| Primary source | OpenStax College Algebra 2e, Section 2.2: "Linear Equations in One Variable" |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable |
College Algebra 2e Section 2.2 begins on page 139 of the PDF.
An equation is a scale in balance.
Both sides weigh the same. Whatever you do to one side, you must do to the other, or the scale tips and the equation becomes false.
This one principle -- equal operations on both sides preserve equality -- is the entire foundation of algebraic equation-solving. There is nothing else. Every technique you will encounter in this course and in calculus is an application of this principle to more complicated situations.
The goal when solving is isolation: get the variable alone on one side. You achieve this by applying inverse operations -- the operations that "undo" what was done to the variable.
The two properties of equality.
| Property | Statement | Meaning |
|---|---|---|
| Addition property | If $A = B$, then $A + C = B + C$ | Add the same number to both sides |
| Multiplication property | If $A = B$, then $AC = BC$ (for $C \neq 0$) | Multiply both sides by the same nonzero number |
Subtraction is adding a negative. Division is multiplying by a reciprocal. So two properties cover everything.
Order of inverse operations (work from outside in):
One-step equations require one inverse operation.
Example 1. Solve $x + 9 = 14$.
The operation done to $x$ is addition of 9. The inverse is subtraction of 9. Subtract 9 from both sides: \[ x + 9 - 9 = 14 - 9 \] \[ x = 5 \]
Check: $5 + 9 = 14$. $\checkmark$
Example 2. Solve $7x = 42$.
The operation done to $x$ is multiplication by 7. The inverse is division by 7. Divide both sides by 7: \[ \frac{7x}{7} = \frac{42}{7} \] \[ x = 6 \]
Check: $7(6) = 42$. $\checkmark$
Example 3. Solve $x - 4 = -11$.
Add 4 to both sides: \[ x = -7 \]
Check: $-7 - 4 = -11$. $\checkmark$
Example 4. Solve $\dfrac{x}{5} = -3$.
Multiply both sides by 5: \[ x = -15 \]
Check: $-15/5 = -3$. $\checkmark$
Two-step equations require two inverse operations. Work from the outside in: undo addition/subtraction first, then multiplication/division.
Why "outside in"? When the equation was built, someone started with $x$, then multiplied it by something, then added something. To reverse the process, undo the last thing done first (the addition), then undo the first thing done (the multiplication).
Example 5. Solve $4x - 3 = 13$.
Step 1. Add 3 to both sides (undo subtraction): \[ 4x = 16 \]
Step 2. Divide by 4 (undo multiplication): \[ x = 4 \]
Check: $4(4) - 3 = 16 - 3 = 13$. $\checkmark$
Example 6. Solve $\dfrac{x}{2} + 7 = 1$.
Step 1. Subtract 7 from both sides: \[ \frac{x}{2} = -6 \]
Step 2. Multiply both sides by 2: \[ x = -12 \]
Check: $-12/2 + 7 = -6 + 7 = 1$. $\checkmark$
Negative coefficient. Dividing by a negative number is just as valid as dividing by a positive -- the result may change sign.
Example 7. Solve $-8x = 56$.
Divide by $-8$: \[ x = \frac{56}{-8} = -7 \]
Check: $-8(-7) = 56$. $\checkmark$
Note: Multiplying or dividing by a negative in an equation is fine. (This differs from inequalities, where doing so requires flipping the inequality sign.)
Equation with fraction coefficients. Multiply both sides by the reciprocal of the coefficient.
Example 8. Solve $\dfrac{3}{4}x = 9$.
Multiply both sides by $\dfrac{4}{3}$ (the reciprocal of $\dfrac{3}{4}$): \[ x = 9 \cdot \frac{4}{3} = 12 \]
Check: $\dfrac{3}{4}(12) = 9$. $\checkmark$
When more than two operations are present, apply inverse operations in sequence, always working from outside to inside (constants before coefficients).
Example 9. Solve $3(x + 2) = 15$.
Option A (distribute first): \[ 3x + 6 = 15 \implies 3x = 9 \implies x = 3 \]
Option B (divide first, since the entire $(x+2)$ is multiplied by 3): \[ x + 2 = 5 \implies x = 3 \]
Both paths give $x = 3$. Option B is faster when there is no variable on the right side of the parentheses.
Check: $3(3 + 2) = 3(5) = 15$. $\checkmark$
Example 10. Solve $2(3x - 1) + 4 = 18$.
Step 1. Subtract 4: $2(3x - 1) = 14$.
Step 2. Divide by 2: $3x - 1 = 7$.
Step 3. Add 1: $3x = 8$.
Step 4. Divide by 3: $x = \dfrac{8}{3}$.
