One-Step Equations and Checking Solutions

What is a one-step equation, and what is a solution?

A one-step equation can be solved by applying one inverse operation to both sides to isolate its variable. A solution is a value that makes the original equation's two sides equal. For example, \(x+4=11\) becomes \(x=7\) after subtracting 4 from both sides.

The equals sign asserts equality, not “do the next calculation.” The expression \(x+4\) alone describes a value; the equation \(x+4=11\) supplies a condition on that value.

This lesson works with real-number solutions unless a context restricts them. All example situations, self-checks, and the diagram below were independently created.

Why the same operation must be applied to both sides

For real numbers A and B with \(A=B\), the following are true:

Adding a fixed quantity can be undone by subtracting it. Multiplying by a nonzero constant can be undone by dividing by it. These reversible steps produce equivalent equations, with the same solutions.

Multiplying both sides by zero makes \(0=0\) regardless of their earlier values, so it can erase the original condition. Dividing by zero is undefined. Neither is a valid way to isolate a variable while preserving the solution set.

See an equality-preserving step

Subtracting 4 only on the left would produce \(x=11\), which does not solve the original equation: \(11+4=15\ne11\). Both sides must receive the same operation.

Choose the inverse operation

Here a and b are known real constants:

Choose from the operation acting on the variable, not merely the sign of the right-hand number. In \(-4x=20\), the variable is multiplied by \(-4\), so divide both sides by \(-4\).

“Move the number and change its sign” can hide the reasoning. Write the operation on both sides first; cancellation explains the result.

A solve-and-check routine

1. Identify the variable and the operation acting on it.
2. Apply the inverse operation to both complete sides.
3. Simplify to isolate the variable. Write the answer as \(x=\text{value}\), not just a bare number.
4. Substitute that value into the original equation.
5. Calculate the left and right sides separately and confirm they match. Check units and context restrictions too.

One-step refers to the equation transformation; arithmetic and the final check may require several written lines.

Worked example: undo addition

Problem: Solve \(x+9=-5\).

\[x+9-9=-5-9,\qquad x=-14.\]

Check in the original: \(-14+9=-5\), equal to the right side. The solution is \(x=-14\). Adding 9 again would not undo the addition already attached to x.

Worked example: undo subtraction

Problem: Solve \(y-8=-3\).

\[y-8+8=-3+8,\qquad y=5.\]

Check: \(5-8=-3\). Add 8 to undo subtracting 8; do not subtract 8 from both sides again.

Be careful with signed constants: \(x-(-4)=9\) is \(x+4=9\). Subtracting 4 from both sides gives \(x=5\), checked by \(5-(-4)=9\).

Worked example: undo multiplication with a negative coefficient

Problem: Solve \(-4x=20\).

\[\frac{-4x}{-4}=\frac{20}{-4},\qquad x=-5.\]

Check: \((-4)(-5)=20\). The negative coefficient must be included in the divisor. In \(-x=6\), the coefficient is \(-1\); dividing by \(-1\) gives \(x=-6\).

Worked example: undo division

Problem: Solve \(x/6=-4\).

Multiply each side by 6:

\[6\left(\frac{x}{6}\right)=6(-4),\qquad x=-24.\]

Check: \(-24/6=-4\). The divisor 6 is nonzero, so multiplication by 6 reverses this division.

This rule describes a constant denominator. An equation such as \(6/x=3\) has the variable in the denominator and also requires \(x\ne0\); it is not the same form as \(x/6=3\).

Worked example: a fractional coefficient

Problem: Solve \(\frac23x=10\).

Multiply both sides by the reciprocal \(\frac32\):

\[\frac32\cdot\frac23x=10\cdot\frac32,\qquad x=15.\]

The coefficient product is 1, so only x remains. Check: \(\frac23\cdot15=10\). Multiplying by \(\frac23\) again would not undo the original multiplication.

For a negative coefficient, \(-\frac23x=10\) instead gives \(x=-15\), using reciprocal \(-\frac32\). Check: \((-\frac23)(-15)=10\).

Worked example: adding a fraction to both sides

Problem: Solve \(y-\frac5{12}=\frac14\).

\[y=\frac14+\frac5{12}=\frac3{12}+\frac5{12}=\frac8{12}=\frac23.\]

The first line adds \(5/12\) to both sides; then a common denominator makes the arithmetic possible. Check: \(\frac23-\frac5{12}=\frac8{12}-\frac5{12}=\frac3{12}=\frac14\).

Worked example: a decimal coefficient

Problem: Solve \(0.4x=3.6\).

\[x=\frac{3.6}{0.4}=9.\]

Divide both sides by 0.4, not by 4. Check: \(0.4(9)=3.6\). Keep exact decimals or fractions during the solution whenever possible rather than rounding a candidate before checking it.

Checking a proposed value without solving first

Problem: Determine whether 6 or 7 is a solution of \(3x=21\).

For \(x=6\), the left side is \(3(6)=18\), while the right side is 21. Since \(18\ne21\), 6 is not a solution.

