6.2: Solving Linear Equations - One or Two-Steps
- Page ID
- 173478
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Definitions and Theorems
Solving a linear equation in one variable is the most basic equation-solving task in algebra, and the technique behind it—isolating the variable by applying the properties of equality—underlies nearly every method that follows. In this section you review the one- and two-step cases.
A linear equation in one variable is an equation that can be written in the standard form\[ax + b = 0,\nonumber\]where \(a\) and \(b\) are real numbers and \(a \neq 0\).
A solution of an equation in the variable \(x\) is a value that produces a true statement when substituted for \(x\). The set of all such values is the solution set of the equation.
Two equations are equivalent if they have the same solution set.
Every linear equation in one variable has exactly one solution. To find it, you replace the equation with a chain of equivalent equations, each simpler than the last, until the variable stands alone. The tools that produce these equivalent equations are the properties of equality.
Let \(a\), \(b\), and \(c\) be real numbers. Each of the following operations, applied to both sides of a true equation, produces an equivalent equation.
Addition Property. If \(a = b\), then \(a + c = b + c\).
Subtraction Property. If \(a = b\), then \(a - c = b - c\).
Multiplication Property. If \(a = b\), then \(ac = bc\).
Division Property. If \(a = b\) and \(c \neq 0\), then \(\dfrac{a}{c} = \dfrac{b}{c}\).
Solving a linear equation means isolating the variable—rewriting the equation until the variable sits alone on one side and a single number sits on the other. You reach that form by undoing operations with their inverses: undo addition with subtraction, and undo multiplication with division. Whatever operation you apply to one side, you must apply to the other, so that the two sides remain equal at every step.
When the variable is multiplied by a number, as in \(3x = 18\), you undo the multiplication by dividing both sides by that number, giving \(x = 6\). A frequent error is to subtract the coefficient instead, writing \(x = 18 - 3\). Match the operation to what is actually being done to the variable: division undoes multiplication, not subtraction.
Examples
Solve \(x - 9 = 4\).
- Answer
- The variable is decreased by \(9\), so add \(9\) to both sides to isolate \(x\).\[\begin{aligned} x - 9 &= 4 \\ x - 9 + 9 &= 4 + 9 \\ x &= 13 \end{aligned}\nonumber\]Substituting \(x = 13\) into the original equation gives \(13 - 9 = 4\), a true statement, so the solution is \(x = 13\).
Solve \(x + 6 = 1\).
- Answer
- The variable is increased by \(6\), so subtract \(6\) from both sides.\[\begin{aligned} x + 6 &= 1 \\ x + 6 - 6 &= 1 - 6 \\ x &= -5 \end{aligned}\nonumber\]Substituting \(x = -5\) gives \(-5 + 6 = 1\), which is true, so the solution is \(x = -5\).
Solve \(7x = 42\).
- Answer
- The variable is multiplied by \(7\), so divide both sides by \(7\).\[\begin{aligned} 7x &= 42 \\ \dfrac{7x}{7} &= \dfrac{42}{7} \\ x &= 6 \end{aligned}\nonumber\]Substituting \(x = 6\) gives \(7(6) = 42\), which is true, so the solution is \(x = 6\).
Solve \(\dfrac{x}{5} = 3\).
- Answer
- The variable is divided by \(5\), so multiply both sides by \(5\).\[\begin{aligned} \dfrac{x}{5} &= 3 \\ 5 \cdot \dfrac{x}{5} &= 5 \cdot 3 \\ x &= 15 \end{aligned}\nonumber\]Substituting \(x = 15\) gives \(\dfrac{15}{5} = 3\), which is true, so the solution is \(x = 15\).
Solve \(3x + 4 = 19\).
- Answer
- Undo the operations in reverse order. First subtract \(4\) from both sides to isolate the variable term, then divide both sides by \(3\).\[\begin{aligned} 3x + 4 &= 19 \\ 3x + 4 - 4 &= 19 - 4 \\ 3x &= 15 \\ \dfrac{3x}{3} &= \dfrac{15}{3} \\ x &= 5 \end{aligned}\nonumber\]Substituting \(x = 5\) gives \(3(5) + 4 = 19\), which is true, so the solution is \(x = 5\).
Solve \(5 - 2x = 13\).
- Answer
- First subtract \(5\) from both sides, then divide both sides by \(-2\). Dividing by a negative number is handled the same way as dividing by any nonzero number.\[\begin{aligned} 5 - 2x &= 13 \\ 5 - 2x - 5 &= 13 - 5 \\ -2x &= 8 \\ \dfrac{-2x}{-2} &= \dfrac{8}{-2} \\ x &= -4 \end{aligned}\nonumber\]Substituting \(x = -4\) gives \(5 - 2(-4) = 5 + 8 = 13\), which is true, so the solution is \(x = -4\).
Sources
Several parts of this text use modifications from the following sources:
- Wikipedia article: "Equality (mathematics)"
- Wikipedia article: "Linear equation"
All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.