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Algebra II : Solving Equations

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Example Questions

Example Question #201 : Solving Equations

Solve for .

$\frac{5}{4}x=1200$

Possible Answers:

1500

960

1020

1280

1440

Correct answer:

960

Explanation:

To solve for the variable isolate it on one side of the equation by moving all other constants to the other side. To do this, perform opposite operations to manipulate the equation.

$\frac{5}{4}x=1200$

Multiply $\frac{4}{5}$ on both sides.

x=960

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Example Question #202 : Solving Equations

Solve the equation: 8x-14=37

Possible Answers:

$\frac{23}{4}$

$\frac{13}{8}$

$\frac{51}{4}$

$\frac{23}{8}$

$\frac{51}{8}$

Correct answer:

$\frac{51}{8}$

Explanation:

Add 14 on both sides.

8x-14+14=37+14

Simplify both sides.

8x=51

Divide by eight on both sides.

$\frac{8x}{8}=\frac{51}{8}$

The answer is: $\frac{51}{8}$

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Example Question #203 : Solving Equations

Solve the equation: 2(2x-3)=18

Possible Answers:

Correct answer:

Explanation:

Divide by two on both sides.

$\frac{2(2x-3)}{2}=\frac{18}{2}$

Simplify both sides.

2x-3=9

Add three on both sides.

2x-3+3=9+3

Simplify both sides.

2x=12

Divide by two on both sides.

$\frac{2x}{2}=\frac{12}{2}$

x=6

The answer is:

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Example Question #204 : Solving Equations

Solve the equation: $\frac{2}{11}x+\frac{1}{11} = \frac{1}{33}$

Possible Answers:

$-\frac{1}{3}$

$-\frac{1}{6}$

$-\frac{3}{11}$

$-\frac{2}{33}$

$-\frac{6}{11}$

Correct answer:

$-\frac{1}{3}$

Explanation:

Notice that the denominators on the left side can be converted to 33 by multiplying the top and bottom by a value of three. This will save time from having to convert between steps.

$(\frac{3}{3})(\frac{2}{11}x+\frac{1}{11} )= \frac{1}{33}$

Simplify by distribution.

$\frac{6}{33}x + \frac{3}{33} = \frac{1}{33}$

Multiply by 33 on both sides in order to cancel the denominators.

$33(\frac{6}{33}x + \frac{3}{33}) = 33(\frac{1}{33})$

The equation becomes: 6x+3=1

Subtract three on both sides.

6x=-2

Divide by six on both sides and reduce the fraction.

$\frac{6x}{6}=\frac{-2}{6}$

The answer is: $x=-\frac{1}{3}$

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Example Question #205 : Solving Equations

Solve the equation: $\sqrt{2x+3}= 6$

Possible Answers:

$\frac{39}{2}$

$\frac{13}{2}$

$\frac{11}{2}$

$\frac{33}{2}$

$\frac{5}{3}$

Correct answer:

$\frac{33}{2}$

Explanation:

In order to isolate the variable, first square both sides of the equation to eliminate the radical.

$(\sqrt{2x+3})^2=( 6)^2$

Simplify both sides.

2x+3 = 36

Subtract three from both sides.

2x=33

Divide by two on both sides.

$\frac{2x}{2}=\frac{33}{2}$

The answer is: $\frac{33}{2}$

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Example Question #206 : Solving Equations

Solve the equation: 3x+9=6x-9

Possible Answers:

Correct answer:

Explanation:

Group the constants and variables on a separate side of the equation.

Subtract on both sides.

3x+9-(3x)=6x-9-(3x)

9=3x-9

Add nine on both sides.

9+9=3x-9+9

Simplify both sides.

18=3x

Divide by three on both sides.

The answer is: x=6

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Example Question #201 : Solving Equations

Solve the equation: 2(6x+9) = -36

Possible Answers:

$-\frac{23}{12}$

$-\frac{27}{2}$

$-\frac{9}{2}$

$-\frac{47}{12}$

$-\frac{3}{2}$

Correct answer:

$-\frac{9}{2}$

Explanation:

Divide by two on both sides.

$\frac{2(6x+9) }{2}=\frac{ -36}{2}$

Simplify both sides of the equation.

