When should I use elimination instead of substitution?

Use elimination when both equations are in standard form (ax + by = c) and have simple coefficients; it often requires fewer steps than substitution. Substitution is usually faster if one equation is already solved for a variable, like y = 2x + 3. On the SAT and ACT, elimination is often the quickest method when coefficients are set up nicely.

Do I have to eliminate x? Can I eliminate y first?

Yes, you can eliminate either variable first. Choose whichever requires fewer multiplication steps or simpler arithmetic. The final answer—the values of x and y—will be the same either way, so pick whatever is easiest for you.

What if I get 0 = 0 or 0 = 5 after adding equations?

If you get 0 = 0 (an identity, always true), the equations represent the same line and have infinitely many solutions. If you get 0 = 5 (a contradiction, never true), the lines are parallel and there is no solution. Either outcome means the system is dependent or inconsistent, not a typical problem with one unique solution.

Is elimination the same as the addition-subtraction method?

Yes. Elimination, addition-subtraction method, and linear combination method all mean the same thing: combining equations by adding or subtracting (after multiplying if needed) to remove one variable. Different textbooks use different names, but the process is identical.

Solving by Elimination

Solve a system of two linear equations by adding or subtracting to eliminate one variable, then solve for the remaining variable.

\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

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What each symbol means

What Solving by Elimination takes

Symbol	Meaning
$a_i$	The coefficient of $x$ in equation $i$ ; can be negative or zero, and if it matches the other equation's $x$ coefficient, you'll need to multiply before eliminating.
$b_i$	The coefficient of $y$ in equation $i$ ; can be negative or zero, and tells you $y$ 's contribution to each equation.
$c_i$	The constant term (right-hand side) of equation $i$ ; represents what the left side equals, and misreading it as a coefficient causes sign errors.
$x$	One variable you are solving for; it appears in both equations, and confusing it with $y$ leads to writing the wrong answer.
$y$	The other variable you are solving for; it appears in both equations alongside $x$ , and must be found separately after eliminating one.

When to use it

Use elimination when two equations are already in standard form and you want to remove one variable by combining them.

Level

Usually taught in: Algebra I · Appears on: SAT, ACT

Worked examples

1. Solve a system with opposite y coefficients

Problem

Solve

\begin{cases} 3x + 2y = 11 \\ 5x - 2y = 13 \end{cases}

$3x + 2y = 11 \quad \text{and} \quad 5x - 2y = 13$
The coefficients of $y$ are opposite: $+2$ and $-2$ , so adding the equations will eliminate $y$ .
$(3x + 5x) + (2y - 2y) = 11 + 13$
Add the left sides and right sides of both equations.
$8x = 24$
Combine like terms: $3x + 5x = 8x$ , $2y - 2y = 0$ , and $11 + 13 = 24$ .
$x = 3$
Divide both sides by 8.
$3(3) + 2y = 11$
Substitute $x = 3$ into the first original equation.
$9 + 2y = 11$
Simplify by computing $3 \cdot 3 = 9$ .
$2y = 2$
Subtract 9 from both sides.
$y = 1$
Divide both sides by 2.

Answer: $x = 3 \text{ and } y = 1$

We chose to add the equations because the $y$ coefficients were already opposites, requiring no multiplication. After eliminating $y$ and finding $x = 3$ , we substituted back into the first equation to find $y = 1$ . Both values satisfy both original equations when checked.

2. Solve a system requiring multiplication

Problem

Solve

\begin{cases} 2x + 3y = 7 \\ 3x + 4y = 10 \end{cases}

$2x + 3y = 7 \quad \text{and} \quad 3x + 4y = 10$
Neither variable has opposite coefficients, so we must multiply to create them.
$3(2x + 3y) = 3(7) \quad \text{and} \quad (-2)(3x + 4y) = (-2)(10)$
Multiply the first equation by 3 and the second by $-2$ to make the $x$ coefficients opposite ( $6x$ and $-6x$ ).
$6x + 9y = 21 \quad \text{and} \quad -6x - 8y = -20$
Distribute the multiplier to every term in each equation: $3 \cdot 2x = 6x$ , $3 \cdot 3y = 9y$ , $3 \cdot 7 = 21$ for the first, and similarly for the second.
$(6x - 6x) + (9y - 8y) = 21 + (-20)$
Add the two equations to eliminate $x$ .
$y = 1$
Simplify: $6x - 6x = 0$ , $9y - 8y = y$ , and $21 - 20 = 1$ .
$2x + 3(1) = 7$
Substitute $y = 1$ into the first original equation.
$2x + 3 = 7$
Simplify by computing $3 \cdot 1 = 3$ .
$2x = 4$
Subtract 3 from both sides.
$x = 2$
Divide both sides by 2.

Answer: $x = 2 \text{ and } y = 1$

Since no coefficients were already opposite, we multiplied strategically: the first equation by 3 and the second by $-2$ to eliminate $x$ . This required careful distribution to all terms, including the constant. After finding $y = 1$ , substituting back gave $x = 2$ . Always verify by checking both solutions in both original equations.

3. Find garden dimensions given perimeter and length constraint

Problem

A homeowner is fencing a rectangular garden. The length is 3 meters more than the width. The perimeter is 50 meters. What are the dimensions?

