Practice
Equations with No Solution and Identity Practice
Fifty original questions covering the complete equations with no solution and identity lesson, with explanations and SAT-focused strategy.
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Question 1
Explanation
A nonzero variable coefficient remains. Solving gives \(2x=8\), so \(x=4\). Therefore the equation has exactly one solution.
Question 2
Explanation
The two sides are identical expressions. Subtracting \(4x\) leaves \(1=1\), a true statement for every real \(x\), so the equation is an identity.
Question 3
Explanation
Subtracting \(5x\) leaves \(-7=2\), which is false. No choice of \(x\) changes that contradiction, so there is no solution.
Question 4
Explanation
The variable has canceled and the remaining statement is true. Therefore every valid value satisfies the original equation; it is an identity.
Question 5
Explanation
The variable has disappeared and \(8=-1\) is false. No value of the missing variable can make unequal constants equal, so the equation has no solution.
Question 6
Explanation
For every real \(x\), \(0x=0\). The equation is an identity, so its solution set is all real numbers.
Question 7
Explanation
The left side is always \(0\), so it can never equal \(9\). The equation represents the false statement \(0=9\) and has no solution.
Question 8
Explanation
The false statement \(6=10\) cannot be repaired by choosing \(x\); the variable already canceled. The correct conclusion is no solution.
Question 9
Explanation
An identity has equivalent expressions on both sides. Because they agree for every real input, the solution set is all real numbers.
Question 10
Explanation
The variable coefficients \(3\) and \(5\) differ, so collection leaves a nonzero variable coefficient. Solving gives \(x=5\), exactly one solution.
Question 11
Explanation
Both the variable coefficients and constants match. Subtracting one side's terms leaves \(0=0\), so the equation is an identity.
Question 12
Explanation
The variable coefficients match but the constants differ. Subtracting \(-2x\) leaves \(6=-5\), so no value works.
Question 13
Explanation
A nonzero coefficient remains, so divide by \(-4\) to get \(x=-5\). The equation has exactly one solution.
Question 14
Explanation
One successful substitution proves only that the tested value is a solution. An identity requires equivalent expressions or a proof that every valid value works.
Question 15
Explanation
A true constant statement after complete legal simplification means the original sides were equivalent. The equation is an identity with infinitely many valid solutions.
Question 16
Explanation
Distribution changes the left side to \(3x+12\), exactly matching the right. The final statement is \(12=12\), so every real \(x\) works.
Question 17
Explanation
The left side is \(8x-6\). Subtracting \(8x\) from \(8x-6=8x+5\) leaves \(-6=5\), a contradiction. There is no solution.
Question 18
Explanation
Distribute: \(5x-10=3x+8\). Subtract \(3x\), add \(10\): \(2x=18\), so \(x=9\). Unequal coefficients produce one solution.
Question 19
Explanation
The right side simplifies to \(x/2+3\), matching the left exactly. Multiplying by \(2\) gives \(x+6=x+6\), so it is an identity.
Question 20
Explanation
Multiply by \(3\): \(2x-1=2x+5\). The variable cancels and leaves \(-1=5\), a contradiction, so there is no solution.
Question 21
Explanation
The right side distributes to \(0.4x+1.2\), exactly the left side. The equation is true for every real \(x\), so it is an identity.
Question 22
Explanation
Subtracting \(0.6x\) leaves \(-2.1=0.9\), which is false. Matching coefficients and different constants mean no solution.
Question 23
Explanation
Inside the bracket, \(3(x-1)+4=3x+1\). Multiplying by \(2\) gives \(6x+2\), exactly the right side, so the equation is an identity.
Question 24
Explanation
Inside, \(2-(x-3)=5-x\). The left becomes \(20-4x\), identical to the right, so every real \(x\) satisfies the equation.
Question 25
Explanation
An identity requires matching variable coefficients and constants. The constants already match, so set \(k=4\). Then both sides are \(4x+7\).
Question 26
Explanation
No solution requires the variable coefficients to match while constants differ. Set \(p=6\); the equation becomes \(6x-3=6x+8\), leaving \(-3=8\).
Question 27
Explanation
Exactly one solution occurs when collecting leaves a nonzero coefficient: \((a-5)x=-11\). Therefore \(a-5\ne0\), or \(a\ne5\).
Question 28
Explanation
An identity requires the expressions to match for every \(x\). Therefore their variable coefficients and constants must both match: \(a=c\) and \(b=d\).
Question 29
Explanation
Distribute: \(2rx+10=10x+10\). Constants already match. Match variable coefficients: \(2r=10\), so \(r=5\).
Question 30
Explanation
The left is \(mx-3m+8\). Equal variable coefficients require \(m=4\). Then its constant is \(-12+8=-4\), which differs from \(1\), so the equation has no solution.
Question 31
Explanation
The left simplifies to \(2x+3\). The right simplifies to \(2x+8-5=2x+3\). Matching expressions make the equation an identity.
Question 32
Explanation
The left becomes \(5-2x+2=7-2x\), while the right is \(9-2x\). The variable cancels and leaves \(7=9\), so there is no solution.
