A linear equation in one variable is an equation involving constants and a single variable which only occurs to the first power. The following equations are linear equations in one variable:

$$2x + 1 = 3$$

$$-4x - 5 = 7x + 13$$

$$\dfrac{2}{3}x - 7 = \dfrac{1}{3} - \dfrac{1}{6}x$$

The problem is to solve for the variable. To do this, you may:
• Perform standard arithmetic operations on either side alone, such as multiplying out or combining terms.
• Add, subtract, multiply, or divide both sides by the same thing. You may not multiply or divide by 0.

The primary goal is to try to get all the variable terms on one side and all the number terms on the other. Once you've done this, the equation can be solved easily.

I'll illustrate with the equations I gave above.

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Example. Solve $2x + 1 = 3$ .

$$\matrix{& 2x & + & 1 & = & 3 \cr - & & & 1 & & 1 \cr \noalign{\vskip2pt\hrule\vskip2pt} & 2x & & & = & 2 \cr \div & 2 & & & & 2 \cr \noalign{\vskip2pt\hrule\vskip2pt} & x & & & = & 1 \cr}$$

The answer is $x = 1$ .

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Example. Solve $-4x - 5 = 7x + 13$ .

$$\matrix{& -4x & - & 5 & = & 7x & + & 13 \cr - & & & 5 & & & & 5 \cr \noalign{\vskip2pt\hrule\vskip2pt} & -4x & & & = & 7x & + & 8 \cr - & 7x & & & & 7x & & \cr \noalign{\vskip2pt\hrule\vskip2pt} & -11x & & & = & & & 8 \cr \div & -11 & & & & & & -11 \cr \noalign{\vskip2pt\hrule\vskip2pt} & x & & & = & & & -\dfrac{8}{11} \cr}$$

The answer is $x = -\dfrac{8}{11}$ .

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Note: In the first two examples, I arranged the work "vertically". You can also work line-by-line as in the next example. Use the approach that works best for you.

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Example. Solve $\dfrac{2}{3}x - 7 = \dfrac{1}{3} - \dfrac{1}{6}x$ .

If an equation has fractions, it's often a good idea to multiply both sides by the least common multiple of the denominators to clear the fractions. You're less likely to make a mistake when dealing with whole numbers. In this case, the denominators are 3 and 6, and their least common multiple is 6. Hence, I'll start by multiplying both sides by 6.

$$\eqalign{ \dfrac{2}{3}x - 7 &= \dfrac{1}{3} - \dfrac{1}{6}x \cr 6 \cdot \left(\dfrac{2}{3}x - 7\right) &= 6 \cdot \left(\dfrac{1}{3} - \dfrac{1}{6}x\right) \cr 6 \cdot \dfrac{2}{3}x - 6 \cdot 7 &= 6 \cdot \dfrac{1}{3} - 6 \cdot \dfrac{1}{6}x \cr 4x - 42 &= 2 - x \cr 4x - 42 + 42 + x &= 2 - x + 42 + x \cr 5x &= 44 \cr \dfrac{1}{5} \cdot 5x &= \dfrac{1}{5} \cdot 44 \cr x &= \dfrac{44}{5} \quad\halmos \cr}$$

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Example. Solve $0.23x + 0.4 = 5(6 - 0.7x)$ .

If an equation has decimals, it's a good idea to clear them if possible. The "0.23" has the largest number of digits (2 digits) to the right of the decimal point. To clear decimals, I need to multiply both sides by $10^2 = 100$ (think "2 zeroes").