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Algebra 1 : Linear Equations

Study concepts, example questions & explanations for Algebra 1

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10 Diagnostic Tests 557 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : Proportions

Solve for .

$\frac{1}{3}=\frac{5}{x}$

Possible Answers:

Correct answer:

Explanation:

To solve for , we just cross-multiply.

$\frac{1}{3}=\frac{5}{x}$

We have 3*5=1*x .

We get 15=x .

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Example Question #1162 : Linear Equations

Solve for .

$\frac{5}{4}=\frac{25}{2x}$

Possible Answers:

100

Correct answer:

Explanation:

To solve for , we just cross-multiply.

$\frac{5}{4}=\frac{25}{2x}$

We have 4*25=5*2x .

We now have 100=10x .

Divide both sides by , we get 10=x .

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Example Question #1163 : Linear Equations

Solve for .

$\frac{9}{4}=\frac{18}{2x}$

Possible Answers:

Correct answer:

Explanation:

To solve for , we just cross-multiply.

$\frac{9}{4}=\frac{18}{2x}$

We have 4*18=9*2x .

We now have 72=18x .

Divide both sides by , we get 4=x .

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Example Question #1164 : Linear Equations

Solve for .

$\frac{3}{7}=\frac{x}{2}$

Possible Answers:

$\frac{5}{7}$

$\frac{6}{7}$

$\frac{8}{9}$

$\frac{7}{6}$

Correct answer:

$\frac{6}{7}$

Explanation:

To solve for , we just cross-multiply.

$\frac{3}{7}=\frac{x}{2}$

We have 3*2=7*x .

We now have 6=7x .

Divide both sides by , we get $\frac{6}{7}=x$ .

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Example Question #1165 : Linear Equations

Solve for .

$\frac{x}{9}=\frac{4}{13}$

Possible Answers:

$\frac{24}{17}$

$\frac{13}{36}$

$\frac{36}{13}$

Correct answer:

$\frac{36}{13}$

Explanation:

To solve for , we just cross-multiply.

$\frac{x}{9}=\frac{4}{13}$

We have 4*9=13*x .

We now have 36=13x .

Divide both sides by , we get $\frac{36}{13}=x$ .

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Example Question #1166 : Linear Equations

Solve for .

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

Possible Answers:

$\pm6$

$\pm36$

Correct answer:

$\pm6$

Explanation:

To solve for , we just cross-multiply.

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

We have 6*6=x*x .

We now have 36=x^2 .

Take square root on both sides. Remember, when we do that the answer is $\pm \sqrt{36}$ .

Two negative or two positive numbers multiplied can give us a positve value. is the answer to that perfect square and we add the positive and negative sign to get $\pm6$ .

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Example Question #1167 : Linear Equations

$\frac{x}{-3}=\frac{27}{-x}$

Possible Answers:

$\pm81$

$\pm9$

Correct answer:

$\pm9$

Explanation:

To solve for , we just cross-multiply.

$\frac{x}{-3}=\frac{27}{-x}$

We have -3*27=-x*x .

When multiplying a negative and positive number, our answer is negative.

We now have -81=-x^2 .

We can divide both sides by to get 81=x^2 .

Take square root on both sides. Remember, when we do that the answer is $\pm \sqrt{81}$ . Two negative or two positive numbers multiplied can give us a positve value. is the answer to that perfect square and we add the positive and negative sign to get $\pm9$ .

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Example Question #31 : Proportions

Solve for .

$\frac{1}{2}=\frac{5}{x+4}$

Possible Answers:

Correct answer:

Explanation:

To solve for , we just cross-multiply.

$\frac{1}{2}=\frac{5}{x+4}$

We have 2*5=1*(x+4) .

Remember we treat the expression and place into a parantheses. We take care of the parantheses first and distribute the .

We have 10=x+4 .

Subtract both sides by .

We get 6=x .

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Example Question #32 : Proportions

Solve for .

$\frac{6}{7}=\frac{12}{x-2}$

Possible Answers:

Correct answer:

Explanation:

To solve for , we just cross-multiply.

$\frac{6}{7}=\frac{12}{x-2}$

We have 7*12=6*(x-2) .

Remember we treat the expression and place into a parantheses. We take care of the parantheses first and distribute the .

We have 84=6x-12 .

Add both sides by .

We have 6x=96 .

Then divide both sides by .

We get 16=x .

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Example Question #41 : Proportions

Solve for .

$\frac{x+2}{3}=4$

Possible Answers:

Correct answer:

Explanation:

To solve for we want to isolate it on one side of the equation with all other constants on the other side.

$\frac{x+2}{3}=4$

Multiply both sides by .

We have x+2=12 .

Subtract both sides by .

We have x=10 .

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Algebra 1 : Linear Equations

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #31 : Proportions

Example Question #1162 : Linear Equations

Example Question #1163 : Linear Equations

Example Question #1164 : Linear Equations

Example Question #1165 : Linear Equations

Example Question #1166 : Linear Equations

Example Question #1167 : Linear Equations

Example Question #31 : Proportions

Example Question #32 : Proportions

Example Question #41 : Proportions

All Algebra 1 Resources

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