Solve one equation for one unknown, put that expression into the other equation, and a system in two variables becomes an equation in one. The method always works; the choice of what to isolate decides whether it is pleasant.
Core ideas
One equation of the system is turned into a name for a variable, and that name is then used everywhere else. The reason this is allowed is worth stating once, plainly.
If x = y + 1 holds, then x and y + 1 are two names for the same number, and either may be written where the other stands. That is the entire justification, and it is why the expression must go into the equation you have not used.
Everything else is bookkeeping. Linear Equations and Inequalities already solves any linear equation in one variable; this lesson is about manufacturing one.
A variable whose coefficient is 1 or −1 isolates with no division. Any other choice produces a fraction, which is survivable — multiply through and it clears — but it is extra work chosen, not imposed.
The material
Four moves, the theorem that licenses them, and the two ways the variable can vanish.
The method has four moves. Solve one of the equations for one of its variables. Substitute the resulting expression into the other equation, which then contains one variable. Solve it. Put that value back into the isolated equation to get the partner, and check the pair in both originals.
Let a system consist of x = E, where E is an expression in y, together with a second equation. Replacing every x in the second equation by E produces a system with exactly the same solutions, provided x = E is kept.
Take a pair (x, y) satisfying both original equations. It satisfies x = E, so x and E are the same number, and putting one in place of the other in a true equation leaves a true equation. So the pair satisfies the substituted equation as well.
Now take a pair satisfying x = E and the substituted equation. The same argument, run with E replaced by x, returns the original second equation. Neither direction loses a pair, so the two systems have the same solution set.
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The proviso is not decoration. The isolated equation stays part of the system, and it is what produces the value of x at the end. Dropping it once the substitution is made leaves a single equation in one variable, which has a different solution set from the system you started with.
Look for a coefficient of 1 or −1. In 5x − 2y = 4 and 3x + y = 9, the y in the second equation has coefficient 1, so y = 9 − 3x arrives without a division. If no coefficient is 1 or −1, pick whichever looks smallest and accept the fraction.
Substituting into the first equation is not a preference. Putting x = (9 + 5y)/2 back into 2x − 5y = 9 gives 9 = 9: true, and completely uninformative, because that equation was used to build the expression in the first place. Information comes from the equation that has not yet been spent.
Sometimes the substitution removes both variables and leaves a statement about numbers. Linear Equations and Inequalities met these as contradictions and identities, and they mean here exactly what they meant there: a false statement means no pair works, a true one means the equation placed no new restriction and the whole line survives.
{(2t + 3, t)} is a recipe: choose t, get a pair. t = 0 gives (3, 0), t = 1 gives (5, 1), t = 5 gives (13, 5). Each satisfies x − 2y = 3 and 2x − 4y = 6.
The letter is arbitrary and the description is not unique: {(s, (s − 3)/2)} describes the same line. Any correct description of the set is a correct answer.
The check at the end is a check of both original equations. This is the same discipline as “Systems of Two Linear Equations” and for the same reason: an arithmetic slip in the substitution produces a pair that fits the equation it came out of.
Interactive
Isolate one unknown, put it in the other equation, and check the answer
Method
Four moves. The third is where the sign errors are.
It isolates without division and keeps the next line clean. If no coefficient qualifies, choose any variable and expect one fraction, which will clear when you multiply through.
Write y = 9 − 3x down on its own line. It is still one of the two equations of the system, and it is the one that supplies the second coordinate at the end.
The whole expression replaces the variable. In 5x − 2y = 4 the y becomes (9 − 3x) inside brackets, so the −2 multiplies both terms: −2(9 − 3x) = −18 + 6x.
Put the number into the isolated equation to get the partner. Then test the pair in the two equations you were given, not in anything you have derived.
The second equation has a y with coefficient 1, so that is what to isolate. Everything after that is one-variable algebra from Linear Equations and Inequalities.
The second line of the substitution is the one that goes wrong. −2 times −3x is +6x; writing 5x − 18 − 6x = 4 instead gives −x = 22 and x = −22, which fails the very first check. The bracket is what prevents it, and the bracket costs nothing.
Isolating x from the first equation instead is legitimate and gives x = (4 + 2y)/5. The same pair comes out. The fifths are the price of ignoring the coefficient of 1 sitting in the other equation.
Faded rehearsal: solve 4x − y = 11 and 2x + 3y = 9. The cheap first move is supplied: isolate y from the first equation. Put the whole expression into the second in brackets, finish the pair, and use both original equations — not a derived one — for the check.
Isolating a variable whose coefficient is 2 or −5 produces an expression such as (9 + 5y)/2, and the method carries on unchanged: multiply through and the denominator goes. The standard is a system in which no coefficient is 1, solved in exact fractions, with the pair checked in both original equations.
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