Subtracting the Second Equation From the First: The Elimination Method, Step by Step

§ 01

The instruction, decoded

A worksheet or a solved-answer page hands you that sentence — "subtract the second equation from the first" — without saying what it authorizes you to do. It means this: take the left side of the first equation minus the left side of the second, and the right side of the first minus the right side of the second, term by term. The goal is always the same: make one variable's combined coefficient hit zero.

Here's a typical textbook pair: $8x+3y=14$ and $4x+3y=8$.

The $y$-terms cancel because both coefficients are $+3$. That leaves $4x=6$, so $x=1.5$. Substitute back into the first equation: $8(1.5)+$$3y=$$14\to$$12+$$3y=$$14\to$$y=$$8(1.5)+3y=14 \rightarrow 12+3y=14 \rightarrow y=\dfrac23$. Check the second equation: $4(1.5)+$$3\left({\displaystyle \frac{2}{3}}\right)=$$6+$$2=$$4(1.5)+3\left(\dfrac23\right)=6+2=8$ — it holds. This is exactly what Common Core standard HSA-REI.C.5 expects an Algebra 1 student to be able to do with a linear combination of two equations. If the systems unit ends in a timed test, prep for a timed algebra test builds practice sets around this exact move; if a twenty-system worksheet is due tonight, you can send the full assignment and get every line worked.

§ 02

Line by line: the subtraction, with the signs kept honest

Most answers online state the rule and stop; the mechanics of why it's legal, and where it goes wrong, are the part worth slowing down for.

Subtracting like terms only

Subtracting equations from each other only ever combines matching terms: $x$-terms with $x$-terms, $y$-terms with $y$-terms, constants with constants. You never subtract an $x$-term from a $y$-term — each column of the stacked equations stays in its own lane, the same way subtracting like terms works in any algebra problem.

Distributing the minus sign across the whole second equation

The trap sits here. Subtracting the second equation from the first means distributing a minus sign across every term on both sides of that second equation, not just the first one you see. If the second equation carries a negative coefficient — say $-3y$ — then subtracting it means $4y-$$(-3y)=$$4y+$$3y=$$4y-(-3y)=4y+3y=7y$, not $4y-3y$. Dropping that sign flip is the single most common way this technique goes wrong; the full worked case appears below under "a case that needs a sign flip."

The result is still a true equation - why you are allowed to do this

Subtracting equal quantities from both sides of a true equation always produces another true equation — that's the same rule that lets you add or subtract any number from both sides of a one-variable equation. Here, you're subtracting an entire equation (itself a claim about equal quantities) instead of a single number, which is why the result is still a valid, solvable equation.

§ 03

Interactive: subtract your own two equations

Enter the coefficients

Type in the six coefficients of a $2\times2$ system — $a_1, b_1, c_1$ for the first equation, $a_2, b_2, c_2$ for the second — and the stepper lines them up the way a textbook prints them, ready to subtract.

Watch the sign on each term

Each term's sign is tracked separately as the subtraction runs, so a coefficient that flips from negative to positive (or the reverse) is highlighted instead of silently corrected.

See whether adding would have been the shorter route

The stepper also states, in plain words, whether adding the two equations would have eliminated a variable in fewer moves than subtracting did. That add-or-subtract verdict is the one feature no static answer page provides.

§ 04

When to subtract and when to add

Same sign, same coefficient: subtract

When both equations carry the same coefficient with the same sign — like the $+3y$ and $+3y$ above — subtracting cancels that variable directly.

Opposite signs, same magnitude: add

When the coefficients are equal in size but opposite in sign — like $+3$ and $-3$ — adding is what cancels them; subtracting would double the term instead of removing it.

The one-sentence decision rule

If the coefficients you're targeting share a sign, subtract; if they don't, add — either move exists to force that variable's combined coefficient to zero.

Adding and subtracting algebraic equations side by side on the same system

Take $3x+2y=7$ and $-3x+y=-1$, which share the solution $x=1,\ y=2$. Adding them: $(3x+(-3x))+$$(2y+y)=$$7+$$(-1)\to$$3y=$$6\to$$y=$$(3x+(-3x))+(2y+y)=7+(-1) \rightarrow 3y=6 \rightarrow y=2$; then $3x+$$2(2)=$$7\to$$x=$$3x+2(2)=7 \rightarrow x=1$ — solved in two lines. Subtracting them instead: $(3x-(-3x))+$$(2y-y)=$$7-$$(-1)\to$$6x+$$y=$$(3x-(-3x))+(2y-y)=7-(-1) \rightarrow 6x+y=8$ — nothing cancels, because the $x$-coefficients were opposite in sign to begin with. Run both on the same system and the correct choice is obvious immediately.

