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Without the order of operations, arithmetic would be ambiguous — it would be like trying to speak a language that has no grammar. The order of operations provides a common framework for everyone to use so that we know we are all calculating comparably. Any time you use math, you will rely on these rules! Below we walk you through 12 examples of arithmetic and manipulating operations.

**1. Solve the expression** \((3 + 5) * 10 \)

Parentheses come first in the order of operations, so start by adding 3 and 5.

\(8 * 10\)

Now multiply.

\(80\)

**2. Solve** \((3x + 3) * 10 = 40\)** for x**

In this case, the variable is inside the parentheses. To isolate the variable, first get the expression in the parentheses alone. This allows you to drop the parentheses and continue isolating the variable.

\((3x + 3) = \frac{40}{10} = 4\)

\(3x + 3 = 4\)

Subtract the 3 on the left-hand side of the equals sign, and then divide both sides by 3.

\(3x = 1\)

\(x = \frac{1}{3} \)

**3. Solve** \(\frac{x^2}{4} = 5 – 1\) **for x.**

First, simplify the right-hand side.

\(\frac{x^2}{4} = 4\)

Next, multiply both sides by 4 to isolate the \(x^2.\)

\(x^2 = 16\)

To undo the exponent, take the square root of both sides. When you’re undoing an even exponent, remember to add \( \pm \) in front of your solution. If you’re undoing an odd exponent, keep the existing sign.

\(\sqrt{x^2} = \sqrt{16}\)

\(x = \pm 4\)

**4. Simplify** \(\frac{3x + 18}{18}\)**.**

Each term in this fraction is a multiple of 3, so we can divide the entire thing by 3.

\(=\frac{3(x + 6)}{3 * 6}\)

\(=\frac{x + 6}{6}\)

Since the x doesn’t have a factor of 6, we can’t simplify the 6’s on the top and bottom any further. If the expression had been \(\frac{6x}{6}\), we could simplify to x.

**5. Solve** \(\frac{x^7 *x^{-4}}{x^6+4x^6} = \frac{-1}{135}\)

The numerator can be simplified using the rules of exponents: \(x^7 * x^{-4} = x^{7-4} = x^3\)

The denominator only requires addition to simplify: \(x^6 + 4x^6 = 5x^6\)

So the left hand side of the equation becomes \(\frac{x^3}{5x^6}\)

This fraction can be simplified by further applying exponent rules:

\(\frac{x^3}{5x^6} = \frac{x^3*x^{-6}}{5} = \frac{x^{-3}}{5} = \frac{1}{5x^3}\)

Our initial equation has now become:

\(\frac{1}{5x^3} = \frac{-1}{135}\)

Cross multiply:

\(-135 = 5x^3\)

Divide by 5:

\(-27 = x^3\)

Take the cube root of both sides:

\(\sqrt[3]{-27} = \sqrt[3]{x^3}\)

\(x = -3\)

**6. Simplify** \(\frac{x^{\frac{3}{2}}}{x^{-4}}\)**.**

We are going to eventually combine the exponents; if you encounter a problem that has multiple exponents, and at least one of the exponents is negative, address the negative exponent(s) before the others.

To simplify the negative exponent, use the inverse rule \((x^{-1} = \frac{1}{x}, \frac{1}{x^{-1}} = x)\).

\(x^{\frac{3}{2}} * x^4\)

Combine the powers using the product rule for exponents:

\(x^{\frac{3}{2}} * x^4 = x^{\frac{3}{2}} * x^{\frac{8}{2}} = x^{\frac{3}{2}+\frac{8}{2}} = x^{\frac{11}{2}}\)

To simplify the fractional exponent, use the power over root rule:

\(x^{\frac{11}{2}} = \sqrt{x^{11}}\)

**7. Solve** \(5x + 3 \geq 28\) **for x.**

We will treat the inequality the same way we would an equation (see question 7 for an exception). First, subtract 3 from the left-hand side.

\(5x \geq 25\)

Then, divide both sides by 5:

\(x \geq 5\)

**8. Solve** \(31 < -2x – 1\) **for x.**

The major thing to watch out for with inequalities is that you will need to flip the direction of the inequality sign if you multiply or divide by a negative number.

First, add 1 to the left-hand side of the inequality:

\(32 < -2x\)

Then divide by -2 and **flip the direction of the inequality sign **(see herefor a brief explanation of why):

\(-16 > x \)

**9. Solve the expression** \(3 log_2 \, (4)\, – 5 log_2 \, (8)\)

First, use the log rule that allows you to turn the coefficient into an exponent:

\(log_2 \,(4^3)\, – log_2 \, (8^5)\)

Now combine the terms using the difference of two logs rule:

\(log_2 \, (4^3)\, – log_2 \, (8^5) = log_2 \, (\frac{4^3}{8^5})\)

We can express the numerator and denominator as powers of 2 to help with simplification.

\(4= 2^2\) and \(8 = 2^3\)

\(log_2\, (\frac{4^3}{8^5}) = log_2\, (\frac{(2^2)^3}{(2^3)^5}) = log_2\, (\frac{2^6}{2^{15}})\)

Using rules of exponents, we can simplify the fraction inside the log:

\(\frac{2^6 }{ 2^{15} }= 2^6 * 2^{-15} = 2^{6-15} = 2^{-9}\)

Our initial expression is now \(log_2\, (2^{-9})\)

Now using the definition of a logarithm, we get the answer:

\(log_2\, (2^{-9}) = -9 \)

**10. A logarithm with base e (2.718…) is called the Natural Logarithm and written as** \(ln\,(x)\)

**Simplify** \(ln\,(4) + ln\,(5)\)

Using the addition rule of logs:

\(ln\,(4) + ln\,(5) = ln\,(4*5) = ln\,(20)\)

\(ln\,(20)\) is our final answer.

**11. Simplify the equation** \(|x+5| + 3*4 = 22\)**.**

First isolate the absolute value term:

\(|x+5| + 12 = 22\)

\(|x+5| = 10\)

Now we can lift the absolute value bars with the understanding that our final answer is 10 away from zero on the number line in both the positive and negative directions. We will therefore need two equations to fully solve for the values of x.

\(x+5 = 10\)

and

\(x+5 = -10\)

Solving for x in both of these equations give us our answers:

\(x = 5\) and \(x = -15\)

**12. Simplify the expression** \((3x+6)(2x-2)\)**.**

FOIL the expression

(F)irst → \(3x*2x = 6x^2\)

(O)utside → \(3x*-2 = -6x\)

(I)nside → \(6*2x = 12x\)

(L)ast → \(6*-2 = -12\)

Combine these four terms:

\(6x^2-6x+12x-12\)

Combine like terms:

\(6x^2+6x-12\)

**13. Simplify the expression** \((x + 2)(x – 1)(x + 4)\)**.**

FOIL the first two terms:

\((x^2 + x – 2)(x + 4)\)

Now distribute each term in the first parentheses to both the the terms in the second parentheses:

\((x^3 + 4x^2) + (x^2 + 4x) + (-2x – 8)\)

Then combine like terms:

\(x^3 + 5x^2 – 2x – 8\)

Feel like you’ve got the hang of it? Try our 5-question Words to Math quiz below.