Algebra 1 : Linear Equations

Study concepts, example questions & explanations for Algebra 1

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Example Questions

Example Question #161 : How To Solve Two Step Equations

Solve the equation for \(\displaystyle x\):

\(\displaystyle 3x-5=13\)

Possible Answers:

\(\displaystyle x=5\)

\(\displaystyle x=-6\)

\(\displaystyle x=2.67\)

\(\displaystyle x=10\)

\(\displaystyle x=6\)

Correct answer:

\(\displaystyle x=6\)

Explanation:

Start by adding \(\displaystyle 5\) on both sides of the equation to yield:

\(\displaystyle 3x=18\)

In order to get \(\displaystyle x\) by itself, we need to divide both sides of the equation by \(\displaystyle 3\)

\(\displaystyle \frac{3x}{3}=\frac{18}{3}\).

This gives us

\(\displaystyle x=6\)

Example Question #170 : How To Solve Two Step Equations

\(\displaystyle 4x+7=-29\)

Possible Answers:

\(\displaystyle x=-9\)

\(\displaystyle x=7\)

\(\displaystyle x=-5.5\)

\(\displaystyle x=9\)

\(\displaystyle x=-7\)

Correct answer:

\(\displaystyle x=-9\)

Explanation:

First, isolate variables and constants on either side of the equal sign. In this case, subtract \(\displaystyle 7\) from both sides. This gives us:

\(\displaystyle 4x=-36\).

Now divide both sides by \(\displaystyle 4\) to give us:

\(\displaystyle x=-9\).

Example Question #701 : Algebra 1

Solve for \(\displaystyle x: (2x+4) = 64\)

Possible Answers:

\(\displaystyle x = 66\)

\(\displaystyle x = 34\)

\(\displaystyle x = 30\)

\(\displaystyle x = 62\)

Correct answer:

\(\displaystyle x = 30\)

Explanation:

\(\displaystyle (2x+4) = 64\) State the equation.

\(\displaystyle 2x = 60\) Subtract \(\displaystyle 4\) from both sides. (Substitution property of equality)

\(\displaystyle x = 30\) Divide both sides by \(\displaystyle 2\). (Division property of equality)

Example Question #702 : Algebra 1

Solve the equation: \(\displaystyle \sqrt{2x} = 6\)

Possible Answers:

\(\displaystyle x = 38\)

\(\displaystyle x = 9\)

\(\displaystyle x = 18\)

\(\displaystyle x = 34\)

\(\displaystyle x = 72\)

Correct answer:

\(\displaystyle x = 18\)

Explanation:

\(\displaystyle \sqrt{2x} = 6\) State the equation.

\(\displaystyle 2x = 36\) Square both sides.

\(\displaystyle x = 18\) Divide both sides by \(\displaystyle 2\).

Example Question #703 : Algebra 1

Solve the equation: \(\displaystyle (x+3)^3 = 27\)

Possible Answers:

\(\displaystyle x = 6\)

\(\displaystyle x = 3\)

\(\displaystyle x = 88\)

\(\displaystyle x= 0\)

\(\displaystyle x = 94\)

Correct answer:

\(\displaystyle x= 0\)

Explanation:

\(\displaystyle (x+3)^3 = 27\) State the equation.

\(\displaystyle x+3 = 3\) Cube root both sides.

\(\displaystyle x = 0\) Subtract \(\displaystyle 3\) from both sides.

Example Question #704 : Algebra 1

What order of operations would result in solving the equation for \(\displaystyle s\)?

\(\displaystyle (4s+7)^3 = 125\)

Possible Answers:

Taking the root, then division, then addition

Cubing, then division, then addition

Multiplication, then taking the root, then subtraction

Division, then taking the root, then subtraction

Taking the root, then subtraction, then division

Correct answer:

Taking the root, then subtraction, then division

Explanation:

\(\displaystyle (4s+7)^3 = 125\)

\(\displaystyle (4s+7) = 5\) Take the root.

\(\displaystyle 4s = -2\) Subtract \(\displaystyle 7\) from both sides.

\(\displaystyle s = -\frac{1}{2}\) Divide both sides by \(\displaystyle 4\) and simplify.

Example Question #701 : Linear Equations

What order of operations would be used to solve the equation \(\displaystyle \frac{x-3}{3} = 3\)?

Possible Answers:

Multiplication, then subtraction

Division, then subtraction

Addition, then multiplication

Division, then addition

Multiplication, then addition

Correct answer:

Multiplication, then addition

Explanation:

\(\displaystyle \frac{x-3}{3} = 3\) State the problem (not an actual step, but useful!)

\(\displaystyle x-3 = 9\) Multiply both sides by \(\displaystyle 3\).

\(\displaystyle x = 12\) Add \(\displaystyle 3\) to both sides.

Example Question #706 : Algebra 1

What is the least number of operations (not counting simplification of fractions) needed to solve for \(\displaystyle z\)?

\(\displaystyle (3z)^5 = 32z^5\)

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle 3\)

\(\displaystyle 5\)

\(\displaystyle 2\)

\(\displaystyle 4\)

Correct answer:

\(\displaystyle 2\)

Explanation:

\(\displaystyle (3z)^5 = 32z^5\) State the problem (not a required step, but it helps!)

\(\displaystyle 3z = 2z\) Take the fifth root of both sides. (Step 1)

\(\displaystyle z = 0\) Subtract \(\displaystyle 2z\) from both sides. (Step 2)

Example Question #702 : Algebra 1

Solve:  \(\displaystyle \frac{10}{x} = 40\)

Possible Answers:

\(\displaystyle \frac{1}{40}\)

\(\displaystyle \frac{2}{5}\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle 40\)

\(\displaystyle 400\)

Correct answer:

\(\displaystyle \frac{1}{4}\)

Explanation:

To solve, multiply both sides by \(\displaystyle x\).

\(\displaystyle \frac{10}{x} \cdot x= 40 \cdot x\)

Simplify the left side by cancelling the \(\displaystyle x\) and multiplying the right side.

\(\displaystyle 10=40x\)

Divide both sides by \(\displaystyle 40\) to isolate \(\displaystyle x\).

\(\displaystyle \frac{10}{40}=\frac{40x}{40}\)

\(\displaystyle x=\frac{10}{40} = \frac{1}{4}\)

The answer is:  \(\displaystyle \frac{1}{4}\)

Example Question #702 : Algebra 1

Solve for \(\displaystyle x\).

\(\displaystyle 6x+24=48\)

Possible Answers:

\(\displaystyle 4\)

\(\displaystyle 8\)

\(\displaystyle 5\)

\(\displaystyle 7\)

\(\displaystyle 6\)

Correct answer:

\(\displaystyle 4\)

Explanation:

\(\displaystyle 6x+24=48\) Subtract \(\displaystyle 24\) on both sides.

\(\displaystyle 6x=24\) Divide \(\displaystyle 6\) on both sides.

\(\displaystyle x=4\)

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