Example

Classifying {2x+y=−3,;x−5y=5}\{2x + y = -3,; x - 5y = 5\} Without Graphing

Determine the number of solutions and classify the system {2x+y=−3x−5y=5\left\{\begin{array}{l} 2x + y = -3 x - 5y = 5 \end{array}\right. without graphing, by comparing slopes and y-intercepts.

First equation: Solve 2x+y=−32x + y = -3 for yy:

y=−2x−3y = -2x - 3

Second equation: Solve x−5y=5x - 5y = 5 for yy:

−5y=−x+5-5y = -x + 5

−5y−5=−x+5−5\frac{-5y}{-5} = \frac{-x + 5}{-5}

y=15x−1y = \frac{1}{5}x - 1

Compare slopes and y-intercepts:

  • First line: m=−2m = -2, b=−3b = -3
  • Second line: m=15m = \frac{1}{5}, b=−1b = -1

Because the slopes are different (−2≠15-2 \neq \frac{1}{5}), the two lines must intersect at exactly one point.

The system has 1 solution and is consistent and independent.

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Updated 2026-04-24

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