One-Step Equations — Free Pre-Algebra Review Games.

Subtract 7 from both sides: x = 12 - 7 = 5.

Add 3 to both sides: x = 10 + 3 = 13.

Divide both sides by 4: x = 20/4 = 5.

Multiply both sides by 3: x = 6 * 3 = 18.

Subtract 15 from both sides: x = 0.

x + 4 = 9, so x = 9 - 4 = 5.

Divide both sides by -5: x = 35 / (-5) = -7.

Multiply both sides by -2: x = 8 * (-2) = -16.

Subtract 3.5: x = 10 - 3.5 = 6.5.

Multiply both sides by 4: x = -3 * 4 = -12.

Multiply both sides by 3/2: x = 8 * 3/2 = 12.

Multiply both sides by -1: x = 15.

Divide both sides by 0.5: x = 7.5 / 0.5 = 15.

Add 2.8 to both sides: x = -1.3 + 2.8 = 1.5.

Multiply both sides by -5: x = 6 * (-5) = -30.

To isolate \(x\), subtract \(9\) from both sides, giving \(x = 14 - 9 = 5\), since subtraction is the inverse operation of addition. The choice \(x = 23\) is wrong because it comes from adding \(9\) instead of subtracting it. Always undo addition with subtraction to keep the equation balanced.

Adding \(6\) to both sides undoes the subtraction, so \(x = 11 + 6 = 17\). The distractor \(x = 5\) results from subtracting \(6\) from \(11\) instead of adding, which is the incorrect inverse operation. Remember that the inverse of subtraction is addition when solving one-step equations.

Dividing both sides by \(3\) isolates \(x\), giving \(x = 21 / 3 = 7\). The option \(x = 63\) comes from multiplying \(21\) by \(3\) instead of dividing, which reverses the correct inverse operation. Division undoes multiplication, so always divide by the coefficient of \(x\).

Multiplying both sides by \(5\) undoes the division, giving \(x = 4 \times 5 = 20\). The choice \(x = 1.25\) mistakenly divides \(4\) by \(5\) instead of multiplying. To solve a division equation, multiply both sides by the divisor.

Subtracting \(20\) from both sides gives \(x = 20 - 20 = 0\), which is a valid solution even though it equals zero. The choice \(x = 40\) incorrectly adds \(20\) instead of subtracting. Zero is a perfectly valid solution to an equation and should not be dismissed.

Adding \(8\) to both sides isolates \(x\), so \(x = 0 + 8 = 8\). The choice \(x = 0\) ignores the \(-8\) term and just copies the right side of the equation. Always perform the inverse operation on both sides rather than leaving a term unaddressed.

Dividing both sides by \(7\) gives \(x = 0 / 7 = 0\), since zero divided by any nonzero number is zero. The choice \(x = 7\) wrongly assumes the coefficient becomes the answer instead of dividing it out. Any equation of the form \(ax = 0\) with \(a \neq 0\) always has \(x = 0\) as its solution.

Multiplying both sides by \(2\) undoes the division, giving \(x = 9 \times 2 = 18\). The choice \(x = 4.5\) divides \(9\) by \(2\) instead of multiplying, which is the wrong inverse operation. To clear a division equation, multiply both sides by the denominator.

Subtracting \(2\) from both sides gives \(x = 2 - 2 = 0\), showing that the solution can be zero when the constants match. The choice \(x = 4\) mistakenly adds instead of subtracting \(2\). Always check whether the numbers on each side cancel out completely before assuming a nonzero answer.

Adding \(10\) to both sides isolates \(x\), giving \(x = 5 + 10 = 15\). The choice \(x = -5\) incorrectly subtracts \(10\) from \(5\) instead of adding it. To undo subtraction in an equation, add the same value to both sides.

Dividing both sides by \(6\) gives \(x = 6 / 6 = 1\). The choice \(x = 6\) mistakenly treats the coefficient as unnecessary and just copies the constant term. Whenever the coefficient and constant are equal, dividing them yields \(x = 1\).

Multiplying both sides by \(6\) undoes the division, giving \(x = 1 \times 6 = 6\). The choice \(x = 1\) incorrectly leaves \(x\) equal to the result of the division rather than solving for the numerator. To solve \(x/a = b\), multiply both sides by \(a\).

Subtracting \(11\) from both sides isolates \(x\), giving \(x = 18 - 11 = 7\). The choice \(x = 29\) comes from adding \(11\) to \(18\) instead of subtracting, which reverses the needed inverse operation. Always subtract the added constant to solve an addition equation.

Since subtracting a negative is the same as adding, the equation becomes \(x + 7 = 15\), so \(x = 15 - 7 = 8\). The choice \(x = 22\) incorrectly adds \(7\) to \(15\) instead of subtracting it after simplifying the double negative. Remember that \(a - (-b) = a + b\) before applying the inverse operation.

