Equations with No Solution or Infinite Solutions

Learn to identify equations with no solution or infinitely many solutions on the Digital SAT. Understand contradictions and identities.

Most linear equations have exactly one solution. But some equations are special: they either have no solution (a contradiction) or infinitely many solutions (an identity). The Digital SAT tests your ability to recognise these cases — both in plain algebra and in problems that ask you to find the value of a constant that creates one of these special situations.

This topic connects to systems of equations (parallel lines = no solution, same line = infinite solutions) and is a favourite for medium-difficulty SAT questions.

Core Concepts

Three Possible Outcomes

When you solve a linear equation, you reach one of three outcomes:

Outcome What Happens Example
One solution You get x=a numberx = \text{a number} 2x+3=7x=22x + 3 = 7 \Rightarrow x = 2
No solution You get a false statement 2x+3=2x+73=72x + 3 = 2x + 7 \Rightarrow 3 = 7
Infinite solutions You get a true statement 2x+3=2x+33=32x + 3 = 2x + 3 \Rightarrow 3 = 3

No Solution (Contradiction)

An equation has no solution when simplifying leads to a statement that is always false, such as 0=50 = 5 or 3=13 = -1.

This happens when both sides have the same variable terms but different constants.

Example: 3x+4=3x+93x + 4 = 3x + 9

Subtract 3x3x: 4=94 = 9 ✗ — this is false, so there is no solution.

Graphically, this corresponds to two parallel lines that never intersect.

Infinite Solutions (Identity)

An equation has infinitely many solutions when simplifying leads to a statement that is always true, such as 0=00 = 0 or 5=55 = 5.

This happens when both sides are identical after simplification.

Example: 2(x+3)=2x+62(x + 3) = 2x + 6

Expand: 2x+6=2x+62x + 6 = 2x + 6

Subtract 2x2x: 6=66 = 6 ✓ — this is always true, so there are infinitely many solutions.

Graphically, this is the same line drawn twice — every point is an intersection.

Finding a Constant for No Solution

SAT questions often ask: "For what value of kk does the equation have no solution?"

Example: For what value of kk does 3x+5=kx+83x + 5 = kx + 8 have no solution?

For no solution, the variable terms must be equal (coefficients of xx match) but the constants must differ.

Coefficients match when k=3k = 3. Then: 3x+5=3x+83x + 5 = 3x + 8, which gives 5=85 = 8 — no solution.

Answer: k=3k = 3

Finding a Constant for Infinite Solutions

Example: For what value of aa does a(x+2)=3x+6a(x + 2) = 3x + 6 have infinitely many solutions?

Expand: ax+2a=3x+6ax + 2a = 3x + 6

For infinite solutions, both sides must be identical:

  • Coefficient of xx: a=3a = 3
  • Constant: 2a=6a=32a = 6 \Rightarrow a = 3

Answer: a=3a = 3

Checking Both Conditions

When a question involves a parameter and asks for "no solution" or "infinite solutions", you typically need to:

  1. Match the coefficients of the variable terms.
  2. Check whether the constant terms also match (infinite solutions) or don't match (no solution).

Strategy Tips

Tip 1: Always Simplify Completely

Distribute and combine like terms on both sides. Only then can you tell whether the equation is a contradiction, identity, or regular equation.

Tip 2: Compare Coefficients

Once simplified to the form ax+b=cx+dax + b = cx + d:

  • If aca \neq c: one solution (x=dbacx = \frac{d - b}{a - c})
  • If a=ca = c and bdb \neq d: no solution
  • If a=ca = c and b=db = d: infinitely many solutions

Tip 3: On the SAT, Look for Matching Variable Terms

If you see the same xx-coefficient on both sides of the equation, immediately check the constant terms. That tells you whether it's no solution or infinite solutions.

Tip 4: Plug In the Given Constant

For questions asking "what value of kk gives no solution", substitute your candidate value back in and verify the equation indeed has no solution.

Tip 5: Watch for Tricky Formatting

Sometimes the SAT writes these as multiple-choice questions where one answer gives no solution and another gives infinite solutions. Read carefully.

Worked Example: Example 1

Problem

How many solutions does 4(x1)=4x+54(x - 1) = 4x + 5 have?