For \(x=7\), the left side is \(3(7)=21\), matching the right side. Therefore 7 is a solution. Checking tests a supplied value; solving finds the value from the equation.

Do not check only the rewritten final statement \(x=7\). A copied sign or arithmetic error earlier could give a wrong candidate that still satisfies that final statement. Return to the original condition.

Worked example: the variable on the right

Problem: Solve \(18=3n\).

Divide both sides by 3: \(18/3=3n/3\), so \(6=n\), equivalently \(n=6\). Equality works in either direction.

Check in the original: \(18=3(6)=18\). The variable need not begin on the left for an equation to be solved.

A one-step quantity model

Problem: Four identical packets contain 36 cards in total. How many cards are in each packet?

Let n be the number of cards per packet, a nonnegative whole number. Then \(4n=36\), so \(n=36/4=9\).

Check the situation: four packets with nine cards each contain 36 cards. The answer is 9 cards per packet, not just 9 with no quantity identified. If a count problem produced a noninteger answer, check whether the stated assumptions allow such a count.

A zero coefficient is a special case

The formula \(x=b/a\) for \(ax=b\) requires \(a\ne0\). If the coefficient is zero, inspect the resulting statement instead:

• \(0x=0\) is true for every real x: all real numbers are solutions.
• \(0x=5\) is false for every real x: there is no solution.

Do not divide either equation by zero. The usual nonzero-coefficient one-step equations in this lesson have a unique real solution; these degenerate cases do not.

Common mistakes

• Operating on only one side: Apply the same inverse operation to both complete sides.
• Undoing subtraction with subtraction: Use addition to undo subtraction.
• Dropping a negative coefficient: Divide by the whole signed coefficient.
• Confusing addition with multiplication: \(x+4\) needs subtraction; \(4x\) needs division.
• Using the same fraction instead of its reciprocal: To undo multiplication by \(2/3\), multiply by \(3/2\).
• Losing decimal scale: Dividing by 0.4 is different from dividing by 4.
• Treating zero as a permitted divisor: Check the nonzero condition before dividing.
• Checking only the final rewritten equation: Substitute into the original equation and compare both sides.
• Ignoring the meaning of the variable: A real-number solution may fail a counting or measurement restriction.

Check your understanding

Solve and check each answer in the original equation. These self-checks do not modify the separate Quizzes or Practice tabs.

1. \(x+7=-2\).
2. \(y-6=-10\).
3. \(-5a=30\).
4. \(b/8=-3\).
5. \(\frac34c=9\).
6. \(d-\frac23=\frac16\).
7. \(0.25m=2.5\).
8. \(-n=11\).
9. \(28=4p\).
10. Is \(q=5\) a solution of \(q+9=15\)? Find the actual solution.
11. Five identical boxes hold 45 pencils. Define a variable, write a one-step equation, and solve.
12. Compare the solution sets of \(0x=0\) and \(0x=-2\).

Answers and checks

1. Subtract 7: \(x=-9\). Check: \(-9+7=-2\).
2. Add 6: \(y=-4\). Check: \(-4-6=-10\).
3. Divide by \(-5\): \(a=-6\). Check: \((-5)(-6)=30\).
4. Multiply by 8: \(b=-24\). Check: \(-24/8=-3\).
5. Multiply by \(4/3\): \(c=12\). Check: \(\frac34(12)=9\).
6. Add \(2/3\): \(d=\frac16+\frac46=\frac56\). Check: \(\frac56-\frac46=\frac16\).
7. Divide by 0.25: \(m=10\). Check: \(0.25(10)=2.5\).
8. Divide by \(-1\): \(n=-11\). Check: \(-(-11)=11\).
9. Divide both sides by 4: \(p=7\). Check: \(28=4(7)\).
10. No: \(5+9=14\ne15\). The solution is \(q=6\), checked by \(6+9=15\).
11. Let k count pencils per box. \(5k=45\), so \(k=9\) pencils per box. Check: \(5(9)=45\).
12. \(0x=0\) has every real x as a solution; \(0x=-2\) has no solution because \(0\ne-2\).

Related lessons

• Variables and Algebraic Expressions: substitution and the meaning of an equation
• Commutative, Associative, and Identity Properties: identities and inverse operations
• Multiplying and Dividing Fractions: use a reciprocal coefficient
• Calculating with Decimals: divide decimal coefficients accurately
• Two-Step and Multi-Step Equations: extend the same equality principle
• Word-Problem Models: define variables and check contextual solutions

Sources and authorship

The publicly accessible educational resources below were consulted for topic coverage and mathematical checks. This lesson's explanations, example settings, tables, self-checks, and equation diagram were independently created; source prose, exercises, and artwork are not reproduced. Public access does not imply unrestricted reuse: check each resource's current license before reproducing material.

• OpenStax Prealgebra 2e, 8.1: Solve Equations Using the Subtraction and Addition Properties of Equality
• OpenStax Prealgebra 2e, 8.2: Solve Equations Using the Division and Multiplication Properties of Equality
• OpenStax Prealgebra 2e: preface and reuse conditions