Dividing both sides by \(-3\) gives \(x = 27 / (-3) = -9\), and dividing a positive by a negative produces a negative result. The choice \(x = 9\) ignores the negative sign on the coefficient and treats it as positive. When solving multiplication equations, carefully track the sign of the coefficient to determine the sign of the solution.

Multiplying both sides by \(-5\) gives \(x = 4 \times (-5) = -20\), since a positive times a negative yields a negative. The choice \(x = 20\) mistakenly drops the negative sign from the divisor. Always multiply by the exact divisor, including its sign, to reverse a division equation correctly.

Subtracting \(4.2\) from both sides gives \(x = 9 - 4.2 = 4.8\). The choice \(x = 13.2\) mistakenly adds \(4.2\) to \(9\) rather than subtracting it. Decimal equations follow the same inverse-operation rules as whole-number equations, so line up the decimal points carefully when subtracting.

Multiplying both sides by \(3\) gives \(x = -7 \times 3 = -21\), since a negative times a positive is negative. The choice \(x = 21\) incorrectly drops the negative sign from \(-7\). When multiplying to undo division, preserve the sign of the number on the right side of the equation.

Subtracting \(3\) from both sides gives \(x = -8 - 3 = -11\), and the equation can be read the same way even though the constant is written first. The choice \(x = -5\) mistakenly adds \(3\) instead of subtracting it from \(-8\). Equations are symmetric, so it does not matter which side the variable expression is on.

Adding \(5.5\) to both sides isolates \(x\), giving \(x = 2.5 + 5.5 = 8\). The choice \(x = -3\) incorrectly subtracts \(5.5\) from \(2.5\) instead of adding it. To solve subtraction equations with decimals, add the subtracted value to both sides just as you would with whole numbers.

Dividing both sides by \(-6\) gives \(x = -42 / (-6) = 7\), since a negative divided by a negative yields a positive. The choice \(x = -7\) incorrectly assumes the signs do not cancel. When both the coefficient and constant are negative, their quotient is always positive.

Multiplying both sides by \(-3\) gives \(x = -9 \times (-3) = 27\), since two negatives multiply to a positive. The choice \(x = -27\) incorrectly treats the product of two negatives as negative. Remember that a negative divided by a negative is positive, so the original quotient being negative already signals a sign relationship to check.

Subtracting \(\frac{1}{2}\) from both sides gives \(x = \frac{3}{2} - \frac{1}{2} = 1\), since the fractions share a common denominator. The choice \(x = 2\) mistakenly adds the fractions instead of subtracting them. When fractions share a denominator, simply subtract the numerators to isolate the variable.

Adding \(\frac{1}{4}\) to both sides gives \(x = \frac{3}{4} + \frac{1}{4} = 1\), combining the like fractions into a whole number. The choice \(x = \frac{1}{2}\) results from incorrectly subtracting instead of adding the fractions. Fractions with common denominators combine just like whole numbers once you apply the correct inverse operation.

Multiplying both sides by the reciprocal \(\frac{3}{2}\) gives \(x = 4 \times \frac{3}{2} = 6\), which isolates \(x\) correctly. The choice \(x = \frac{8}{3}\) comes from multiplying instead of using the reciprocal to cancel the fraction. When a fraction multiplies \(x\), multiply both sides by its reciprocal to solve.

Multiplying both sides by \(7\) gives \(x = -2 \times 7 = -14\), keeping the negative sign intact. The choice \(x = 14\) drops the negative sign from \(-2\). Always carry negative signs through the multiplication step when undoing a division equation.

Since \(x + (-9)\) is the same as \(x - 9\), adding \(9\) to both sides gives \(x = 5 + 9 = 14\). The choice \(x = -4\) mistakenly subtracts \(9\) from \(5\) instead of adding it. Recognizing that adding a negative is equivalent to subtracting helps simplify equations before solving.

Dividing both sides by \(-4\) gives \(x = 0 / (-4) = 0\), since zero divided by any nonzero number remains zero. The choice \(x = -4\) incorrectly treats the coefficient as the solution instead of dividing it out. Any nonzero coefficient times zero always equals zero, so \(x = 0\) is always the answer in such equations.

Adding \(12\) to both sides gives \(x = -12 + 12 = 0\), since the two values are opposites that cancel to zero. The choice \(x = -24\) mistakenly subtracts \(12\) from \(-12\) instead of adding it. When a constant and its opposite appear on both sides after isolating, the result is often zero.

Multiplying both sides by \(2\) gives \(x = 100 \times 2 = 200\), undoing the division regardless of which side the variable is on. The choice \(x = 50\) incorrectly divides \(100\) by \(2\) instead of multiplying. Equations remain valid no matter which side the variable expression appears on, so apply the same inverse operation.

Subtracting \(3.75\) from both sides gives \(x = 10 - 3.75 = 6.25\). The choice \(x = 13.75\) incorrectly adds the two decimal values instead of subtracting them. Careful decimal subtraction, aligning place values, is essential to avoid small arithmetic errors in one-step equations.

