Example

Solving {6x+15y=10,  2x5y=5}\{-6x + 15y = 10,\; 2x - 5y = -5\} by Elimination

Solve the system {6x+15y=102x5y=5\left\{\begin{array}{l} -6x + 15y = 10 \\ 2x - 5y = -5 \end{array}\right. using the elimination method.

Step 1 — Write both equations in standard form. Both equations are already in the form Ax+By=CAx + By = C.

Step 2 — Make the coefficients of one variable opposites. To eliminate xx, multiply the second equation by 33 so that the xx-coefficients become 6-6 and +6+6:

3(2x5y)=3(5)    6x15y=153(2x - 5y) = 3(-5) \implies 6x - 15y = -15

The system becomes {6x+15y=106x15y=15\left\{\begin{array}{l} -6x + 15y = 10 \\ 6x - 15y = -15 \end{array}\right.

Step 3 — Add the equations.

6x+15y+6x15y=10+(15)-6x + 15y + 6x - 15y = 10 + (-15)

050 \neq -5

Because 0=50 = -5 is a false statement and both variables have been completely eliminated, the equations are inconsistent. Their graphs are parallel lines that never intersect. The system has no solution.

This outcome is the opposite of the dependent case: when all variable terms cancel during elimination and the resulting numerical statement is false (such as 0=50 = -5), no ordered pair can satisfy both equations simultaneously. When the statement is true (such as 0=00 = 0), the system has infinitely many solutions instead.

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Updated 2026-04-21

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