Subtract 14 from each side.

6n+14-14=38-14

6n=24

Divide both sides by 6.

$\frac{6n}{6}=\frac{24}{6}$

n=4

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Example Question #261 : Pre Algebra

Solve for if, $5x^{2}= 245$ .

Possible Answers:

$x=\pm5$

$x=\pm9$

$x=\pm8$

$x=\pm7$

Correct answer:

$x=\pm7$

Explanation:

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

When solving a simple two-step equation we must remove the numbers that are interacting with in this order: Addition/Subtraction, Multiplication/Division, Exponents.

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

In this case the number is so we divide each side of the equation by to make it look like this $\frac{5x^{2}}{5}= \frac{245}{5}$

Our new problem will look like this $x^{2}=49$ .

We then must solve the exponential equation.

To solve an equation with we must take the square root of each side of the equation to get by itself.

Doing this makes our problem look like $\sqrt{x^{2}}=\sqrt{49}$

Then we perform the square root to get the answer $x=\pm7$

Remember that $(-7)^{2}$ can also equal so our answer will be both positive and negative.

The final answer looks like $x=\pm7$ .

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Example Question #262 : Pre Algebra

Solve for the value of $\small e$ .

6e+8=14

Possible Answers:

e=36

e=1

e=132

$e=\frac{22}{6}$

Correct answer:

e=1

Explanation:

6e+8=14

We need ot work to isolate the variable using inverse operations.

Subtract $\small 8$ from both sides.

6e+8-8=14-8

6e=6

Divide each side by $\small 6$ .

$\frac{6e}{6}=\frac{6}{6}$

e=1

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Example Question #11 : How To Solve Two Step Equations With Integers In Pre Algebra

x+7=12*2

Solve for .

Possible Answers:

Correct answer:

Explanation:

To solve x+7=12*2 , start by simplifying the right side.

x+7=12*2

x+7=24

Now isolate . This means we need to subtract from both sides.

x+7-7=24-7

x=24-7

x=17

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Example Question #13 : How To Solve Two Step Equations With Integers In Pre Algebra

Solve for X in the following equation: 3X+4=6X-11

Possible Answers:

Correct answer:

Explanation:

To solve for X, we must isolate X on one side of the equation. In this case, we do this by subtracting 3X and adding 11 to both sides.

(3X-3X)+(4+11)=(6X-3X)+(-11+11)

15=3X

We now have isolated X on one side of the equation, so we need only to reduce 3X to 1X. We can do this by dividing both sides by 3:

(15/3=3X/3)=(5=X)

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High School Math : Pre-Algebra

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #2 : How To Solve Two Step Equations With Integers In Pre Algebra

Example Question #3 : How To Solve Two Step Equations With Integers In Pre Algebra

Example Question #4 : How To Solve Two Step Equations With Integers In Pre Algebra

Example Question #6 : How To Solve Two Step Equations With Integers In Pre Algebra

Example Question #421 : High School Math

Example Question #2 : How To Solve Two Step Equations With Integers In Pre Algebra

Example Question #261 : Pre Algebra

Example Question #262 : Pre Algebra

Example Question #11 : How To Solve Two Step Equations With Integers In Pre Algebra

Example Question #13 : How To Solve Two Step Equations With Integers In Pre Algebra

All High School Math Resources

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