Precalculus Examples

$x + y = 3$ , $x + y = 6$

Solve the system of equations.

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Multiply each equation by the value that makes the coefficients of $x$ opposite.

$x + y = 3$

$(- 1) \cdot (x + y) = (- 1) (6)$

Simplify.

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Simplify the left side.

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Simplify $(- 1) \cdot (x + y)$ .

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Apply the distributive property.

$x + y = 3$

$- 1 x - 1 y = (- 1) (6)$

Rewrite negatives.

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Rewrite $- 1 x$ as $- x$ .

$x + y = 3$

$- x - 1 y = (- 1) (6)$

Rewrite $- 1 y$ as $- y$ .

$x + y = 3$

$- x - y = (- 1) (6)$

$x + y = 3$

$- x - y = (- 1) (6)$

$x + y = 3$

$- x - y = (- 1) (6)$

$x + y = 3$

$- x - y = (- 1) (6)$

Simplify the right side.

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Multiply $- 1$ by $6$ .

$x + y = 3$

$- x - y = - 6$

$x + y = 3$

$- x - y = - 6$

$x + y = 3$

$- x - y = - 6$

Add the two equations together to eliminate $x$ from the system.

Since $0 \neq - 3$ , there are no solutions.

No solution

Since the system has no solution, the equations and graphs are parallel and do not intersect. Thus, the system is inconsistent.

Inconsistent

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