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Solving Equations with Variables on Both Sides Practice
Fifty original questions covering the complete solving equations with variables on both sides lesson, with explanations and SAT-focused strategy.
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Question 1
Explanation
Subtract \(3x\) from both sides: \(2x+2=10\). Subtract \(2\): \(2x=8\). Divide by \(2\): \(x=4\). Checking gives \(22\) on both sides.
Question 2
Explanation
Subtract \(5x\): \(2x-4=8\). Add \(4\): \(2x=12\). Divide by \(2\): \(x=6\). Both original sides then equal \(38\).
Question 3
Explanation
Subtract \(x\) from each side to get \(x+9=15\). Subtract \(9\): \(x=6\). This demonstrates that a variable term may be removed from the right by a legal equality operation.
Question 4
Explanation
Subtract \(4x\) from both sides: \(1=2x-7\). Add \(7\): \(8=2x\). Divide by \(2\): \(x=4\).
Question 5
Explanation
Add \(2x\) to both sides: \(9=3x\). Divide by \(3\): \(x=3\). The check gives \(9-6=3\).
Question 6
Explanation
Subtract \(2x\) from both sides: \(x=-11\). Substitution gives \(-33\) on both sides, so the negative solution is valid.
Question 7
Explanation
Subtract \(x\): \(12=3x\). Divide by \(3\): \(x=4\). Both original sides equal \(16\).
Question 8
Explanation
Subtract \(6x\): \(-5=2x+3\). Subtract \(3\): \(-8=2x\). Divide by \(2\): \(x=-4\).
Question 9
Explanation
Subtracting \(5x\) from both sides gives \(4x+4=28\), leaving a positive coefficient. Subtracting \(9x\) is legal but produces \(-4x\) on the right.
Question 10
Explanation
Subtracting \(3x\) or subtracting \(7x\) from both sides are both valid. They create different-looking intermediate equations but preserve the same solution set.
Question 11
Explanation
Subtracting \(2x\) from both complete sides yields \(4x-2x-3=11\), or \(2x-3=11\). The other options do not apply a valid operation to both sides.
Question 12
Explanation
A term changes sides only as the visible result of an equality operation. Subtracting \(3x\) from both sides preserves equality and explains why an opposite term appears.
Question 13
Explanation
Solving gives \(3x=15\), so \(x=5\). Substitution produces \(24\) on the left and \(24\) on the right, directly verifying the solution.
Question 14
Explanation
Three more than twice \(n\) is \(2n+3\). Seven less than four times \(n\) is \(4n-7\). Equality gives \(2n+3=4n-7\).
Question 15
Explanation
Adding \(3x\) to both sides gives \(11=8x-13\), leaving positive \(8x\). Subtracting \(5x\) is legal but leaves \(-8x\) on the left.
Question 16
Explanation
Distribute: \(3x+6=2x+11\). Subtract \(2x\): \(x+6=11\). Subtract \(6\): \(x=5\).
Question 17
Explanation
Distribute and combine: \(4x-12+2=2x+10\Rightarrow4x-10=2x+10\). Subtract \(2x\), add \(10\), and divide by \(2\): \(x=10\).
Question 18
Explanation
Distribute: \(10x-5=3x+16\). Subtract \(3x\): \(7x-5=16\). Add \(5\): \(7x=21\), so \(x=3\).
Question 19
Explanation
Distribute: \(-6x+8=5-x\). Add \(6x\): \(8=5+5x\). Subtract \(5\): \(3=5x\), so \(x=3/5\).
Question 20
Explanation
Distribute: \(7-3x+6=2x+8\), so \(13-3x=2x+8\). Add \(3x\), subtract \(8\): \(5=5x\), giving \(x=1\).
Question 21
Explanation
Distribute and combine on the left: \(2x+10-3x+3=8\Rightarrow-x+13=8\). Subtract \(13\): \(-x=-5\), so \(x=5\).
Question 22
Explanation
Subtract \(\frac14x\): \(\frac14x+3=8\). Subtract \(3\): \(\frac14x=5\). Multiply by \(4\): \(x=20\).
Question 23
Explanation
Multiply every term by \(6\): \(2x-12=x+30\). Subtract \(x\), then add \(12\): \(x=42\).
Question 24
Explanation
Subtract \(\frac12x\): \(\frac14x-1=4\). Add \(1\): \(\frac14x=5\). Multiply by \(4\): \(x=20\).
Question 25
Explanation
Subtract \(0.3x\): \(0.5x+1.4=5.9\). Subtract \(1.4\): \(0.5x=4.5\). Divide by \(0.5\): \(x=9\).
Question 26
Explanation
Subtract \(0.4x\): \(0.8x-3.6=2.8\). Add \(3.6\): \(0.8x=6.4\). Divide by \(0.8\): \(x=8\).
Question 27
Explanation
Distribute: \(0.25x+1=0.5x-2\). Subtract \(0.25x\) and add \(2\): \(3=0.25x\). Divide by \(0.25\): \(x=12\).
Question 28
Explanation
Multiplying each coefficient and constant by \(10\) yields \(6x-12=2x+28\). Every term on both sides must receive the factor.
Question 29
Explanation
The LCD \(12\) gives \(12(x/4)+12(1/3)=12(x/6)+12(5/2)\), so \(3x+4=2x+30\).
Question 30
Explanation
Subtracting \(2x\) leaves \(17=7x-4\), with positive coefficient \(7\). Subtracting \(9x\) is valid but leaves \(-7x\) on the left.
Question 31
Explanation
Simplify both sides: \(3x-6+4x=2x+10+9\Rightarrow7x-6=2x+19\). Subtract \(2x\), add \(6\): \(5x=25\), so \(x=5\).
