When should I use elimination instead of substitution?

Use elimination when both equations are in standard form Ax + By = C and no variable already has a coefficient of 1. Substitution shines when a variable is alone, like y = 3x - 5, because you can plug it straight in. If one equation has a variable with coefficient 1, either method is quick; if all four coefficients are 2 or larger, elimination usually avoids messy fractions.

Do I have to add the equations, or can I subtract?

Both are valid. Adding works when the matching coefficients are opposites; subtracting works when they are equal. Many students choose to always add, and get there by multiplying one equation by a negative constant. Subtraction requires distributing the minus sign across every term, including the constant, and that is where sign errors creep in.

What does it mean if I end up with 0 = 5?

It means the system has no solution. Both variables canceled and what remains is false for every possible x and y, so no ordered pair satisfies both equations. Graphically the lines are parallel with different y-intercepts. Write "no solution" as your answer — do not write 0 or leave it blank.

Can elimination handle a system with fractions or decimals?

Yes, and the first move is to clear them. Multiply each equation by the least common denominator, or by a power of ten for decimals, to get whole-number coefficients. For example, 1/2x + 2/3y = 5 becomes 3x + 4y = 30 after multiplying every term by 6. Then eliminate as usual.

ALG1-5.3

Solving Systems by Elimination

Learn to solve linear systems by elimination in Algebra 1: add or subtract equations, scale first when needed, and read what 0 = 0 or 0 = 12 really means.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Solving Systems by Elimination, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Solving Systems by Elimination free

What this lesson covers

You already know two ways to find where two lines meet: graphing shows the point, and substitution isolates a variable. Elimination is the third method, and it is usually the fastest when both equations are written in the form

Ax + By = C

with no variable conveniently alone.

The idea is simple and a little sneaky: if you add the two equations together in just the right way, one variable disappears completely, leaving a one-variable equation you can solve in a single step. Sometimes the coefficients line up on their own. More often you multiply one or both equations by a constant first so that the coefficients of one variable become opposites. This lesson shows you how to choose that multiplier, how to finish and check, and how to read the strange-looking results like

0 = 0

and

0 = 12

that tell you the lines are identical or parallel.

Why Adding Two Equations Is Legal

Elimination rests on one property: if

a = b

and

c = d

, then

a + c = b + d

. You are adding equal amounts to both sides of an equation, which never changes the solution set.

Here is what that means for a system. Suppose the true solution is the point

(x, y)

that satisfies both equations at once. At that point, the left side of equation 2 and the right side of equation 2 are literally the same number. So adding the left side of equation 2 to the left side of equation 1, and the right side to the right side, adds the same number to both sides of equation 1. The point

(x, y)

still works in the new equation.

The payoff comes when the

y

-terms are opposites. Consider

5x + 2y = 1

3x - 2y = 7

Adding straight down gives

8x + 0y = 8

, so

x = 1

. The

y

-terms canceled because

2y + (-2y) = 0

. Substituting

x = 1

into either original equation gives

5(1) + 2y = 1

, so

2y = -4

and

y = -2

. The solution is

(1, -2)

.

Two cautions. First, you must add the entire equations, every term including the constants — a very common slip is adding the variable terms but forgetting the right-hand sides. Second, adding only helps when a pair of coefficients are opposites. If the coefficients are equal instead, subtract, or multiply one equation by

-1

and then add, which many students find safer because subtraction sign errors are so easy to make.

Scaling One Equation to Force a Match

Most systems do not hand you opposite coefficients. You create them. Multiplying an entire equation by a nonzero constant produces an equivalent equation — the same line, the same solutions — but with new coefficients you can control.

Take

3x + 4y = 10

5x - 2y = 8

The

y

-coefficients are

4

and

-2

. Multiply the second equation by

2

to get

10x - 4y = 16

. Now the

y

-coefficients are

4

and

-4

. Adding gives

13x = 26

, so

x = 2

, and back-substituting into

3x + 4y = 10

gives

4y = 4

, so

y = 1

. The solution is

(2, 1)

.

The decision you have to make is which variable to eliminate and what to multiply by. Scan for a coefficient that divides evenly into the other one in the same column. In the example above,

2

divides into

4

, so one multiplication was enough.

Situation	What to do
Coefficients already opposites, like $2y$ and $-2y$	Add the equations
Coefficients already equal, like $3x$ and $3x$	Subtract, or multiply one by $-1$ and add
One coefficient divides the other, like $2$ and $6$	Scale the smaller equation only
Neither divides the other, like $2$ and $3$	Scale both, using the least common multiple

The single biggest error here is multiplying only part of an equation. If you multiply

5x - 2y = 8

2

, the

8

must become

16

. Write the scaled equation out on its own line rather than editing terms in place, and the mistake mostly disappears.

Scaling Both Equations

When neither coefficient divides the other, use the least common multiple. For

4x + 3y = 6

3x + 5y = 10

the

x

-coefficients are

4

and

3

, with

\mathrm{LCM} = 12

. Multiply the first equation by

3

and the second by

-4

12x + 9y = 18

-12x - 20y = -40

Adding gives

-11y = -22

, so

y = 2

. Then

4x + 3(2) = 6

gives

4x = 0

and

x = 0

. The solution is

(0, 2)

, and you should check it in the other original equation:

3(0) + 5(2) = 10

. True.

