Equality and one-step equations

Read the equals sign in both directions

The statement 3 + 4 = 7 says that the expression on the left and the expression on the right have the same value. The reversed statement 7 = 3 + 4 is also true. Equality is symmetric: swapping the sides does not change the relationship. The equals sign does not mean that an answer must always appear on its right.

Keep every written equality true. The chain 3 + 4 = 7 = 14 is false because seven and fourteen are different numbers. If you want to double seven, write a separate calculation, 2 × 7 = 14. In an equation such as x + 3 = 5, the task is to find which permitted values of x make the two sides equal.

Ask which inputs make the equation true

A candidate is a value you are considering. It becomes a solution only if it belongs to the stated domain and makes the original equality true. For the scalar equation x + 3 = 5 with all real numbers permitted, x = 2 is a solution because 2 + 3 = 5. The candidate x = 5 is not: its left side is eight while its right side is five. The solution set contains every permitted value that works.

An identity is true at every input in its domain. For example, 0x = 0 is true for every real x because zero times any real number is zero. By contrast, 0x = 5 has no real solutions. Equations need not have exactly one solution. Distinguish an equation with one solution, an identity with all permitted inputs as solutions, and an equation with no permitted solutions.

Preserve the whole solution set

Two equations are equivalent on the same stated domain when they have exactly the same solutions. A reversible step gives a reason that no solution is lost or added: each solution of the original equation satisfies the new one, and reversing the step takes every solution of the new equation back to the original.

For x + 3 = 5, subtracting three from both sides gives x = 2. Adding three to both sides reverses that step. This establishes the single solution over the real numbers; it is stronger than trying a few inputs. Keep the original domain throughout. A rearrangement does not make an excluded input admissible.

Add or subtract the same amount on both sides

If two real-number expressions have equal values, adding the same known number to both preserves equality. Subtracting that same number reverses the change. Subtraction is also reversible: adding back the subtracted number recovers the original statement. Write the operation on both complete sides rather than altering only one side.

Subtracting three from x + 3 = 5 produces (x + 3) − 3 = 5 − 3, which simplifies to x = 2. Subtracting three only from the left would incorrectly produce x = 5. Negative numbers follow the same rule: adding three to x − 3 = −5 gives (x − 3) + 3 = −5 + 3, hence x = −2.

Choose the operation that undoes the stated change

For x + c = b, subtract the known number c from both sides to obtain x = b − c. For x − c = b, add c to both sides to obtain x = b + c. This is the reason behind the shorthand sometimes described as moving a term across the equals sign: the actual justification is the same operation on both sides.

The known numbers may be negative or fractional. From x + 1/2 = −3/2, subtract 1/2 on both sides: x = −3/2 − 1/2 = −2. From x − (−3) = 5, add −3 on both sides: x = 5 + (−3) = 2. Preserve parentheses around a signed number and check the operation that the original equation actually specifies.

Use a nonzero factor for a reversible multiplication step

Multiplying both sides by the same known nonzero real number preserves equality and can be reversed by dividing by that number. Dividing both sides by a known nonzero number can likewise be reversed by multiplication. A negative factor is permitted for equations: equal values remain equal after both are multiplied by −2, for example.

Zero is different. Multiplying x = 2 by zero produces 0x = 0, but the new equation is true for every real x. The step has added solutions, so it does not give an equivalent equation. Division by zero is undefined and cannot reverse it. Equality-preserving alone is therefore not enough; a solving step must also preserve the solution set in the reverse direction.

Divide by the whole nonzero coefficient

In ax = b with a known nonzero coefficient a, divide both sides by a to obtain x = b/a. For 3x = 3/2, this gives x = (3/2) ÷ 3 = 1/2. For −2x = −10, divide both sides by −2 to get x = (−10) ÷ (−2) = 5. In each case, multiplication by the original coefficient reverses the step.