Question 33
Explanation
Multiply by \(6\): \(4(x+6)=3x+30\). Then \(4x+24=3x+30\), so \(x=6\). Unequal simplified coefficients produce one solution.
Question 34
Explanation
The two sides are equal for every real \(x\), not only at zero. Subtracting \(7x\) leaves the true statement \(4=4\), so the equation is an identity.
Question 35
Explanation
Choice B simplifies on the left to \(4x-4+7=4x+3\), exactly its right side. The other options yield a contradiction or one numerical solution.
Question 36
Explanation
Choice C simplifies to \(8x-2=8x+3\), leaving \(-2=3\). Choice A is an identity, while B and D have one solution.
Question 37
Explanation
Changing only the right constant produces \(4x+9=4x+7\). Equal variable coefficients and unequal constants leave \(9=7\), so no solution.
Question 38
Explanation
The second expression distributes to \(6n+14\), exactly the first. The equality is an identity, so every real \(n\) makes the descriptions agree.
Question 39
Explanation
Equating the rules and distributing gives \(8h+5=8h+12\). The variable cancels and leaves \(5=12\), so no value of \(h\) makes the totals equal.
Question 40
Explanation
The left expands to \(ax+2a+b\). Match the \(x\)-coefficient: \(a=5\). Then match constants: \(2(5)+b=13\), so \(b=3\).
Question 41
Explanation
Multiply by \(3\): \(kx-6=6x-6\). The constants already match, so require \(k=6\) for equal variable coefficients.
Question 42
Explanation
The left simplifies to \(0.4px+2\). Match variable coefficients for cancellation: \(0.4p=1.2\), so \(p=3\). Constants \(2\) and \(3\) differ, giving no solution.
Question 43
Explanation
Expand the left: \(2ax-a+4\). Match variable coefficients: \(2a=6\), so \(a=3\). Match constants: \(-3+4=b\), so \(b=1\). Therefore \(a+b=4\).
Question 44
Explanation
The left is \(qx+4q\). At \(q=7\), it becomes \(7x+28\), matching the right exactly. For \(q\ne7\), unequal coefficients produce one solution.
Question 45
Explanation
Simplify the left: \(3(2x+2c-4)=6x+6c-12\). Match constants with \(30\): \(6c-12=30\), so \(6c=42\) and \(c=7\).
Question 46
Explanation
At \(r=5\), the equation becomes \(0x=12\), or \(0=12\), which is false and has no solution. For \(r\ne5\), division gives one solution.
Question 47
Explanation
Inside the bracket, \(4x-12+10=4x-2\). Half of that is \(2x-1\), exactly the right side, so the equation is an identity.
Question 48
Explanation
An identity is true for every value in the original domain. Algebraic cancellation does not restore values at which the original expression was undefined.
Question 49
Explanation
The left is \(6x-12+3=6x-9\). The right is also \(6x-9\), so the equation is an identity and has infinitely many solutions.
Question 50
Explanation
Variable cancellation alone does not decide the classification. The remaining statement \(-5=-5\) is true, so every valid value solves the original equation; it is an identity.
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Questions to review
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- Question 1Solution classificationEasy
- Question 2Identity recognitionEasy
- Question 3No-solution recognitionEasy
- Question 4Identity recognitionEasy
- Question 5No-solution recognitionEasy
- Question 6Identity recognitionEasy
- Question 7No-solution recognitionEasy
- Question 8Misconception analysisEasy
- Question 9Identity definitionEasy
- Question 10Coefficient comparisonEasy
- Question 11Coefficient comparisonEasy
- Question 12Coefficient comparisonEasy
- Question 13One-solution recognitionEasy
- Question 14Identity reasoningEasy
- Question 15Solution classificationEasy
- Question 16Identity with distributionMedium
- Question 17No solution with distributionMedium
- Question 18One solution with distributionMedium
- Question 19Identity with fractionsMedium
- Question 20No solution with fractionsMedium
- Question 21Identity with decimalsMedium
- Question 22No solution with decimalsMedium
- Question 23Nested identityMedium
- Question 24Nested identityMedium
- Question 25Identity parametersMedium
- Question 26No-solution parametersMedium
- Question 27One-solution parametersMedium
- Question 28General classification ruleMedium
- Question 29Identity parametersMedium
- Question 30No-solution parametersMedium
- Question 31Identity with like termsMedium
- Question 32No solution with distributionMedium
- Question 33One solution with fractionsMedium
- Question 34Misconception analysisMedium
- Question 35Identity recognitionMedium
- Question 36No-solution recognitionMedium
- Question 37Structural reasoningMedium
- Question 38Identity in contextMedium
- Question 39No solution in contextMedium
- Question 40Multiple parametersMedium
- Question 41Identity parameters with fractionsHard
- Question 42No-solution parameters with decimalsHard
- Question 43Multiple parametersHard
- Question 44Parameter classificationHard
- Question 45Identity parametersHard
- Question 46Parameter classificationHard
- Question 47Complex identityHard
- Question 48Identity domain restrictionsHard
- Question 49Complex classificationHard
- Question 50Misconception analysisHard