Systems with No Solution or Infinitely Many Solutions — SAT Math

Introduction

A system of two linear equations can behave in exactly three ways: one solution, no solution, or infinitely many solutions. The SAT tests whether you can identify which case applies — purely from the algebra — and find the constant that forces a specific case. This skill falls in the Algebra domain (≈35% of the section) and appears in both multiple-choice and student-produced response questions. It is purely non-contextual here: the focus is on the equations and their graphs, not word problems (those are covered in Systems of Equations in Context).

Core Concept

Write each equation in slope-intercept form $y = mx + b$ , then compare slopes and y-intercepts.

Slopes	Intercepts	Lines	Solutions
Different ( $m_1 \neq m_2$ )	Either	Intersecting	Exactly one
Equal ( $m_1 = m_2$ )	Different ( $b_1 \neq b_2$ )	Parallel	No solution
Equal ( $m_1 = m_2$ )	Equal ( $b_1 = b_2$ )	Identical (same line)	Infinitely many

Why this works graphically: A solution is a point where the two lines cross. Parallel lines never cross. Identical lines share every point.

The ratio test (standard form): When equations are written as $ax + by = c$ , you can compare ratios of coefficients directly without converting to slope-intercept form:

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \implies \text{no solution}

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \text{infinitely many solutions}

\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \implies \text{exactly one solution}

Short illustration:

3x - 6y = 9 \quad \text{and} \quad x - 2y = 3

Divide the first equation by 3: $x - 2y = 3$ . The equations are identical → infinitely many solutions.

Compare to $3x - 6y = 9$ and $x - 2y = 5$ : ratios $\frac{3}{1} = \frac{-6}{-2} = 3$ , but $\frac{9}{5} \neq 3$ → no solution.

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Introduction

Core Concept

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