Solving a System by Substitution and Elimination

Two lines on a graph cross at exactly one point, unless they're parallel or identical. That crossing point is the only (x, y) pair that lies on both lines, so it's the only pair that satisfies both equations simultaneously. Any other point might satisfy one equation but not the other. Solving a system algebraically just finds that crossing point without drawing anything.

Elimination works by adding or subtracting the two equations so one variable cancels out. If you have 2x + y = 11 and x - y = 1, adding them straight gives 3x = 12, because +y and -y cancel. Sometimes neither variable cancels right away, so you scale one or both equations first: for 3x + 4y = 26 and 5x - 2y = 26, multiply the second equation by 2 to get 10x - 4y = 52, then add that to the first equation so the y terms cancel. Substitution works differently: solve one equation for a variable, then plug that expression into the other equation. It wins when a variable is already alone, like y = 4x - 3, since there's nothing to rearrange before substituting.

Once you find one variable, don't stop and bubble it in. Many questions ask for x + y, 2x - y, or some other combination, not the bare value of x. Systems sit right next to linear equations and inequalities in this domain: the same skill of isolating and combining variables carries over, and systems word problems often disguise themselves as two-variable setups you have to write yourself before solving.