Multiplying both sides by \(-6\) gives \(x = -3 \times (-6) = 18\), since two negatives multiply to a positive. The choice \(x = -18\) incorrectly keeps the result negative despite the sign cancellation. Whenever both sides of a proportion-like equation are negative, the negatives cancel to produce a positive solution.

Multiplying both sides by the reciprocal \(\frac{4}{3}\) gives \(x = 9 \times \frac{4}{3} = 12\), correctly clearing the fractional coefficient in one step. The choice \(x = 6.75\) comes from multiplying \(9\) by \(\frac{3}{4}\) instead of by its reciprocal \(\frac{4}{3}\). When solving equations with a fraction times \(x\), always multiply both sides by the reciprocal of that fraction.

Since \(-x = 22\) means \(-1 \cdot x = 22\), dividing both sides by \(-1\) gives \(x = 22 / (-1) = -22\). The choice \(x = 22\) mistakenly assumes the negative sign disappears without changing the value. Whenever an equation has a lone negative variable, multiply or divide by \(-1\) to isolate the positive variable correctly.

Dividing both sides by \(0.25\) gives \(x = 5 / 0.25 = 20\), since dividing by a decimal less than one increases the result. The choice \(x = 1.25\) comes from multiplying instead of dividing by \(0.25\). When the coefficient is a decimal less than \(1\), dividing by it will always produce a larger number than the original constant.

Adding \(3.6\) to both sides gives \(x = -2.1 + 3.6 = 1.5\), combining a negative and a larger positive decimal. The choice \(x = -5.7\) incorrectly adds the two decimals as if both were negative. When adding a positive and negative decimal, subtract their absolute values and keep the sign of the larger magnitude number.

Multiplying both sides by \(-7\) gives \(x = 5 \times (-7) = -35\), since a positive times a negative is negative. The choice \(x = 35\) incorrectly drops the negative sign attached to the fraction bar. When \(x\) has a negative sign in front of a fraction, treat the entire coefficient of \(x\) as negative when multiplying to solve.

Multiplying both sides by \(\frac{2}{5}\) gives \(x = 15 \times \frac{2}{5} = 6\), clearing the fraction in one step. The choice \(x = 37.5\) comes from multiplying \(15\) by \(\frac{5}{2}\) instead of its reciprocal \(\frac{2}{5}\). Always multiply by the reciprocal of the coefficient fraction, not the fraction itself, to isolate \(x\).

Subtracting \(8.9\) from both sides gives \(x = 2.4 - 8.9 = -6.5\), since subtracting a larger decimal from a smaller one yields a negative result. The choice \(x = 11.3\) incorrectly adds the two decimals instead of subtracting them. When the constant being subtracted is larger than the value on the other side, expect a negative solution.

Multiplying both sides by \(-3\) gives \(x = -11 \times (-3) = 33\), since multiplying two negatives produces a positive value. The choice \(x = -33\) incorrectly keeps the result negative despite both sides being negative. When both the coefficient and constant carry negative signs, their product or quotient becomes positive.

Multiplying both sides by the reciprocal \(\frac{8}{7}\) gives \(x = 21 \times \frac{8}{7} = 24\), since \(21\) divided by \(7\) is \(3\) and \(3 \times 8 = 24\). The choice \(x = 18.375\) comes from multiplying \(21\) by \(\frac{7}{8}\) instead of the reciprocal. Simplifying the fraction multiplication before finishing the arithmetic can prevent errors with the reciprocal method.

Since subtracting a negative is the same as adding, the equation becomes \(x + 5.5 = 3.2\), so \(x = 3.2 - 5.5 = -2.3\). The choice \(x = 8.7\) incorrectly adds \(3.2\) and \(5.5\) instead of subtracting after simplifying the double negative. Always simplify double negatives to addition before applying the correct inverse operation.

Multiplying both sides by \(-4\) gives \(x = -12 \times (-4) = 48\), since multiplying two negatives yields a positive number. The choice \(x = -48\) incorrectly assumes the negatives do not cancel out. When an equation has negative signs on both sides of a division relationship, they cancel to produce a positive solution.

Multiplying both sides by \(\frac{5}{9}\) gives \(x = 18 \times \frac{5}{9} = 10\), since \(18\) divided by \(9\) is \(2\) and \(2 \times 5 = 10\). The choice \(x = 32.4\) results from multiplying \(18\) by \(\frac{9}{5}\) instead of the correct reciprocal \(\frac{5}{9}\). Simplifying before multiplying can help avoid confusing which fraction to use as the reciprocal.

Multiplying both sides by \(-4\) gives \(x = -3.5 \times (-4) = 14\), since a negative times a negative produces a positive result. The choice \(x = -14\) incorrectly assumes the negative signs do not cancel out. Whenever both the divisor and the quotient are negative, the resulting numerator must be positive.