A fractional coefficient is still one complete factor. From (3/4)x = −3/2, divide by 3/4, equivalently multiply both sides by its reciprocal 4/3: x = (−3/2) × (4/3) = −2. The coefficient 3/4 is nonzero. Check by substitution: (3/4) × (−2) = −3/2. Exact fractions keep the equality exact without rounding.

Multiply to undo a quotient

In x/a = b, the known divisor a must be nonzero. Multiply both sides by a to obtain x = ab. For x/4 = 0, multiply both sides by four: x = 4 × 0 = 0. The original check is 0/4 = 0; a zero numerator is allowed because the divisor is four.

For x/(−2) = 3/2, multiplying both sides by −2 gives x = (−2) × (3/2) = −3. Check the original quotient: (−3) ÷ (−2) = 3/2. These equations divide x by a known nonzero number. A denominator that itself contains the unknown requires additional domain reasoning and is outside this one-step sequence.

Inspect a zero coefficient without dividing by it

When the coefficient is zero, evaluate what the equation says before choosing an inverse operation. In 0x = 0, the left side is zero for every real input. The equation reduces to the true statement 0 = 0, so all real numbers are solutions. There is no unique value to isolate and no legitimate division by zero.

In 0x = 5, the left side is still always zero, while the right side is five. No real input can make 0 = 5 true. More generally, 0x = b has all real solutions when b = 0 and none when b ≠ 0. If the original context restricts x, the all-solution case means all inputs in that permitted domain.

Check the original statement and the meaning of the unknown

Substitute a proposed solution into the original equation and evaluate the two sides independently. For x − 3 = −5 at x = −2, the left side is (−2) − 3 = −5 and the right side is −5. Their equality confirms this candidate. The reversible solving steps explain why it is the only real solution; the substitution check alone confirms the supplied case.

Keep units compatible when the unknown names a quantity. If an exact length L satisfies L + 3 cm = 5 cm, subtract 3 cm from both sides to obtain L = 2 cm. A context can also exclude an algebraic value. If n counts complete groups, each containing four objects, 4n = 10 has real solution n = 5/2, but no nonnegative integer solution. Two and a half complete groups do not satisfy that count model.

Compare both sides at a candidate, then justify the solution

Start with x + 3 = 5 and candidate x = 2. The explorer evaluates the original left side as five and the right side as five. One fixed scalar number axis places their numerical values on separate marker rows. Matching positions show equal values at this candidate; the diagram locates numbers and does not represent a physical balance.

Choose one of the seven equations and one of the five exact candidates −2, 0, 1/2, 2 and 5. The 7 × 5 = 35 states show whether the selected candidate satisfies the original equality. Signed and fractional values remain exact. Alongside the original equation, a reversible inverse step gives the solution for each of the five equations with a unique solution. The two zero-coefficient cases instead explain all real solutions or no real solutions without dividing by zero.

Use the samples to inspect a candidate and catch a mismatch. Five selected inputs do not exhaust the real domain. For 0x = 0, the reason every real input works is the rule that zero times any real number is zero; for 0x = 5, that same rule excludes every real input. The full solution claim rests on the arithmetic structure and reversible reasoning, not just the plotted examples.

Finish with a justified answer, then connect it to formulas

Evaluation starts with a supplied input and finds an expression’s value. Checking a candidate compares both sides of an equation at that input. Solving identifies all permitted inputs that make the equation true and justifies that the answer is complete. For x + 3 = 5, evaluating x + 3 at two gives five; solving explains why two is the only real input that satisfies the equality.

Read the original operation, state the domain, apply the same reversible operation to both sides, and check the result in the original equation. Report a value with its units or an all-solution or no-solution conclusion as appropriate. The next lesson, Formulas and rearranging simple relationships, connects these ideas to rules with named quantities and units.

Worked examples

Try it yourself

1. Do x + 3 = 5 and 5 = x + 3 have the same real solutions? Check x = 2 and x = 5 in the original statement. What is wrong with writing 2 + 3 = 5 = 8?

