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Algebra 1 : Algebra 1

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 #361 : Algebra 1

Solve for $\displaystyle x$.

$\frac{x}{-27}=9$ $\displaystyle \frac{x}{-27}=9$

Possible Answers:

343 $\displaystyle 343$

$\displaystyle -343$

243 $\displaystyle 243$

279 $\displaystyle 279$

$\displaystyle -243$

Correct answer:

$\displaystyle -243$

Explanation:

$\frac{x}{-27}=9$ $\displaystyle \frac{x}{-27}=9$ Multiply $\displaystyle -27$ on both sides. When we multiply with a positive number, our answer is negative.

x=-243 $\displaystyle x=-243$

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Example Question #362 : Linear Equations

Solve for $\displaystyle x$.

$\frac{x}{-16}=-88$ $\displaystyle \frac{x}{-16}=-88$

Possible Answers:

$\displaystyle -1408$

$\displaystyle -1418$

1418 $\displaystyle 1418$

1428 $\displaystyle 1428$

1408 $\displaystyle 1408$

Correct answer:

1408 $\displaystyle 1408$

Explanation:

$\frac{x}{-16}=-88$ $\displaystyle \frac{x}{-16}=-88$ Multiply $\displaystyle -16$ on both sides. When multiplying with another negative number, our answer is positive.

x=1408 $\displaystyle x=1408$

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Example Question #363 : Linear Equations

Solve: $\frac{x}{10}= 20$ $\displaystyle \frac{x}{10}= 20$

Possible Answers:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

200 $\displaystyle 200$

$\displaystyle 20$

$\displaystyle 2$

$\displaystyle 30$

Correct answer:

200 $\displaystyle 200$

Explanation:

In order to isolate the unknown variable $\displaystyle x$, we need to multiply both sides by ten.

$\frac{x}{10} \cdot 10= 20 \cdot 10$ $\displaystyle \frac{x}{10} \cdot 10= 20 \cdot 10$

Simplify both sides.

x=200 $\displaystyle x=200$

The answer is: 200 $\displaystyle 200$

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Example Question #361 : Algebra 1

Solve: $\frac{x}{5} = 7$ $\displaystyle \frac{x}{5} = 7$

Possible Answers:

$\displaystyle 35$

$\frac{7}{5}$ $\displaystyle \frac{7}{5}$

$\displaystyle 12$

$\frac{5}{7}$ $\displaystyle \frac{5}{7}$

$\displaystyle 2$

Correct answer:

$\displaystyle 35$

Explanation:

Multiply by five on both sides.

$\frac{x}{5} \times 5= 7\times 5$ $\displaystyle \frac{x}{5} \times 5= 7\times 5$

Simplify both sides.

x=35 $\displaystyle x=35$

The answer is $\displaystyle 35$.

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Example Question #365 : Linear Equations

Solve the following one-step equation for $\displaystyle f$

49f=343 $\displaystyle 49f=343$

Possible Answers:

$\displaystyle 41$

$\displaystyle 16$

$\displaystyle 49$

$\displaystyle 7$

$\displaystyle 1$

Correct answer:

$\displaystyle 7$

Explanation:

Solve the following one-step equation for $\displaystyle f$

49f=343 $\displaystyle 49f=343$

One-step equations can be solved in one step.

In this case, we simply need to divide by $\displaystyle 49$. That will get rid of the $\displaystyle 49$ on the left and show us what value of $\displaystyle f$ is the solution.

$\frac{49f}{49}=\frac{343}{49}$ $\displaystyle \frac{49f}{49}=\frac{343}{49}$

f=7 $\displaystyle f=7$

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Example Question #366 : Linear Equations

Solve the following equation: 6x = 78 $\displaystyle 6x = 78$

Possible Answers:

$\displaystyle 13$

$\frac{1}{13}$ $\displaystyle \frac{1}{13}$

$\displaystyle 84$

$\displaystyle 71$

$\frac{1}{71}$ $\displaystyle \frac{1}{71}$

Correct answer:

$\displaystyle 13$

Explanation:

Divide by six on both sides in order to isolate the unknown variable.

