Systems of Equations

Two machines, one input where they agree

A5 left two phone plans side by side — $B(g) = 3g + 20$ and $L(g) = 5g + 10$ — and could only compare them one gigabyte at a time: is $B(4) < L(4)$? ($32$ versus $30$ — the pricier-per-gig plan wins at four gigs.) But the real question was never about four gigabytes specifically. It was where does the winner change? That’s a single input $g$ where both machines give the same output, and finding it means writing both rules down at once:

$B(g) = 3g + 20 \qquad\qquad L(g) = 5g + 10$

Two equations, considered together, are a system. Solving a system means finding the input (or inputs) that make every equation in it true simultaneously — not one at a time, both. A4 called a two-variable equation a membership test for points; a system just runs two membership tests on the same point and keeps only what passes both.

The easiest system is already solved for you

Here both rules are already solved for the same output, so the system practically writes its own first move: if $B(g)$ and $L(g)$ are ever equal, then whatever they equal, they equal each other.

At exactly $5$ gigabytes both plans cost $$35$; below $5$, $L$ (the steeper rate) is cheaper; above it, $B$ wins. That’s substitution in its cleanest form: two expressions for the same output, set equal, solved with tools you already own.

Substitution when nothing’s isolated yet

Most systems don’t arrive pre-solved. $2x + y = 7$ and $x - y = 2$ still hide a single agreeing point, but neither equation is sitting in "$y = \ldots$" form. Substitution still works — you just isolate one variable first. The second equation gives $x$ almost for free: $x = y + 2$. Substitute that expression everywhere the first equation has an $x$:

Always pick whichever variable already has a coefficient of $1$ (or $-1$) to isolate — it costs no fractions and no extra work.

Elimination: two level scales, added together

A1 built solving on one rule: an equation is a balance scale, and whatever you do to one side you must do to the other. A system hands you two balanced scales at once — and here’s the move substitution doesn’t show you: if scale one is level and scale two is level, stacking their pans keeps the result level. Add left side to left side, right side to right side, and you get a third true equation, for free.

That’s only useful if a variable disappears in the process, which happens when its coefficients in the two equations are equal or opposite. Take $2x + 3y = 13$ and $2x - y = 1$: both have $2x$, so subtracting cancels it directly.

Then $2x + 3(3) = 13$ gives $x = 2$ — the point $(2, 3)$. When the coefficients aren’t already equal or opposite, multiply one or both entire equations by a constant first — legal for the same balance-scale reason A1 gave: multiplying both sides of a level scale by the same number keeps it level.

Predict first: with the default $2x + y = 7$ and $x - y = 2$, guess $(x, y)$ before reading the steps — you already solved something like it above. Then flip the toggle between substitution and elimination on the same system: same point, different path. Try the chip that reads $x + 2y = 6$ and $2x + 4y = 20$ — it looks like the second equation is just the first one doubled, but check the constants.

Graphing it: where the lines actually meet

Every equation in a system is a line (A4), so a system is just two lines on one grid, and the solution is wherever they physically cross. That geometric picture explains the three outcomes algebra can produce:

• One crossing — different slopes. The ordinary case: one solution.
• No crossing — same slope, different intercept: the lines are parallel, and algebra echoes it exactly the way A1 warned you it would — every variable cancels, leaving a false statement like $0 = 8$.
• Every point shared — same slope and same intercept: it’s the same line, drawn twice. Algebra again cancels everything, but lands on a true statement like $0 = 0$ — infinitely many solutions.

Start on the parallel chip and predict: will dragging either intercept ever make these two lines meet? (No — only a slope change can.) Then try same line — notice the two equations printed in the info panel are identical, not just similar-looking.

Misconceptions worth naming

“Substitution and elimination give different answers” — they can’t. Both start from the same two true statements about $x$ and $y$; they just retire a variable in a different order. If your two methods disagree, one of them has an arithmetic slip, not a genuinely different system.

Elimination hides a sign trap. Subtracting one whole equation from another means flipping the sign of every term on the side you subtract — not just the first term your eye lands on. $(2x + 3y) - (2x - y)$ is not $2x + 3y - 2x - y$: the second equation’s $-y$ flips too, becoming $+y$, which is exactly why the $y$‘s add instead of vanish. Treat the subtracted equation as one parenthesized block and distribute the minus across all of it, the way A1 taught you to distribute a negative through parentheses.

Solving for one variable and stopping is the most common half-answer of all: a system’s solution is a point, and reporting $x = 3$ without also stating $y$ answers half the question. Always back-substitute.

Finally, “no solution” is not the same as “a coordinate happens to be $0$.” A system genuinely has no solution only when the lines are strictly parallel — never touching anywhere on the infinite grid, not just outside the little window you happened to graph.

The one thing to remember

A system is two (or more) equations that must be true together; solving it means finding the input(s) where they agree. Substitution and elimination are two roads to the same point — pick whichever leaves the least arithmetic — and the number of solutions is really a question about two lines: cross once, run parallel forever, or lie on top of each other.

Two methods, side by side

Both always land on the same point — pick whichever costs less arithmetic for the system in front of you.

Reading the number of solutions off the lines