6x+9=-18

Subtract nine from both sides.

6x+9-9=-18-9

Simplify both sides.

6x=-27

Divide by six on both sides.

$\frac{6x}{6}=\frac{-27}{6}$

Reduce the fractions.

$x=\frac{-3\times 9}{3\times 2} = -\frac{9}{2}$

The answer is: $-\frac{9}{2}$

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Example Question #331 : Equations

Solve the following equation:

$\frac{3}{x} + \frac{6}{x} - \frac{1}{x} = 2x$

Possible Answers:

x=-2

x=2

x=2,-2

x=1,-1

x=2,1

Correct answer:

x=2,-2

Explanation:

First, multiply both sides of the equation by to eliminate the denominators of the terms in the left-hand side:

$(x)(\frac{3}{x} + \frac{6}{x} - \frac{1}{x}) = 2x(x)$

$\Rightarrow 3+6-1= 2x^2$

$\Rightarrow$ 8=2x^2

Next, divide both sides by to yield a power of on the right-hand side:

$\frac{8}{2} = \frac{2x^2}{2}$

$\Rightarrow4=x^2$

Take the square root of both sides to solve for the unknown variable :

$\sqrt{4}=\sqrt{x^2}$

$\Rightarrow\pm2=x$

x=2,-2

We now have two solutions for the unknown variable. Substitute and into the original equation to verify that both solutions yield an identity:

If x=2 , then

$\frac{3}{2} + \frac{6}{2} - \frac{1}{2} = 2(2)$

$\Rightarrow \frac{8}{2} = 4$

which is true.

If x=-2 , then

$-\frac{3}{2} - \frac{6}{2} + \frac{1}{2} = 2(-2)$

$\Rightarrow -\frac{8}{2} = -4$

which is also true.

Hence, the two solutions for the given equation are

x=2,-2

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Example Question #209 : Solving Equations

Solve the following equation:

$1+\frac{5}{3x-7}=2$

Possible Answers:

x=4

x=2

x=-1

x=5

x=7

Correct answer:

x=4

Explanation:

First, subtract from both sides of the equation:

$1-1+\frac{5}{3x-7}=2-1$

$\Rightarrow \frac{5}{3x-7}=1$

Next, multiply both sides by 3x-7 to eliminate the denominator of the term on the left-hand side of the equation:

$\frac{5}{3x-7}(3x-7)=1(3x-7)$

$\Rightarrow5=3x-7$

Add to both side of the equation to isolate the term containing the unknown variable on the right-hand side of the equation:

5+7=3x-7+7

$\Rightarrow12=3x$

Divide both sides by :

$\frac{12}{3}=\frac{3x}{3}$

$\Rightarrow x=4$

Checking the solution x=4 by substituting it into the original equation yields:

$1+\frac{5}{3(4)-7}=2$

$\Rightarrow 1+\frac{5}{5}=2$

$\Rightarrow 1+1=2$

which is true. Hence, the solution to this equation is x=4 .

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Example Question #201 : Solving Equations

Solve the equation: 9x-3=-9x-6

Possible Answers:

$-\frac{1}{9}$

$\textup{No solution.}$

$-\frac{1}{2}$

$-\frac{1}{3}$

$-\frac{1}{6}$

Correct answer:

$-\frac{1}{6}$

Explanation:

Group like terms. Start with the x-variable.

Add on both sides.

9x-3+(9x)=-9x-6+(9x)

18x-3=-6

Add three on both sides.

18x-3+3=-6+3

Simplify both sides.

18x=-3

Divide by 18 on both sides.

$\frac{18x}{18}=\frac{-3}{18}$

Simplify both fractions.

$x=-\frac{1}{6}$

The answer is: $-\frac{1}{6}$

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Algebra II : Solving Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #201 : Solving Equations

Example Question #202 : Solving Equations

Example Question #203 : Solving Equations

Example Question #204 : Solving Equations

Example Question #205 : Solving Equations

Example Question #206 : Solving Equations

Example Question #201 : Solving Equations

Example Question #331 : Equations

Example Question #209 : Solving Equations

Example Question #201 : Solving Equations

All Algebra II Resources

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