   Show the answer and reasoning

   Swapping sides preserves the solutions. At x = 2 both sides are five; at x = 5 they are eight and five, so five is not a solution. The chain is false because five is not eight.

   The equals sign states equal values in either direction. Every equality in a written chain must be true, and a candidate must satisfy the original statement.

2. Solve x + 1/2 = −3/2 and x − (−3) = 5 over the real numbers. Name the same operation applied to both sides and check each original equation.

   Show the answer and reasoning

   Subtracting 1/2 gives x = −2; the check is −2 + 1/2 = −3/2. Adding −3 to both sides of the second equation gives x = 2; the check is 2 − (−3) = 5.

   Undo the written addition or subtraction. A signed number is one operand, so preserve its parentheses and use the same reversible operation on both sides.

3. Solve −2x = −10 and 3x = 3/2 over the real numbers. Does the negative coefficient prevent division? Check both answers in their original equations.

   Show the answer and reasoning

   Division by −2 gives x = 5, checked by (−2) × 5 = −10. Division by three gives x = 1/2, checked by 3 × (1/2) = 3/2. A negative nonzero coefficient is permitted.

   The essential restriction is that the divisor is nonzero. Dividing by the entire coefficient is reversible and keeps an exact fractional result exact.

4. Solve x/4 = 0 and x/(−2) = 3/2 over the real numbers. Why is the zero in the first equation allowed, and why would replacing its divisor with zero be a different problem?

   Show the answer and reasoning

   Multiplying by four gives x = 0, and 0/4 = 0. Multiplying by −2 gives x = −3, and (−3)/(−2) = 3/2. A zero numerator is allowed with a nonzero divisor; division by zero is undefined.

   Undo a quotient by multiplying both sides by its known nonzero divisor. Preserve the difference between a zero numerator and a zero denominator.

5. Give the real solution sets of 0x = 0 and 0x = 5. If x = 2 is multiplied by zero to produce 0x = 0, do the two equations still have the same solutions? Can division by zero recover x = 2?

   Show the answer and reasoning

   0x = 0 has all real numbers as solutions; 0x = 5 has none. Multiplying x = 2 by zero adds solutions, so the new equation is not equivalent. Division by zero is undefined and cannot recover the unique original solution.

   Zero times every real input is zero. A step used to solve must preserve solutions in both directions, rather than merely turn a true equality into another true equality.

6. An exact length L satisfies L + 3 cm = 5 cm and must be positive. Solve the equation with units, check it, and decide whether L = 2 m is the same candidate.

   Show the answer and reasoning

   Subtract 3 cm from both sides to obtain L = 2 cm. The check is 2 cm + 3 cm = 5 cm, and this positive length is admissible. L = 2 m is 200 cm, so it is a different candidate and does not satisfy the equation.

   The unknown names a length quantity, and addition requires compatible units. A matching numeral without a matching unit does not identify the same quantity.

7. In the explorer, x = 1/2 makes 3x = 3/2 true. Does checking that candidate alone establish that no other real input works? How can you justify the full solution set? Why are the five working samples for 0x = 0 not by themselves a proof for every real input?

   Show the answer and reasoning

   The check confirms 1/2 as a solution. Reversible division of 3x = 3/2 by three gives x = 1/2 and establishes uniqueness. For 0x = 0, every real input works because zero times any real number is zero, not merely because five samples agree.

   Separate evaluating at a supplied input, checking a candidate, and solving over the whole stated domain. The completeness claim needs reasoning that covers every permitted input.

What to carry forward

• Equality states that two values agree and can be read in either direction; a solution must satisfy the original equation in its stated domain.
• Equivalent equations have the same solution set. Adding or subtracting the same known number on both sides is reversible.
• Multiplying or dividing both sides by the same known nonzero number is reversible, including negative numbers and fractions.
• A zero coefficient requires an all-solution or no-solution check; division by zero is never an inverse solving step.
• Check the original equality, units and context, and use reversible reasoning to justify the full solution set rather than relying only on selected candidates.