Question 32
Explanation
Simplify inside: \(3(x+1)-4=3x-1\). Then \(6x-2=5x+6\). Subtract \(5x\) and add \(2\): \(x=8\).
Question 33
Explanation
Distribute: \(5-2+2x=3x+7\), so \(3+2x=3x+7\). Subtract \(2x\): \(3=x+7\). Subtract \(7\): \(x=-4\).
Question 34
Explanation
Add \(7x\) to both sides: \(3x+9=-6\). Subtract \(9\): \(3x=-15\). Divide by \(3\): \(x=-5\).
Question 35
Explanation
Add \(5x\): \(12=2+2x\). Subtract \(2\): \(10=2x\). Divide by \(2\): \(x=5\).
Question 36
Explanation
Let \(n\) be the number. Write \(2n+9=5n-12\). Subtract \(2n\) and add \(12\): \(21=3n\), so \(n=7\).
Question 37
Explanation
Set equal costs: \(18+6m=42+3m\). Subtract \(3m\) and \(18\): \(3m=24\), so \(m=8\). Both plans then cost \(66\) dollars.
Question 38
Explanation
Set the equal length expressions together: \(3w+2=5w-10\). Subtract \(3w\) and add \(10\): \(12=2w\), so \(w=6\).
Question 39
Explanation
Solve first: subtract \(4x\) and add \(18\) to get \(21=3x\), so \(x=7\). Then \(2x-5=14-5=9\).
Question 40
Explanation
The equal sign separates two expressions. To put the variable terms on one side, subtract \(2x\) from both sides, producing \(4x-5=19\). Directly combining across \(=\) is not legal.
Question 41
Explanation
Multiply by \(6\): \(2(2x-1)+3(x+4)=5x-2\). Simplify: \(4x-2+3x+12=5x-2\), so \(7x+10=5x-2\), \(2x=-12\), and \(x=-6\).
Question 42
Explanation
Simplify left: \(2-(x-5)=7-x\), so \(21-3x=2x+15\). Subtract \(15\), add \(3x\): \(6=5x\), giving \(x=6/5\).
Question 43
Explanation
Distribute: \(0.45x-1.7=0.15x+0.9+2.6=0.15x+3.5\). Subtract \(0.15x\), add \(1.7\): \(0.30x=5.2\), so \(x=52/3\).
Question 44
Explanation
Multiply by \(10\): \(2x+7=5x-8\). Add \(8\), subtract \(2x\): \(15=3x\), so \(x=5\).
Question 45
Explanation
Substitute \(a=3\): \(3(2x-5)=4x+9\). Distribute: \(6x-15=4x+9\). Then \(2x=24\), so \(x=12\).
Question 46
Explanation
Subtracting \(6x\) from both sides gives \(2x-6x+5=-7\), so the variable term is \(-4x\), not \(+4x\). The operation is legal; the coefficient subtraction was wrong.
Question 47
Explanation
Choice A simplifies left to \(9x-6-8=9x-14\) and right to \(4x+10+11=4x+21\). The other choices produce different constants or coefficients.
Question 48
Explanation
The integers are \(n,n+1,n+2\). Write \(n+(n+1)+(n+2)=2(n+2)+18\). Simplifying gives \(3n+3=2n+22\), so \(n=19\).
Question 49
Explanation
Simplify left: \(3(x-2)+6=3x\), then \(\frac23(3x)=2x\). Right: \(x+5-7=x-2\). Thus \(2x=x-2\), so \(x=-2\).
Question 50
Explanation
Simplify: \(10x-15+4=3x+21+2x\), so \(10x-11=5x+21\). Then \(5x=32\), \(x=32/5\), and \(x+6=62/5\).
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Questions to review
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- Question 1Variables on both sidesEasy
- Question 2Variables on both sidesEasy
- Question 3Collect variable termsEasy
- Question 4Strategic collectionEasy
- Question 5Signed variable termsEasy
- Question 6Variables on both sidesEasy
- Question 7Collect variable termsEasy
- Question 8Strategic collectionEasy
- Question 9Strategic first stepsEasy
- Question 10Equivalent solution pathsEasy
- Question 11Equality propertiesEasy
- Question 12Legal algebra reasoningEasy
- Question 13Solution verificationEasy
- Question 14Verbal equationsEasy
- Question 15Strategic collectionEasy
- Question 16DistributionMedium
- Question 17Distribution and like termsMedium
- Question 18DistributionMedium
- Question 19Negative distributionMedium
- Question 20Negative distributionMedium
- Question 21Multiple groupingMedium
- Question 22Fractional coefficientsMedium
- Question 23Clearing denominatorsMedium
- Question 24Fractional coefficientsMedium
- Question 25Decimal coefficientsMedium
- Question 26Decimal coefficientsMedium
- Question 27Decimal distributionMedium
- Question 28Clearing decimalsMedium
- Question 29Clearing denominatorsMedium
- Question 30Strategic collectionMedium
- Question 31Multi-step both sidesMedium
- Question 32Nested distributionMedium
- Question 33Negative distributionMedium
- Question 34Negative coefficientsMedium
- Question 35Negative coefficientsMedium
- Question 36Verbal equationsMedium
- Question 37Break-even modelsMedium
- Question 38Geometric relationshipsMedium
- Question 39Solve then evaluateMedium
- Question 40Error analysisMedium
- Question 41Complex fraction equationsHard
- Question 42Nested groupingHard
- Question 43Decimal distributionHard
- Question 44Fraction equationsHard
- Question 45Parameter substitutionHard
- Question 46Error analysisHard
- Question 47Reverse reasoningHard
- Question 48Consecutive-number equationsHard
- Question 49Complex simplificationHard
- Question 50Solve then evaluateHard