Linear equations and equivalent transformations

Solve for all permitted values, rather than one successful guess

An equation asserts that two expressions have the same value. A solution is a permitted value of the unknown that makes this assertion true. Unless a context says otherwise, this lesson works over the real numbers. Solving means finding every such value; checking one candidate only tells you about that candidate.

We now extend one-step equations and rearranging formulas to repeated unknowns, brackets and unknowns on both sides. For instance, 3x − 4 = 2x + 5 is not a sequence of operations applied to one unknown on only one side. Both expressions depend on the same value of x.

Collect terms with the same variable part

A term is a part separated by addition or subtraction; its sign belongs with it. In 3x + 2 − x + 4, the variable terms are 3x and −x, and the constant terms are 2 and 4. Since 3x − x = (3 − 1)x = 2x, the expression becomes 2x + 6 for every real x.

Collecting like terms uses the distributive property in reverse. It does not permit 3x + 2 = 5x: the constant 2 is not two copies of x. A coefficient of 1 is usually unwritten, so x = 1x; similarly, −x = (−1)x. Keep a minus sign attached to its entire term when reorganising a sum.

Apply a bracket’s multiplier to every term

The distributive property gives 2(x + 3) = 2x + 6. A negative multiplier also acts on every term: −2(x − 3) = −2x + 6. To simplify 5 − (2x + 1), regard subtraction of the bracket as adding its negative: 5 − 2x − 1 = 4 − 2x.

Replacing an expression by an equal expression is valid for every permitted input. This differs from changing an equation by applying an operation to both sides. Expanding only one side is legitimate when that side’s value stays identical; subtracting a term from only one side generally changes the equation.

Choose transformations that work in both directions

Equivalent equations have exactly the same solutions in the stated domain. Adding or subtracting the same expression on both sides preserves the solutions when that expression is defined throughout the domain: the operation can be undone. Multiplying or dividing both sides by the same fixed nonzero number is also reversible.

Multiplying by zero can destroy information. Multiplying x = 3 by zero produces 0 = 0, which accepts every real x rather than only 3. Dividing by x is not an automatically permitted step, because x might be zero; in 2x = x, that step would discard the valid solution zero. This lesson clears numerical denominators only. Variable denominators and squaring need additional domain and implication checks in later lessons.

Bring variable terms together by an explicit equal operation

For 3x − 4 = 2x + 5, subtract 2x from both sides. Simplification gives x − 4 = 5. Add 4 to both sides to obtain x = 9. Substituting in the original gives 3 × 9 − 4 = 23 and 2 × 9 + 5 = 23, so the candidate is a solution.

A term does not move across the equals sign by itself. The familiar change of sign records addition or subtraction performed on both sides. Another valid route subtracts 3x first, yielding −4 = −x + 5, then −9 = −x, and finally 9 = x. Choose a convenient route and record enough steps to explain why it preserves the solutions.

Clear nonzero numerical denominators across the whole equation

To solve (x − 2)/3 = (x + 4)/6, multiply each complete side by 6, a nonzero common multiple. The left side becomes 2(x − 2) and the right side becomes x + 4. Expand to 2x − 4 = x + 4, subtract x and add 4 to obtain x = 8. The original sides both equal 2.

In an equation with several terms, the multiplier applies to every term, including those without a denominator. For x/2 + 1 = x/3 + 4, multiplying by 6 gives 3x + 6 = 2x + 24, and thus x = 18. Multiplying only the fraction terms would create a different equation. Keeping exact fractions avoids introducing rounding errors into exact algebra.

Check the coefficient before dividing

After simplification, these equations have the form ax + b = cx + d, with fixed real coefficients. Subtract cx and then b from both sides to obtain (a − c)x = d − b. Write A = a − c and B = d − b, so the final relation is Ax = B.

If A is nonzero, dividing by A gives the unique real solution x = B/A. If A is zero, division is unavailable. When B is also zero, 0 = 0 is true for every real x. When B is nonzero, 0 = B is false for every real x. These three cases follow from the reduced equation, not from the number of sample candidates that happened to pass.

Inspect the same candidates through a reversible chain

The model below begins with a stated original equation, expands it where needed, then collects the variable and constant terms. Choose a candidate independently of the transformation stage. At every stage, the truth of the equality for that candidate stays the same; the actual numerical values of the two sides can change.

For a zero coefficient, the chain ends with a true or false numerical statement instead of a division by zero. The finite candidate list illustrates the transformations; the symbolic classification identifies the complete real solution set. All displayed arithmetic is exact.

Check the original relation, units and permitted values

An exact algebraic check substitutes the result into the original equation, including its brackets and denominators. It can catch an arithmetic or copying error. The reversible derivation explains why no other permitted solutions were missed. If you used an approximate display, keep the exact result for this check.

A context can restrict an otherwise real unknown. Suppose two stated total-cost models are C = 5 + 2n and C = 2 + 4n, where n counts indivisible items. Equality gives n = 3/2. This is the unique real algebraic solution, but it is not a permitted whole-item count, so the models have no equal total at an allowed n. In a continuous-quantity model the same numerical value could be allowed. Name the quantity, units, assumptions and domain before interpreting the answer.

Worked examples

Try it yourself

1. Simplify 7x − 3 + 2 − 4x. Explain why 3x − 1 is valid.

   Show the answer and reasoning

   3x − 1

   The variable coefficients combine as 7 − 4 = 3; the constants combine as −3 + 2 = −1. This preserves the expression’s value for every real x.

2. Solve −2(x − 3) = x and check the original.

   Show the answer and reasoning

   x = 2

   Distribute to get −2x + 6 = x, then subtract x and 6 to get −3x = −6. Divide by −3. The original has −2(2 − 3) = 2 on the left and 2 on the right.

3. Solve x/2 + 1 = x/3 + 4 using a nonzero common multiplier.

   Show the answer and reasoning

   x = 18

   Multiply every term by 6: 3x + 6 = 2x + 24. Subtract 2x and 6. In the original, 18/2 + 1 = 10 and 18/3 + 4 = 10.

4. Solve 2x = x. Why would dividing by x lose an answer?

   Show the answer and reasoning

   The unique solution is x = 0.

   Subtract x from both sides to get x = 0. Dividing by x assumes x is nonzero, so it removes precisely the permitted value that satisfies the original equation.

5. Classify 3(x + 2) = 3x + 6 and 3(x + 2) = 3x + 5 over the real numbers.

   Show the answer and reasoning

   The first has every real number as a solution; the second has none.

   Expanding and subtracting 3x gives 6 = 6 for the first and 6 = 5 for the second. A true identity does not force x to be zero; a contradiction does not depend on x.

6. For whole-item counts n ≥ 0, can 5 + 2n equal 2 + 4n?

   Show the answer and reasoning

   No permitted whole-item count makes the totals equal.

   The equation gives 2n = 3, hence n = 3/2. It passes the real algebraic check, but it is not an integer count. The original models and the contextual domain must both be satisfied.

What to carry forward

• Collect like terms and distribute to every term in a bracket before choosing an equation-solving step.
• Equivalent transformations preserve the entire solution set; explain the equal operation and its reversibility.
• After reducing to Ax = B, inspect A before division: the result can be one solution, all real values or no solution.
• Check the original equation and interpret the result within the problem’s units, assumptions and permitted values.