The Exact Crossing
Substitution and elimination find a system's exact solution: replace a variable with the expression it equals, or add scaled equations so a variable cancels — then back-substitute and check in both originals. · 12 min
Last folio you solved a system by graphing: draw both lines, read the crossing. That works beautifully — until the crossing lands between grid lines. If two lines meet at x = 2.5, your eye can get close, but close is not an answer you can check. This folio gives you two algebraic methods — substitution and elimination — that find the crossing exactly, no graph required. Both end the same way: back-substitute to find the second coordinate, then check the pair in both original equations.
Guess before you learn
The system: y = 2x and x + y = 12. One pair (x, y) satisfies both equations. Which is it?
Since y = 2x, wherever y appears you may write 2x instead: x + 2x = 12, so 3x = 12 and x = 4, making y = 8. If you picked (6, 6) or (2, 4), keep the pencil mark — each satisfies one equation and fails the other, which is exactly why the final check uses both. The swap you just watched — replacing y with the expression it equals — is substitution.
9–12
3–5
A riddle: a cat and a dog weigh 15 pounds together, and the dog weighs the same as two cats. Swap in what you know — the dog counts as two cats — so three cats weigh 15 pounds. One cat: 5 pounds. The dog: 10. Trading a thing for what it equals turns two clues into one.
There is a second trick. If two statements are both balanced, you may add them — left sides together, right sides together — and the result is still balanced. Sometimes the adding makes a whole letter vanish. Then the riddle is easy.
6–8
Substitution uses the fact that equal things are interchangeable. If y = 2x, then y and 2x name the same number, so in x + y = 12 you may write x + 2x = 12. One variable disappears; a one-variable equation remains, and you already know how to solve those.
Elimination uses the fact that adding equals to equals preserves equality. If 3x + 2y = 16 and 5x − 2y = 8 are both true, their sum 8x = 24 is true too — and the y-terms, +2y and −2y, cancel on the way. Either method turns two equations in two unknowns into one equation in one.
9–12
Both methods are licensed by one principle: an equation's two sides name the same number, so you may substitute an expression for the variable it equals, and you may add equal quantities to equal quantities. Multiplying an equation through by a nonzero constant changes neither its truth nor its solutions — that is what lets elimination scale an equation until a pair of coefficients opposes.
Each move produces an equivalent system — same solution set, easier shape. Choose by shape: an equation already solved for a variable invites substitution; aligned or opposable coefficients invite elimination. Either way, back-substitute for the second coordinate and check the pair in both originals — the check catches the arithmetic slips each method quietly permits.
K–2
Two toys balance one big block on a scale. So anywhere the big block sits, the two toys can sit instead. The scale stays fair. Swapping equal things keeps things true.
Two scales are both fair. Pour everything from both onto one big scale. Still fair. Adding fair to fair stays fair.
Undergrad
Elimination is the germ of Gaussian elimination: a system is replaced by the result of row operations — swap two equations, scale one by a nonzero constant, add a multiple of one to another — each invertible, hence each preserving the solution set exactly. Solving a 2×2 system by hand is echelon reduction in miniature.
Substitution is the same idea seen through functions: y = 2x parametrizes the first line, and imposing the second equation on that parametrization computes the intersection of the two solution sets. Both methods intersect affine subspaces; the graphical crossing is that intersection made visible.
Postgrad
Row operations are left-multiplications by elementary matrices, each invertible, so A ↦ EA acts on systems without disturbing {x : Ax = b}. Uniqueness of the crossing is the condition det A ≠ 0; parallel and coincident lines are precisely the singular cases, with the rank of the augmented matrix deciding between empty and infinite solution sets.
Substitution generalizes to elimination theory: solving one relation for a variable and rewriting the rest projects the solution set onto a coordinate subspace. For linear systems the projection is exact; for polynomial systems it becomes resultants and Gröbner bases — the honest descendants of this folio's two tricks.
back-substitute
Once one coordinate is known, feed it into either original equation to find the other. Find x, then back-substitute for y.
Substitution, as a procedure. First, take an equation already solved for a variable — or solve one for whichever variable has coefficient 1. Second, in the other equation, replace that variable with its expression, wrapped in parentheses. Third, solve the one-variable equation that results. Fourth, back-substitute to get the second coordinate. Fifth, check the pair in both originals. The parentheses in step two are not decoration: they keep the swapped-in expression intact when a coefficient multiplies it.
Solve by substitution: y = x + 3 and 2x + y = 12 — the steps fade as you master them
2x + (x + 3) = 12
3x + 3 = 12
3x = 9
x = 3
y = 6
6 + 6 = 12, which matches
Substitution shines when an equation hands you a variable already isolated. When both equations arrive as ax + by = c, elimination is usually faster. If two equations are true, their sum is true: add left sides together, right sides together. Arrange first for one variable's coefficients to be opposites, and the sum drops that variable. In 3x + 2y = 12 and 5x − 2y = 4, the y-terms already oppose: adding gives 8x = 16, so x = 2, and back-substituting gives y = 3.
Coefficients rarely oppose on their own. You may multiply an entire equation — every term, both sides — by any nonzero number without changing its solutions. Scale one equation, or both, until a pair of coefficients becomes opposites, then add. Choosing what to scale is the only judgment call; the rest is arithmetic.
Solve by elimination: 2x + 3y = 12 and x − y = 1 — the steps fade as you master them
3x − 3y = 3
5x = 15
x = 3
3 − y = 1, so y = 2
6 + 6 = 12, which matches
Why is this true?
Why must the solution be checked in both original equations?
Each method works with combinations of the equations, so an arithmetic slip can produce a pair that satisfies one equation but not the other. Only the pair that makes both originals true is the crossing.
You now hold three routes to the same point: graph and estimate, substitute and be exact, eliminate and be exact. And when the variables all vanish along the way, folio 7's rule still governs: a false sentence like 0 = 5 means parallel lines — no solution; a true one like 0 = 0 means both equations drew one line — every point on it works. Next unit: exponents, and then an unknown that multiplies itself.
Practice — new ink and old, interleaved
1.The lines y = 2x and y = x + 3 cross where 2x = x + 3. What is the x-coordinate of the solution?
2.Put the moves of the substitution method in working order.
- Replace y in the other equation with (x + 3)
- Solve one equation for y
- Solve the one-variable equation for x
- Back-substitute to find y
- Check the pair in both original equations
3.To eliminate y from 4x + 3y = 18 and 2x − y = 4, multiply the second equation by:
4.In y = −3x + 4, which point is on the line for certain — no computation needed?
5.Without looking back: what does it mean, in graph terms, when elimination ends in 0 = 7?
The variables vanished and left a false sentence, so the lines are parallel and the system has no solution.
How close were you? Grade yourself honestly — it sets your review date.
6.Match each equation to its solution.
7.Solve y = 2x − 1 and x + y = 8. Give y.
8.You solved a system and got (2, 5). In one sentence: what must be true of (2, 5) before you write it down as the answer?
9.6x + 1 = 6x + 1. How many solutions?