§ 05

When the coefficients do not match: multiply first

This is where most students actually stall, and it's the step the elimination method needs before subtraction (or addition) can do anything.

Scaling one equation

For $2x+y=7$ and $x+3y=11$, the $x$-coefficients are 2 and 1 — not equal, so subtracting as-is does nothing useful. Scale the second equation by 2: $2x+6y=22$. Now subtract the first: $(2x-2x)+$$(6y-y)=$$22-$$7\to$$5y=$$15\to$$y=$$(2x-2x)+(6y-y)=22-7 \rightarrow 5y=15 \rightarrow y=3$. Back-substitute: $2x+3=7 \rightarrow x=2$. Check: $2+3(3)=11$ — confirmed.

Scaling both (least common multiple of the coefficients)

For $3x+2y=18$ and $5x-3y=11$, neither coefficient matches the other on either variable. Scale to the least common multiple of the $y$-coefficients, 6: multiply the first equation by 3 ($9x+6y=54$) and the second by 2 ($10x-6y=22$). The $y$-terms are now opposite in sign and equal in magnitude, so add: $19x=76 \rightarrow x=4$. Back-substitute: $3(4)+$$2y=$$18\to$$y=$$3(4)+2y=18 \rightarrow y=3$. Check: $5(4)-$$3(3)=$$20-$$9=$$5(4)-3(3)=20-9=11$ — confirmed.

A worked 3x + 2y / 5x - 3y system

That's the full worked $3x+$$2y=$$18,5x-$$3y=$$3x+2y=18,\ 5x-3y=11$ system above: scale both equations to a common $y$-coefficient of 6, then combine. The linear combination that results — $19x=76$ — is a direct application of what mathematicians call Gaussian elimination, run by hand on a two-variable case. A circuit with two loop currents produces the same shape of system, and the physics desk works a two-loop circuit solved with Kirchhoff's laws with real resistances and voltages in place of arbitrary coefficients.

§ 06

Worked 2x2 systems, start to finish

Solving systems algebraically always ends the same way: one variable eliminated, the other found by back-substitution, and both checked in the original equations — not the simplified ones.

A clean subtraction case

$5x+2y=16$ and $5x-3y=1$. The $x$-coefficients match exactly, so subtract:

$y=3$. Back-substitute: $5x+$$2(3)=$$16\to$$5x=$$10\to$$x=$$5x+2(3)=16 \rightarrow 5x=10 \rightarrow x=2$. Check both: $5(2)+$$2(3)=$$5(2)+2(3)=16$ ✓ and $5(2)-$$3(3)=$$10-$$9=$$5(2)-3(3)=10-9=1$ ✓.

A case that needs a sign flip

$5x+4y=7$ and $5x-3y=21$. The $x$-coefficients match, so subtract — carefully, since the second equation's $y$-term is negative:

Note the $y$-row: $4y-(-3y)=7y$, not $4y-3y$ — the sign flip is the whole trick. So $y=-2$. Back-substitute: $5x+$$4(-2)=$$7\to$$5x=$$15\to$$x=$$5x+4(-2)=7 \rightarrow 5x=15 \rightarrow x=3$. Check both: $5(3)+$$4(-2)=$$15-$$8=$$5(3)+4(-2)=15-8=7$ ✓ and $5(3)-$$3(-2)=$$15+$$6=$$5(3)-3(-2)=15+6=21$ ✓.

A case that needs scaling

$2x+3y=13$ and $4x-y=19$. Neither variable's coefficients match, so scale the second equation by $-3$ first: $-12x+3y=-57$. Now the $y$-coefficients match, so subtract:

$x=5$. Back-substitute: $2(5)+$$3y=$$13\to$$3y=$$3\to$$y=$$2(5)+3y=13 \rightarrow 3y=3 \rightarrow y=1$. Check both: $2(5)+$$3(1)=$$2(5)+3(1)=13$ ✓ and $4(5)-1=19$ ✓.

