GradeDraftSubtracting the Second Equation From the First: Solving Systems by Elimination
Subtracting the second equation from the first means subtracting left side from left side and right side from right side, term by term, so that matching coefficients cancel and one variable disappears. For minus , the -terms cancel and the result is , leaving one variable to solve directly.
On this page
- The instruction, decoded
- Line by line: the subtraction, with the signs kept honest
- Interactive: subtract your own two equations
- When to subtract and when to add
- When the coefficients do not match: multiply first
- Worked 2x2 systems, start to finish
- Elimination vs substitution vs graphing: which to use on a test
- Sign errors that cost marks (and how to catch them)
- No solution and infinitely many: what elimination tells you
- Practice: ten systems with answers
- If the whole problem set is elimination
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 the exact pair currently sitting on Wyzant and Brainly answers: and .
8x+3y&=14\\ -(4x+3y&=8)\\ \hline 4x+0y&=6 \end{aligned}$$ 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 \rightarrow 12+3y=14 \rightarrow y=\dfrac23$. Check the second equation: $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. ## 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=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. ## 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.| Coefficients on the target variable | Move |
|---|---|
| Same sign, same magnitude | Subtract |
| Opposite signs, same magnitude | Add |
Same sign, same coefficient: subtract
When both equations carry the same coefficient with the same sign — like the and above — subtracting cancels that variable directly.
Opposite signs, same magnitude: add
When the coefficients are equal in size but opposite in sign — like and — 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 and , which share the solution . Adding them: ; then — solved in two lines. Subtracting them instead: — nothing cancels, because the -coefficients were opposite in sign to begin with. Run both on the same system and the correct choice is obvious immediately.
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 and , the -coefficients are 2 and 1 — not equal, so subtracting as-is does nothing useful. Scale the second equation by 2: . Now subtract the first: . Back-substitute: . Check: — confirmed.
Scaling both (least common multiple of the coefficients)
For and , neither coefficient matches the other on either variable. Scale to the least common multiple of the -coefficients, 6: multiply the first equation by 3 () and the second by 2 (). The -terms are now opposite in sign and equal in magnitude, so add: . Back-substitute: . Check: — confirmed.
A worked 3x + 2y / 5x - 3y system
That's the full worked system above: scale both equations to a common -coefficient of 6, then combine. The linear combination that results — — 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; see physics homework help for a worked example using Kirchhoff's laws instead of arbitrary coefficients.
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
and . The -coefficients match exactly, so subtract:
5x+2y&=16\\ -(5x-3y&=1)\\ \hline 0x+5y&=15 \end{aligned}$$ $y=3$. Back-substitute: $5x+2(3)=16 \rightarrow 5x=10 \rightarrow x=2$. Check both: $5(2)+2(3)=16$ ✓ and $5(2)-3(3)=10-9=1$ ✓. ### 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: $$\begin{aligned} 5x+4y&=7\\ -(5x-3y&=21)\\ \hline 0x+7y&=-14 \end{aligned}$$ Note the $y$-row: $4y-(-3y)=7y$, not $4y-3y$ — the sign flip is the whole trick. So $y=-2$. Back-substitute: $5x+4(-2)=7 \rightarrow 5x=15 \rightarrow x=3$. Check both: $5(3)+4(-2)=15-8=7$ ✓ and $5(3)-3(-2)=15+6=21$ ✓. ### 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: $$\begin{aligned} 2x+3y&=13\\ -(-12x+3y&=-57)\\ \hline 14x+0y&=70 \end{aligned}$$ $x=5$. Back-substitute: $2(5)+3y=13 \rightarrow 3y=3 \rightarrow y=1$. Check both: $2(5)+3(1)=13$ ✓ 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. ## Elimination vs substitution vs graphing: which to use on a test| Situation | Best method |
|---|---|
| Coefficients on one variable already match or scale easily | Elimination |
| One equation already has a variable isolated (like ) | Substitution |
| You just need a rough visual check, not a graded answer | Graphing |
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 and , the substitution method is faster than elimination: substitute directly, , then .
Graphing: fine for a check, slow for a grade
For and , the graphing method shows two lines crossing at — 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 coordinate-geometry skill covered in geometry help.
Sign errors that cost marks (and how to catch them)
| Error | What it looks like on paper | Wrong result it produces | How to catch it |
|---|---|---|---|
| Forgetting to subtract the constant | Keeps the first equation's constant unchanged instead of subtracting the second | instead of | Substitute the "solution" back — it fails one original equation |
| Subtracting only the left side | Cancels the matching variable but leaves the other variable's coefficient untouched | instead of | The variable that should have vanished is still sitting in the result |
| Losing the minus on a negative coefficient | Writes instead of | instead of | Re-check the sign of every term in the second equation before subtracting |
| Distributing the minus sign across only part of the equation | Flips the sign on the variable terms but not the constant, or the reverse | A result off by exactly twice the missed term | Rewrite the subtraction as adding the negative of every term, one at a time |
Scroll horizontally to compare all columns.
Forgetting to subtract the constant
Using and : subtracting correctly gives , so . Forgetting to subtract algebraic equations' constants correctly — leaving — hands back , which fails the check against either original equation.
Subtracting only the left side
The same pair, subtracted incorrectly by canceling only the -terms and leaving the -terms untouched, produces — 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: must become , not . 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.
No solution and infinitely many: what elimination tells you
When both variables vanish and you get 0 = 5
and . Subtracting: — 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
and . Scale the first equation by 2: . Subtracting: — 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, and statistics homework help for the least-squares systems behind a regression line.
Practice: ten systems with answers
| # | System | Type | Answer |
|---|---|---|---|
| 1 | Clean subtraction | ||
| 2 | Clean subtraction | ||
| 3 | Sign flip | ||
| 4 | Clean subtraction | ||
| 5 | Clean subtraction | ||
| 6 | Needs adding | ||
| 7 | Dependent | Infinitely many solutions | |
| 8 | Inconsistent | No solution | |
| 9 | Needs adding | ||
| 10 | Needs scaling |
Scroll horizontally to compare all columns.
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 to do my math homework instead of rechecking every sign by hand.
GradeDraft's math homework desk 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 shows up in one of the equations, see forms of quadratic equations. 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.
Common questions about adding and subtracting equations
What is the result of subtracting the second equation from the first?
A new, simpler equation in which one variable's coefficient has been reduced to zero — the matching terms cancel, leaving a single-variable equation you can solve directly and then back-substitute.
When do you add equations and when do you subtract them?
Subtract when the coefficients on the variable you want to eliminate have the same sign; add when they have opposite signs and the same magnitude. Either move is aimed at the same goal: forcing that variable's combined coefficient to zero.
How do you subtract two equations?
Subtract the left side of the second equation from the left side of the first, and the right side of the second from the right side of the first, term by term, keeping every sign — including the ones on negative coefficients and constants.
Why does subtracting eliminate a variable?
Because subtracting equal coefficients from each other always produces zero. If both equations share the exact same coefficient on a variable, subtracting cancels that variable's term entirely, leaving one equation with only the other variable in it.
What if the coefficients do not match?
Scale one or both equations first — multiply every term in an equation by the same number — until the target variable's coefficients match in magnitude. Then subtract or add as usual, as shown in the "multiply first" section above.
Is elimination faster than substitution on a test?
Usually, when the coefficients already match or scale in one step. When a variable is already isolated — like — substitution is typically faster, since elimination would need an unnecessary scaling step first.