Systems with No Solution or Infinitely Many Solutions

Identify when systems of equations have no solution or infinitely many solutions on the Digital SAT. Understand parallel and coincident lines.

Most systems of two linear equations have exactly one solution — the point where the two lines intersect. But some systems are special: the lines may be parallel (no intersection = no solution) or the same line (infinite intersections = infinitely many solutions).

The Digital SAT loves testing this concept, often by giving you a system with a parameter and asking you to find the value that produces no solution or infinitely many solutions.

Core Concepts

Three Possibilities for Linear Systems

Type Lines Solutions Algebraic Result
Consistent & independent Intersect at one point Exactly one x=ax = a, y=by = b
Inconsistent Parallel (never meet) None 0=k0 = k (false statement)
Consistent & dependent Same line (coincident) Infinitely many 0=00 = 0 (true statement)

Parallel Lines — No Solution

Two lines are parallel when they have the same slope but different y-intercepts.

y=3x+2andy=3x+7y = 3x + 2 \quad \text{and} \quad y = 3x + 7

Both have slope 3 but different y-intercepts. They never intersect → no solution.

In standard form: Ax+By=C1Ax + By = C_1 and Ax+By=C2Ax + By = C_2 (where C1C2C_1 \neq C_2) → no solution.

Same Line — Infinitely Many Solutions

Two equations represent the same line when one is a multiple of the other.

y=3x+2and2y=6x+4y = 3x + 2 \quad \text{and} \quad 2y = 6x + 4

The second equation is just 2× the first. Every point on the line satisfies both equations → infinitely many solutions.

Using the Ratio Test

For the system a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2:

  • If a1a2b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}: one solution
  • If a1a2=b1b2c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}: no solution (parallel)
  • If a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}: infinitely many solutions (same line)

Finding a Parameter for No Solution

Example: For what value of kk does the system have no solution?

2x+3y=72x + 3y = 7 kx+6y=10kx + 6y = 10

For no solution, the lines must be parallel: same slope, different y-intercepts.

Slope of first: 23-\frac{2}{3}

Slope of second: k6-\frac{k}{6}

For parallel: k6=23-\frac{k}{6} = -\frac{2}{3}k=4k = 4

Check constants: 71036\frac{7}{10} \neq \frac{3}{6}, so k=4k = 4 gives no solution ✓

Finding a Parameter for Infinite Solutions

Example: For what values of aa and bb does the system have infinitely many solutions?

3x+6y=153x + 6y = 15 ax+by=5ax + by = 5

For infinite solutions: 3a=6b=155\frac{3}{a} = \frac{6}{b} = \frac{15}{5}

155=3\frac{15}{5} = 3, so 3a=3\frac{3}{a} = 3a=1a = 1 and 6b=3\frac{6}{b} = 3b=2b = 2.

Strategy Tips

Tip 1: Convert to Slope-Intercept Form

The quickest way to spot parallel or identical lines is to compare slopes and y-intercepts.

Tip 2: Use the Ratio Test for Standard Form

When both equations are in standard form, compare the ratios of corresponding coefficients.

Tip 3: If You're Solving and Variables Cancel

If you're working through substitution or elimination and the variable terms cancel:

  • False statement remaining → no solution
  • True statement remaining → infinitely many solutions

Tip 4: The SAT Loves Parameter Questions

Expect questions like "For what value of kk..." Be ready to set up slope equality conditions.

Tip 5: Graphically Think About It

Parallel = same direction, never meet. Same line = perfect overlap. One solution = they cross once.

Worked Example: Example 1

Problem

How many solutions does this system have?

4x6y=124x - 6y = 12 2x3y=82x - 3y = 8

Multiply the second by 2: 4x6y=164x - 6y = 16

Compare: 4x6y=124x - 6y = 12 vs 4x6y=164x - 6y = 16 → same LHS, different RHS → no solution.

Solution

Worked Example: Example 2

Problem

How many solutions?

5x+10y=205x + 10y = 20 x+2y=4x + 2y = 4

Divide the first by 5: x+2y=4x + 2y = 4

Both equations are identical → infinitely many solutions.

