How To Solve For X

Algebra: It's a word that has struck fear into the heart of many a student, and with good reason. Algebra can be hard. You're dealing with unknown amounts, and math suddenly become less concrete. But, as with all math skills, you have to start out with the basic foundation and then build on it. In algebra, solving algebraic equations starts with practicing equations in which you solve for ​x​, which simply means you have to figure out the unknown amount.

1. Algebra's Golden Rule

Learn the golden rule. The first step to solving for ​x​ is going to be getting ​x​ alone on one side of the equation and everything else on the other side. Remember the algebraic golden rule: What you do to one side of the equation, you must do to the other side. That's how the equation stays equal!

2. Start Simple: Solve for x

Start with a simple equation. The most basic algebra equation involves simple addition or subtraction with one unknown quantity, such as

\(2 + x = 7\)

How do you get ​x​ by itself? Subtract 2 from both sides:

\(2 – 2 + x = 7 – 2\)

Now simplify the equation by doing the math:

\(2-2+x=7-2\)
\(0+x=5\)
\(\text{or } x = 5\)

Check your work by substituting the answer, 5, into the equation for ​x​. Does 2 + 5 = 7? Yes, so the correct answer is ​x​ = 5.

3. More Difficult Equation Examples

Increase your level of difficulty. Not every equation is going be simple, so try more difficult equation examples that require more steps. A more difficult equation might be

\(5x – 10 = 5\)

First, get x on one side of the equal sign. To accomplish this, add 10 to both sides:

\(5x – 10 + 10 = 5 + 10\)

That simplifies the equation to

\(5x = 15\)

Now that you've moved the 10, you need to get the 5 away from the ​x​. Divide both sides by 5:

\(\frac{5x}{5} = \frac{15}{5}\)

Simplified, the answer is ​x​ = 3. Check your answer by substituting 3 for ​x​ in the equation. Does (5 × 3) − 10=5? Solving the equation shows (5 × 3) − 10 = 15 − 10 = 5, so the correct answer is ​x​ = 3.

Another level of difficulty happens when a problem when ​x​ has an exponent. For example, consider the problem

\(x^2-11=25\)

You start just like other algebra problems by getting the x term on one side of the equal sign and everything else on the other side. Follow the algebra golden rule by adding 11 to both sides of the equation so that

\(x^2-11+11=25+11\)

Simplifying the equation shows that

\(x^2=36\)

Remembering that ​x​² means ​x​ times ​x​ and reviewing the multiplication tables shows that

\(6 × 6 = 36 \text{ so } x=6\)

Check the answer by replacing x in the equation with 6. Does

\(6^2-11=25 ?\)

Since 6²=36, the equation becomes

\(36-11=25\)

so the correct answer is ​x​ = 6.

4. Equations with Multiple Variables

Continue learning more about algebra. In algebra, you might find some equations that have more than one letter. The equations may work out to where the answer for ​x​ may actually contain another letter itself. An example of this would be

\(5x + 3 = 10y + 18\)

You want to solve for ​x​, just like before, so get ​x​ by itself on one side of the equation. Subtract 3 from both sides:

\(5x + 3 -3 = 10 y + 18 – 3\)

Simplify:

\(5x = 10y + 15\)

Now divide both sides by 5:

\(\frac{5x}{5} = \frac{10y + 15}{5}\)

Simplify:

\(x = 2y + 3\)

And there's your answer!

In this case, checking the answer means substituting the quantity (2y+3) for ​x​ in the equation. The equation becomes

\(5(2y+3)+3=10y+18\)

Multiplying and simplifying the left side of the equation gives you

\(10y+15+3 \text{ or } 10y+18\)

which does equal the right side of the equation, 10​y​+18, so the correct answer is indeed ​x​ = 2​y​ + 3.