$l = w + 3$
Translate "length is 3 meters more than width" into an equation, where $l$ is length and $w$ is width.
$2l + 2w = 50$
Translate "perimeter is 50 meters" using the perimeter formula for a rectangle: $P = 2l + 2w$ .
$l - w = 3$
Rearrange the first equation to standard form by subtracting $w$ from both sides.
$l + w = 25$
Simplify the second equation by dividing all terms by 2.
$\begin{cases} l - w = 3 \\ l + w = 25 \end{cases}$
Write as a system: the $w$ coefficients are now opposite ( $-1$ and $+1$ ).
$(l + l) + (-w + w) = 3 + 25$
Add the equations to eliminate $w$ .
$2l = 28$
Combine like terms: $l + l = 2l$ , $-w + w = 0$ , and $3 + 25 = 28$ .
$l = 14$
Divide both sides by 2.
$14 + w = 25$
Substitute $l = 14$ into the equation $l + w = 25$ .
$w = 11$
Subtract 14 from both sides.
$\text{Check:} \quad 2(14) + 2(11) = 28 + 22 = 50 \quad \checkmark$
Verify the perimeter formula: $2l + 2w = 2(14) + 2(11) = 28 + 22 = 50$ is satisfied, and $l - w = 14 - 11 = 3$ is also satisfied.

Answer: $\text{Length} = 14 \text{ meters, Width} = 11 \text{ meters}$

We translated the word problem into two equations: one relating length to width, and one for the perimeter. After rearranging both to standard form, the width coefficients were already opposite, so adding directly eliminated $w$ . This gave the length, which we substituted back to find the width. Word problems require an extra step of translation before the algebraic elimination method begins.

Common mistakes

Where Solving by Elimination usually goes wrong

The mistake	Why it is wrong	The fix
Adding equations when the coefficients have the same sign instead of opposite signs.	If both $y$ terms are positive or both are negative, adding does not cancel them; you must either multiply one equation by $-1$ first or subtract instead.	Before adding or subtracting, check the signs: if you have $+3y$ and $+2y$ , multiply the second equation by $-1$ to get $-2y$ , then add. Or subtract the second equation from the first.
Multiplying only the variable terms by a scaling factor and forgetting the constant term.	When you multiply an equation by a number, the multiplication must apply to every term, or the equation becomes false.	When multiplying an equation like $2x + 5y = 9$ by 3, write $3(2x + 5y) = 3(9)$ and distribute: $6x + 15y = 27$ , not $6x + 15y = 9$ .
Finding one variable and stopping, reporting only $x$ without finding $y$ .	A system of two equations requires two solutions, one for each variable. Elimination gives you one; you must substitute back into an original equation to find the other.	After eliminating one variable and solving for the first variable, always substitute your answer back into an original equation to find the missing variable.

Tips and when to use something else

Look at the coefficients before you start: if one pair is already opposite (like $3x$ and $-3x$ , or $2y$ and $-2y$ ), you can add or subtract immediately without any multiplication.
Choose which variable to eliminate by picking the one where multiplication is simplest. For example, if one coefficient is $1$ and another is $2$ , multiplying by $2$ is easier than dealing with larger numbers.
Always substitute your first solution back into one of the original equations to find the second variable; this step also acts as a check on your arithmetic.
If elimination requires dealing with messy fractions or large coefficients, consider Solving by Substitution instead, especially if one equation is already solved for a single variable.

Related formulas

Linear EquationSolve for the value of x in any linear equation by rearranging to isolate x, finding the unique solution where the expression equals zero.Solving by SubstitutionSolving by Substitution solves a system of equations by replacing one variable with its equivalent expression, turning two equations into one.Systems of Three EquationsSystems of three equations let you find three unknowns from three relationships, the standard approach when a real-world situation has multiple constraints.Distributive PropertyThe Distributive Property lets you multiply a number by a sum or difference by distributing the multiplication to each term separately.Two-Step EquationsSolve two-step equations of the form ax + b = c by undoing operations in reverse order to isolate the variable using inverse operations.

Frequently asked questions

When should I use elimination instead of substitution?: Use elimination when both equations are in standard form ( $ax + by = c$ ) and have simple coefficients; it often requires fewer steps than substitution. Substitution is usually faster if one equation is already solved for a variable, like $y = 2x + 3$ . On the SAT and ACT, elimination is often the quickest method when coefficients are set up nicely.
Do I have to eliminate x? Can I eliminate y first?: Yes, you can eliminate either variable first. Choose whichever requires fewer multiplication steps or simpler arithmetic. The final answer—the values of $x$ and $y$ —will be the same either way, so pick whatever is easiest for you.
What if I get 0 = 0 or 0 = 5 after adding equations?: If you get $0 = 0$ (an identity, always true), the equations represent the same line and have infinitely many solutions. If you get $0 = 5$ (a contradiction, never true), the lines are parallel and there is no solution. Either outcome means the system is dependent or inconsistent, not a typical problem with one unique solution.
Is elimination the same as the addition-subtraction method?: Yes. Elimination, addition-subtraction method, and linear combination method all mean the same thing: combining equations by adding or subtracting (after multiplying if needed) to remove one variable. Different textbooks use different names, but the process is identical.

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Reviewed 2026-09-18