$\frac{6x}{6} = \frac{78}{6}$ $\displaystyle \frac{6x}{6} = \frac{78}{6}$

x=13 $\displaystyle x=13$

The answer is $\displaystyle 13$.

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Example Question #367 : Linear Equations

Solve the following equation: 6x = 78 $\displaystyle 6x = 78$

Possible Answers:

$\displaystyle 71$

$\frac{1}{71}$ $\displaystyle \frac{1}{71}$

$\displaystyle 13$

$\displaystyle 84$

$\frac{1}{13}$ $\displaystyle \frac{1}{13}$

Correct answer:

$\displaystyle 13$

Explanation:

Divide by six on both sides in order to isolate the unknown variable.

$\frac{6x}{6} = \frac{78}{6}$ $\displaystyle \frac{6x}{6} = \frac{78}{6}$

Simplify both sides.

x=13 $\displaystyle x=13$

The answer is $\displaystyle 13$.

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Example Question #368 : Linear Equations

Solve for x in the following equation:

$\frac{x}{-4} = 3$ $\displaystyle \frac{x}{-4} = 3$

Possible Answers:

x = -12 $\displaystyle x = -12$

x = -4 $\displaystyle x = -4$

x = 12 $\displaystyle x = 12$

x = -1 $\displaystyle x = -1$

x = 7 $\displaystyle x = 7$

Correct answer:

x = -12 $\displaystyle x = -12$

Explanation:

To solve for x, we want x to be by itself. Given the equation

$\frac{x}{-4} = 3$ $\displaystyle \frac{x}{-4} = 3$

we need to multiply both sides of the equation by -4 to get x to stand alone. So,

$\frac{x}{-4} \cdot -4 = 3 \cdot -4$ $\displaystyle \frac{x}{-4} \cdot -4 = 3 \cdot -4$

x = -12 $\displaystyle x = -12$

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Example Question #369 : Linear Equations

Solve for y in the following equation:

$\frac{y}{-6} = 3$ $\displaystyle \frac{y}{-6} = 3$

Possible Answers:

y = 3 $\displaystyle y = 3$

y = -3 $\displaystyle y = -3$

$y = -\frac{1}{2}$ $\displaystyle y = -\frac{1}{2}$

y = 9 $\displaystyle y = 9$

y = -18 $\displaystyle y = -18$

Correct answer:

y = -18 $\displaystyle y = -18$

Explanation:

To solve for y, we want to get y to stand alone. Given the equation

$\frac{y}{-6} = 3$ $\displaystyle \frac{y}{-6} = 3$

to get y to stand alone, we need to multiply both sides by -6. We get

$\frac{y}{-6} \cdot -6 = 3 \cdot -6$ $\displaystyle \frac{y}{-6} \cdot -6 = 3 \cdot -6$

y = -18 $\displaystyle y = -18$

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Example Question #370 : Linear Equations

Solve the equation: $\displaystyle 9+x = 78$

Possible Answers:

$\displaystyle -87$

$\displaystyle 69$

$\displaystyle 72$

$\displaystyle 87$

$\displaystyle -69$

Correct answer:

$\displaystyle 69$

Explanation:

In order to isolate the unknown variable, we will need to subtract nine on both sides of the equation.

$\displaystyle 9+x-9 = 78-9$

The left side simplifies to only x, and the right side of the equation will also simplify.

x=69 $\displaystyle x=69$

The answer is $\displaystyle 69$.

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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #361 : Algebra 1

Example Question #362 : Linear Equations

Example Question #363 : Linear Equations

Example Question #361 : Algebra 1

Example Question #365 : Linear Equations

Example Question #366 : Linear Equations

Example Question #367 : Linear Equations

Example Question #368 : Linear Equations

Example Question #369 : Linear Equations

Example Question #370 : Linear Equations

All Algebra 1 Resources

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