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Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #782 : Basic Single Variable Algebra

Solve: 8(-8x+6) = -x+2

Possible Answers:

$\frac{50}{63}$

$-\frac{16}{9}$

$\textup{The answer is not given.}$

$-\frac{46}{63}$

$\frac{46}{63}$

Correct answer:

$\frac{46}{63}$

Explanation:

Solve the left side by distribution.

8(-8x)+8(6) = -x+2

The equation becomes:

-64x+48 =-x+2

Add 64x on both sides.

-64x+48 +64x=-x+2+64x

48= 63x+2

Subtract 2 from both sides.

48-2= 63x+2-2

46= 63x

Divide by 63 on both sides.

$\frac{46}{63}= \frac{63x}{63}$

The answer is: $\frac{46}{63}$

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Example Question #311 : Solving Equations

Solve: $\frac{7}{3x}+\frac{1}{9} = \frac{1}{12}$

Possible Answers:

$-\frac{7}{108}$

$-\frac{1}{84}$

Correct answer:

Explanation:

Multiply both sides by the least common denominator in order to cancel all fractions.

Each denominator must share an term, and all the denominators must be a factor of the integer in front of the term.

The LCD is: 36x

Multiply this value on both sides.

$36x(\frac{7}{3x}+\frac{1}{9} )= \frac{1}{12} \cdot 36x$

84+4x =3x

Subtract from both sides to group the -terms.

84+4x -4x=3x-4x

84=-x

Divide by negative one on both sides.

$\frac{84}{-1}=\frac{-x}{-1}$

The answer is:

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Example Question #441 : Equations

Solve: $\frac{4}{7}x-\frac{1}{8} = \frac{3}{28}$

Possible Answers:

$-\frac{1}{98}$

$-\frac{1}{32}$

$\frac{13}{32}$

Correct answer:

$\frac{13}{32}$

Explanation:

Determine the least common denominator. Write down the multiples for seven, eight, and twenty eight.

7-[7,14,21,28,...56]

8-[8,16,24,32,...,56]

28-[28,56]

Since all values are common to 56, we will need to multiply 56 on both sides of the equation in order to eliminate the fractions.

$56(\frac{4}{7}x-\frac{1}{8}) = \frac{3}{28} \cdot 56$

32x-7 = 6

Add 7 on both sides.

32x-7 +7= 6+7

Simplify both sides.

32x=13

Divide by 32 on both sides.

$\frac{32x}{32}=\frac{13}{32}$

The answer is: $\frac{13}{32}$

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Example Question #313 : Solving Equations

Solve the equation: $\frac{3}{x}-5 = 10$

Possible Answers:

$\frac{3}{5}$

$\frac{1}{5}$

$\frac{1}{45}$

Correct answer:

$\frac{1}{5}$

Explanation:

Add five on both sides.

$\frac{3}{x}-5+5 = 10+5$

$\frac{3}{x} = 15$

Multiply by on both sides to eliminate the fraction.

$\frac{3}{x} \cdot x= 15 \cdot x$

3=15x

Divide by 15 on both sides.

$\frac{3}{15}=\frac{15x}{15}$

Reduce both fractions.

The answer is: $\frac{1}{5}$

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Example Question #314 : Solving Equations

Solve the equation: 3x-9 = 27-3x

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

Add on both sides.

3x-9+[3x] = 27-3x+[3x]

6x-9 = 27

Add 9 on both sides.

6x-9 +9= 27+9

6x=36

Divide by 6 on both sides.

$\frac{6x}{6}=\frac{36}{6}$

The answer is:

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Example Question #441 : Equations

Solve the equation: $\frac{1}{4x}-\frac{2}{5x} = \frac{1}{10}$

Possible Answers:

$-\frac{1}{6}$

$-\frac{5}{4}$

$-\frac{5}{2}$

$\frac{20}{21}$

$-\frac{3}{2}$

Correct answer:

$-\frac{3}{2}$

Explanation:

Multiply both sides by the least common denominator in order to eliminate all fractions in the equation.

All the terms will need to share an variable, and the value of the LCD must be able to divide evenly among the coefficients of the denominators in the original problem.

This means that the LCD must be: 20x

Multiply this value on both sides.

$20x(\frac{1}{4x}-\frac{2}{5x} )= \frac{1}{10}\cdot 20x$

5-8 = 2x

Simplify this equation.

-3=2x

Divide by two on both sides.

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

The answer is: $-\frac{3}{2}$

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Example Question #316 : Solving Equations

Solve the equation: 9x-17 = -3+4x

Possible Answers:

$\frac{14}{5}$

$\frac{14}{13}$

$\frac{4}{13}$

$-\frac{14}{5}$

$\textup{The answer is not given.}$

Correct answer:

$\frac{14}{5}$

Explanation:

Subtract from both sides.

9x-17 -4x= -3+4x-4x

The equation becomes:

5x-17 = -3

Add 17 on both sides.

5x-17 +17= -3+17

5x=14

Divide by 5 on both sides.

$\frac{5x}{5}=\frac{14}{5}$

The answer is: $\frac{14}{5}$

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Example Question #317 : Solving Equations

Solve: 5x-18 = -(2x+3)

Possible Answers:

$\frac{3}{5}$

$-\frac{15}{7}$

$\frac{15}{7}$

Correct answer:

$\frac{15}{7}$

Explanation:

Divide by negative one on both sides.

$\frac{5x-18 }{-1}= \frac{-(2x+3)}{-1}$

-5x+18 = 2x+3

Add on both sides, and subtract 3 on both sides.

-5x+18 +[5x]-(3)= 2x+3+[5x] -(3)

Simplify both sides.

15=7x

Divide by seven on both sides.

$\frac{15}{7}=\frac{7x}{7}$

The answer is: $\frac{15}{7}$

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Example Question #318 : Solving Equations

Solve the equation: -5(x-9)=-14

Possible Answers:

$\frac{31}{5}$

$\frac{31}{14}$

$-\frac{59}{5}$

$-\frac{31}{5}$

$\frac{59}{5}$

Correct answer:

$\frac{59}{5}$

Explanation:

Multiply and distribute the negative five through the binomial.

-5(x)-(-5)(9)=-14

-5x+45 =-14

Subtract 45 from both sides.

-5x+45 -45=-14-45

-5x =-59

Divide by negative five on both sides.

$\frac{-5x}{-5} =\frac{-59}{-5}$

The answer is: $\frac{59}{5}$

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Example Question #441 : Equations

Solve for x and y.

3x-4y=7

1.5x+7y=21.5

Possible Answers:

$x=-5 \text{ and }y=2$

$x=5 \text{ and }y=2$

$x=4 \text{ and }y=6$

$x=-5 \text{ and }y=-2$

$x=-4 \text{ and }y=6$

Correct answer:

$x=5 \text{ and }y=2$

Explanation:

Solve for x and y by using substitution. Solve the first equation for x.

$x=\frac{7+4y}{3}$

Next substitute x into the second equation and simplify,

$1.5(\frac{7+4y}{3})+7y=21.5$

3.5+2y+7y=21.5

9y=18

y=2

Now use the value for y to solve for x.

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

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Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #782 : Basic Single Variable Algebra

Example Question #311 : Solving Equations

Example Question #441 : Equations

Example Question #313 : Solving Equations

Example Question #314 : Solving Equations

Example Question #441 : Equations

Example Question #316 : Solving Equations

Example Question #317 : Solving Equations

Example Question #318 : Solving Equations

Example Question #441 : Equations

All Algebra II Resources

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