Back-substituting to get the second variable

In every case above, elimination only finds one variable. Take that value straight back into either original equation — whichever has simpler numbers — and solve normally for the second variable, exactly as shown three times over.

Checking the pair in both original equations

The pair isn't confirmed until both original equations accept it, not just the one used for back-substitution. All three worked systems above check clean in both equations — that double check is what separates a finished answer from a guessed one.

§ 07

Elimination vs substitution vs graphing: which to use on a test

Pick elimination when coefficients line up

If a quick glance shows matching or easily-scaled coefficients, elimination is usually the fastest route on paper.

Pick substitution when a variable is already isolated

For $x=y+4$ and $x+y=10$, the substitution method is faster than elimination: substitute directly, $(y+4)+$$y=$$10\to$$2y=$$6\to$$y=$$(y+4)+y=10 \rightarrow 2y=6 \rightarrow y=3$, then $x=7$. The same goes for a system that pairs a line with a direct variation such as y = 3x: for $y=3x$ and $x+y=8$, substitution gives $4x=8$, so $x=2$ and $y=6$.

Graphing: fine for a check, slow for a grade

For $x+y=5$ and $x-y=1$, the graphing method shows two lines crossing at $(3,2)$ — a useful sanity check, but plotting two lines by hand is rarely faster than either algebraic method on a timed test. Finding where two lines cross is also the skill behind coordinate-geometry problems.

§ 08

Sign errors that cost marks (and how to catch them)

Scroll the table sideways to see every column.

Forgetting to subtract the constant

Using $5x+3y=11$ and $5x+y=7$: subtracting correctly gives $2y=4$, so $y=2$. Forgetting to subtract algebraic equations' constants correctly — leaving $2y=11$ — hands back $y=5.5$, which fails the check against either original equation.

Subtracting only the left side

The same pair, subtracted incorrectly by canceling only the $y$-terms and leaving the $x$-terms untouched, produces $5x+2y=4$ — a line, not a solved value, because a variable that should have vanished didn't.

Losing the minus on a negative coefficient

This is the sign-flip trap from the worked system above: $4y-(-3y)$ must become $7y$, not $y$. Losing that minus sign is the single most common way to subtract algebraic equations with variables incorrectly.

The 30-second check that catches all three

Substitute both values back into both original equations, not the simplified ones. If either fails, the error happened during the subtraction step, not the back-substitution — go back and re-check every sign in the second equation before touching anything else.

§ 09

No solution and infinitely many: what elimination tells you

When both variables vanish and you get 0 = 5

$x+2y=8$ and $x+2y=3$. Subtracting: $(x-x)+$$(2y-2y)=$$8-$$3\to$$0=$$(x-x)+(2y-2y)=8-3 \rightarrow 0=5$ — a false statement. Both variables vanished and nothing survived to solve for, so this is an inconsistent system: no solution exists.

When both vanish and you get 0 = 0

$x+2y=8$ and $2x+4y=16$. Scale the first equation by 2: $2x+4y=16$. Subtracting: $(2x-2x)+$$(4y-4y)=$$16-$$16\to$$0=$$(2x-2x)+(4y-4y)=16-16 \rightarrow 0=0$ — always true. This is a consistent, dependent system: the two equations describe the same line, so there are infinitely many solutions. A market's supply and demand curves are the more familiar version of a two-equation system; see accounting, finance and economics homework for that application, while the least-squares systems behind a regression line are statistics-desk work.

§ 10

Practice: ten systems with answers

Scroll the table sideways to see every column.

§ 11

If the whole problem set is elimination

Twenty systems, each one a fresh set of coefficients to check for signs and scaling, is close to ninety minutes of careful arithmetic — and one dropped sign on system 14 costs the same mark as skipping it outright. That's the point where a lot of students go looking for someone to work the whole systems set instead of rechecking every sign by hand.

GradeDraft's math desk, part of our do my homework service, solves systems problem sets, from clean-subtraction items through the scaling and sign-flip cases above, with the elimination shown line by line the same way it's shown here, priced from $25 per problem after a quick look at the set.

If a single equation is the sticking point rather than a pair, start with what value makes a single equation true. Once $x^2$ shows up in one of the equations, see the three forms a quadratic can take. Students working the CPM curriculum should also see CPM Core Connections systems problems, which drills this exact elimination method in that textbook's own notation. Free lessons on CK-12 and Homework.Study.com walk through the same mechanics if you want more practice before checking your own work here.