Check: $2(3 \cdot \frac{8}{3} - 1) + 4 = 2(8 - 1) + 4 = 14 + 4 = 18$. $\checkmark$
Checking a solution is not a formality. It is the only way to be certain your answer is correct.
Three reasons to always check:
How to check: Substitute your solution back into the original (unmodified) equation and verify that both sides equal the same number.
Example 11. A student solved $5x - 3 = 22$ and got $x = 5$.
Check: $5(5) - 3 = 25 - 3 = 22$. $\checkmark$ Correct.
Example 12. A student solved $5x - 3 = 22$ and got $x = 4$.
Check: $5(4) - 3 = 20 - 3 = 17 \neq 22$. Incorrect. The correct answer is $x = 5$.
| Error | Example | Correction |
|---|---|---|
| Adding to one side only | $x - 7 = 12 \to x = 12$ | Add 7 to both sides: $x = 19$ |
| Dividing the wrong side | $4x = 20 \to x = 4 \cdot 20$ | Divide both sides: $x = 5$ |
| Wrong order (coefficient before constant) | $4x - 3 = 13 \to x = (13/4) - 3$ | Remove constant first: $4x = 16$, then $x = 4$ |
| Sign error with negative | $-3x = 12 \to x = 4$ | Divide by $-3$: $x = -4$ |
| Not checking the answer | Solution accepted without substitution | Always check in the original equation |
Problem 1. Solve: $x + 15 = -3$.
Subtract 15: $x = -18$.
Check: $-18 + 15 = -3$. $\checkmark$
Problem 2. Solve: $-6x = 48$.
Divide by $-6$: $x = -8$.
Check: $-6(-8) = 48$. $\checkmark$
Problem 3. Solve: $\dfrac{2}{3}x = 8$.
Multiply by $\dfrac{3}{2}$: $x = 12$.
Check: $\frac{2}{3}(12) = 8$. $\checkmark$
Problem 4. Solve: $5x + 11 = -4$.
Subtract 11: $5x = -15$.
Divide by 5: $x = -3$.
Check: $5(-3) + 11 = -15 + 11 = -4$. $\checkmark$
Problem 5. Solve: $\dfrac{x - 3}{4} = -2$.
Multiply by 4: $x - 3 = -8$.
Add 3: $x = -5$.
Check: $\dfrac{-5 - 3}{4} = \dfrac{-8}{4} = -2$. $\checkmark$
Problem 6. Solve: $2(4x - 5) = 30$.
Divide by 2: $4x - 5 = 15$.
Add 5: $4x = 20$.
Divide by 4: $x = 5$.
Check: $2(4 \cdot 5 - 5) = 2(15) = 30$. $\checkmark$
Problem 7. A number is tripled and then 8 is subtracted. The result is 19. Find the number.
Let $n$ = the number.
$3n - 8 = 19$
$3n = 27$
$n = 9$
Check: $3(9) - 8 = 27 - 8 = 19$. $\checkmark$
Problem 8. Explain why the equation $x + 5 = x + 7$ has no solution without doing any algebra.
The equation claims that adding 5 to $x$ gives the same result as adding 7 to $x$. For any real number $x$, adding 7 always gives a result 2 more than adding 5. No value of $x$ makes these equal. (Algebraically: subtract $x$ from both sides to get $5 = 7$, which is false.)
Problem 9. Find all values of $a$ for which $ax + 3 = 10$ has exactly one solution in $x$, and all values for which it has no solution.
Rearrange: $ax = 7$.
If $a \neq 0$: $x = 7/a$. Exactly one solution for every $a \neq 0$.
If $a = 0$: the equation becomes $0 = 7$, which is false for all $x$. No solution.
Summary: exactly one solution when $a \neq 0$; no solution when $a = 0$.
Think of an equation as a recipe in reverse.
Someone took a number, performed operations on it in a specific order, and arrived at a result. You see only the result and the operations. Your job is to reverse the recipe, step by step, to recover the original ingredient.
If the recipe is "take $x$, multiply by 4, subtract 3, get 13," then to recover $x$:
The key word is "undo" -- every algebraic operation has an inverse, and solving is just applying inverses in reverse order.
For students who find math intimidating: If you can consistently solve two-step equations and check your answers, you have the algebraic foundation needed for this course. This lesson is a reset point -- everything else builds on it.
For students interested in proof: The addition and multiplication properties of equality are consequences of the field axioms. Any structure that satisfies those axioms supports the same equation-solving procedure. Modular arithmetic (used in cryptography) is one example: the same steps solve equations there, with the same inverse-operation logic.
For students interested in careers: Equation solving is the calculation layer beneath every formula you will ever use. Ohm's law ($V = IR$), the ideal gas law ($PV = nRT$), Newton's second law ($F = ma$): in every case, the practical task is "given some values, find the one that is unknown." That is a one-step or two-step equation in each case.