Systems of equations: substitution
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Solve the system: y = 2x + 1 and x + y = 10.
Common mix-up
In this example, a student wrote x + 2x + 1 = 10, so 3x = 10 and x = 10/3.
The student combined x + 2x + 1 = 10 into 3x = 10, dropping the +1, and concluded x = 10/3.
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See why this works
How does y = 2x + 1 help you solve x + y = 10?
Two equations with two unknowns feel harder than one. Substitution turns the pair into a single equation with one letter.
Replace one variable with an equivalent expression, then solve the remaining one-variable equation.
A system of equations just means both equations are true at the same time. Here y = 2x + 1 tells you exactly what y equals in terms of x. Using that information in the other equation is the whole method: wherever y appears in x + y = 10, write the expression 2x + 1 instead.
The result is x + 2x + 1 = 10, or 3x + 1 = 10 after collecting the x terms. The second equation is no longer needed for the algebra; it has become a single equation in one variable, and the y has disappeared entirely. This is the moment substitution does its work.
Solve 3x + 1 = 10 by undoing the operations: subtract 1 from each side to get 3x = 9, then divide by 3 to get x = 3. Now the first equation can pay the value back. Substitute x = 3 into y = 2x + 1 to get y = 7, and the system has its answer: x = 3, y = 7.
Check both original equations. In x + y = 10, the pair gives 3 + 7 = 10. In y = 2x + 1, it gives 7 = 6 + 1. Both hold, so the solution is genuine. Skipping the check is risky: a mistake in the substitution or in the back-substitute looks just like a solution but fails one of the equations.
Keep this idea
Substitute the known expression into the other equation, solve, then back-substitute into either original equation to finish.
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