Solution

Worked Example: SAT-Style

Problem

For what value of cc does the system below have no solution?

3x+5y=73x + 5y = 7 6x+10y=c6x + 10y = c

The second equation is 2×2 \times the first equation's left side. For no solution, the RHS must NOT be 2×7=142 \times 7 = 14.

So no solution when c14c \neq 14.

But if the question asks for infinite solutions, c=14c = 14.

Solution

Worked Example: Example 4

Problem

For what value of kk does the system have infinitely many solutions?

kx4y=10kx - 4y = 10 6x8y=206x - 8y = 20

For infinite solutions, one equation must be a scalar multiple of the other.

Compare: k6=48=1020\frac{k}{6} = \frac{-4}{-8} = \frac{10}{20}

48=12\frac{-4}{-8} = \frac{1}{2} and 1020=12\frac{10}{20} = \frac{1}{2}

So k6=12\frac{k}{6} = \frac{1}{2}k=3k = 3.

Solution

Worked Example: Example 5

Problem

Classify the system: y=12x+3y = \frac{1}{2}x + 3 and y=12x1y = \frac{1}{2}x - 1

Same slope (12\frac{1}{2}), different y-intercepts (3 vs 1-1) → no solution (parallel lines).

Solution

Practice Problems

  1. Problem 1

    How many solutions? y=4x+1y = 4x + 1 and 8x2y=28x - 2y = -2.

    Problem 2

    How many solutions? 3x+9y=273x + 9y = 27 and x+3y=10x + 3y = 10.

    Problem 3

    For what value of aa does the system have no solution? 2x+ay=62x + ay = 6 and 4x+6y=154x + 6y = 15.

    Problem 4

    For what value of kk does the system have infinitely many solutions? 5x2y=k5x - 2y = k and 10x4y=1410x - 4y = 14.

    Problem 5

    Classify: x2+y3=1\frac{x}{2} + \frac{y}{3} = 1 and 3x+2y=63x + 2y = 6.

    Problem 6

    The system ax+2y=8ax + 2y = 8 and 3x+by=123x + by = 12 has infinitely many solutions. Find aa and bb.

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Common Mistakes

  • Confusing no solution with x=0x = 0 or y=0y = 0. No solution means NO ordered pair works, not that xx or yy is zero.
  • Not checking both coefficient ratios and constant ratios. Same coefficient ratio but different constant ratio → no solution. All three ratios equal → infinite solutions.
  • Algebraic errors when multiplying equations. Be careful when scaling equations to compare them.
  • Forgetting that "no solution" questions might have a specific cc value answer. The question asks for the parameter value that creates the special case.
  • Thinking different-looking equations can't be the same line. y=2x+3y = 2x + 3 and 4x2y=64x - 2y = -6 look different but are the same line.

Frequently Asked Questions

Can a system of two linear equations have exactly two solutions?

No. Two distinct lines can intersect at most once. So there's either 0, 1, or infinitely many solutions.

What's the graphical meaning of these cases?

One solution = lines cross. No solution = parallel lines. Infinite solutions = same line (overlapping).

How do I know which case to expect on the SAT?

If the question asks "for what value of kk...", expect a parameter question about special cases. If it just says "solve the system", expect one solution.

Can I use the Desmos calculator?

Yes — graph both equations. If you see two distinct lines crossing, there's one solution. If you see parallel lines, no solution. If you see only one line, they're the same.

Is this related to single-equation no-solution questions?

Yes — the same logic applies. When combining a system into one equation, contradictions mean no solution, identities mean infinite solutions.

Key Takeaways

  • No solution: parallel lines — same slope, different y-intercepts.

  • Infinitely many solutions: same line — one equation is a multiple of the other.

  • Use the ratio test for standard form: compare a1a2\frac{a_1}{a_2}, b1b2\frac{b_1}{b_2}, c1c2\frac{c_1}{c_2}.

  • Parameter questions: set coefficient ratios equal and check constant ratios.

  • If solving and variables cancel: check the remaining statement (true or false).

  • A system of two linear equations can never have exactly two solutions.

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