Distribute: 4x4=4x+54x - 4 = 4x + 5

Subtract 4x4x: 4=5-4 = 5

This is false. No solution.

Solution

Worked Example: Example 2

Problem

How many solutions does 12(6x+10)=3x+5\frac{1}{2}(6x + 10) = 3x + 5 have?

Distribute: 3x+5=3x+53x + 5 = 3x + 5

Both sides are identical. Infinitely many solutions.

Solution

Worked Example: SAT-Style

Problem

If 2(3x+k)=6x+102(3x + k) = 6x + 10, for what value of kk does the equation have infinitely many solutions?

Distribute: 6x+2k=6x+106x + 2k = 6x + 10

The xx-coefficients already match (6=66 = 6). For infinite solutions, constants must match:

2k=10k=52k = 10 \Rightarrow k = 5

Solution

Worked Example: SAT-Style

Problem

For what value of aa does ax+12=5x+3aax + 12 = 5x + 3a have no solution?

For no solution: coefficients of xx must be equal, constants must differ.

Coefficients: a=5a = 5

Constants when a=5a = 5: LHS constant =12= 12, RHS constant =3(5)=15= 3(5) = 15.

121512 \neq 15 ✓ — the constants are indeed different.

Answer: a=5a = 5

Solution

Worked Example: Example 5

Problem

How many solutions does 5x3(x+2)=2(x3)5x - 3(x + 2) = 2(x - 3) have?

LHS: 5x3x6=2x65x - 3x - 6 = 2x - 6

RHS: 2x62x - 6

Both sides: 2x6=2x62x - 6 = 2x - 6. Infinitely many solutions.

Solution

Practice Problems

  1. Problem 1

    How many solutions does 3(2x+1)=6x+43(2x + 1) = 6x + 4 have?

    Problem 2

    How many solutions does 7x2=7(x2)+127x - 2 = 7(x - 2) + 12 have?

    Problem 3

    For what value of mm does mx+6=4x+2mmx + 6 = 4x + 2m have no solution?

    Problem 4

    For what value of cc does 5(x+c)=5x+155(x + c) = 5x + 15 have infinitely many solutions?

    Problem 5

    The equation k(x1)=3xkk(x - 1) = 3x - k has infinitely many solutions. What is kk?

    Problem 6

    Does the equation 2x+42=x+3\frac{2x+4}{2} = x + 3 have one solution, no solution, or infinitely many solutions?

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Common Mistakes

  • Claiming "no solution" when there is one. Always simplify completely first. Don't jump to conclusions based on how the equation looks.
  • Confusing no solution with x=0x = 0. If x=0x = 0 is a valid solution, the equation has one solution — it's not "no solution".
  • Forgetting to check both conditions. For "no solution", coefficients must match AND constants must differ. For infinite solutions, BOTH must match.
  • Arithmetic errors during distribution. These problems require careful expanding, especially with negatives.
  • Misreading the question. The SAT might ask for the value of a constant OR for the number of solutions. Make sure you answer what's asked.

Frequently Asked Questions

How often does this appear on the SAT?

Typically 1–2 questions per test. It's a reliable topic that appears on nearly every SAT administration.

Can a linear equation have exactly two solutions?

No. A linear equation in one variable has either one solution, no solution, or infinitely many solutions. Never two.

How does this relate to systems of equations?

For a system of two linear equations: parallel lines = no solution, same line = infinite solutions, intersecting lines = one solution. The same logic applies.

What if the question gives me specific answer choices for how many solutions?

Simplify the equation. If you get a false statement, choose "no solution" or "zero." If a true statement, choose "infinitely many." If you get x=x = a value, choose "one" or "exactly one."

Can this come up with absolute value or quadratic equations on the SAT?

Yes, but the SAT primarily tests this concept with linear equations and linear systems.

Key Takeaways

  • No solution: simplification produces a false statement (e.g., 5=85 = 8). The variable terms cancel but the constants don't match.

  • Infinitely many solutions: simplification produces a true statement (e.g., 0=00 = 0). Both sides are identical.

  • To find a constant for no solution: match the variable coefficients and ensure the constants differ.

  • To find a constant for infinite solutions: match both the variable coefficients and the constants.

  • Always simplify fully before making a determination.

  • This concept also applies to systems of equations — same underlying logic.

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