Notice the choice to make one multiplier negative. You could instead multiply by

3

and

4

and subtract, but building the negative into the multiplier means you only ever add, and adding produces far fewer sign errors than subtracting a whole row of terms.

You can also choose to eliminate

y

instead:

\mathrm{LCM}(3, 5) = 15

, so multiply the first by

5

and the second by

-3

. You get the same answer. Pick whichever column gives smaller numbers; here the

x

-column does.

One more habit worth building: after you solve for the first variable, substitute back into an original equation, not a scaled one. Scaled equations are correct, but if you made an arithmetic error while scaling, substituting into the original catches it. And always finish by writing the answer as an ordered pair. A solved system has a point as its answer, not a lone number.

When Both Variables Vanish: No Solution or Infinitely Many

Sometimes elimination wipes out both variables at once. What is left tells you everything.

Case one, a false statement. Solve

2x + 6y = 8

x + 3y = 10

Multiply the second by

-2

-2x - 6y = -20

. Adding gives

0 = -12

, which is false for every

x

and

y

. No ordered pair can make it true, so the system has no solution. Geometrically, the two equations describe parallel lines with the same slope and different intercepts — they never intersect. Such a system is called inconsistent.

Case two, a true statement. Solve

x - 2y = 5

-2x + 4y = -10

Multiply the first by

2

2x - 4y = 10

. Adding gives

0 = 0

, true always. Every point on the line

x - 2y = 5

satisfies both equations, so there are infinitely many solutions. The second equation is just the first multiplied by

-2

— the same line written twice.

Result after elimination	Meaning	The graph
$x = a$ (then $y = b$ )	One solution $(a, b)$	Lines cross once
False, like $0 = -12$	No solution	Parallel distinct lines
True, like $0 = 0$	Infinitely many solutions	One line, written twice

Where students go wrong: writing "

x = 0

" or "the answer is 0" after reaching

0 = 0

. Nothing equals zero here — both variables are gone. The honest answer is a sentence about the number of solutions, and for the infinite case it is good practice to name the line, such as "all points on

x - 2y = 5

Key terms

System of linear equations.: Two or more linear equations considered together; a solution must satisfy every equation in the system at the same time.
Elimination method.: A solving strategy that adds or subtracts multiples of the equations so that one variable's terms cancel, leaving a single equation in one variable.
Equivalent equations.: Equations with exactly the same solution set. Multiplying an entire equation by a nonzero constant produces an equivalent equation.
Coefficient.: The number multiplying a variable. In $5x - 2y = 7$ , the coefficients are $5$ and $-2$ .
Least common multiple (LCM).: The smallest positive number both coefficients divide into; used to choose multipliers when both equations must be scaled.
Inconsistent system.: A system with no solution. Elimination produces a false numeric statement such as $0 = -12$ , and the lines are parallel.
Dependent system.: A system whose equations describe the same line, giving infinitely many solutions. Elimination produces $0 = 0$ .
Ordered pair solution.: The point $(x, y)$ that makes both equations true; the standard way to write the answer to a system with one solution.

Worked example

Solve the system by elimination and check your answer:

2x + 3y = 12

and

5x - 2y = 11

Step 1: Decide which variable to eliminate. The

x

-coefficients are

2

and

5

; the

y

-coefficients are

3

and

-2

. Neither column has one number dividing the other, so both equations must be scaled either way. The

y

-column is attractive because the signs are already opposite, so plain addition will finish the job.

Step 2: Find the multipliers.

\mathrm{LCM}(3, 2) = 6

. Multiply the first equation by

2

and the second by

3

2(2x + 3y = 12) \;\longrightarrow\; 4x + 6y = 24

3(5x - 2y = 11) \;\longrightarrow\; 15x - 6y = 33

Every term, including the constant on the right, gets multiplied.

Step 3: Add the scaled equations. The

y

-terms cancel because

6y + (-6y) = 0

19x = 57

Step 4: Solve for

x

. Divide both sides by

19

to get

x = 3

.

Step 5: Back-substitute into an original equation. Using

2x + 3y = 12

2(3) + 3y = 12

, so

6 + 3y = 12

, then

3y = 6

and

y = 2

.

Step 6: Check in the other original equation, the one you did not use:

5(3) - 2(2) = 15 - 4 = 11

. That matches, so both equations are satisfied.

Solution:

(3, 2)

. On a graph, the lines

2x + 3y = 12

and

5x - 2y = 11

cross at exactly this point.

Practice questions

To solve the system

3x + 2y = 7

and

4x - 3y = -2

by eliminating

y

, which pair of multipliers works?

Multiply the first equation by $3$ and the second by $2$
Multiply the first equation by $4$ and the second by $3$
Multiply the first equation by $2$ and the second by $3$
Multiply the first equation by $-1$ and leave the second alone

Answer: Multiply the first equation by $3$ and the second by $2$

To eliminate

y

, focus only on the

y

-coefficients, which are

2

and

-3

. Their least common multiple is

6

. Multiplying the first equation by

3

gives

9x + 6y = 21

; multiplying the second by

2

gives

8x - 6y = -4

. Now

6y

and

-6y

are opposites, so adding gives

17x = 17

and

x = 1

. The choice using

4

and

3

targets the

x

-column instead, which would eliminate

x

, not

y

Solve by elimination:

x - 4y = 6

and

-3x + 12y = -18

. Describe the solution set completely.

Answer: Infinitely many solutions — every point on the line $x - 4y = 6$ .

Multiply the first equation by

3

to get

3x - 12y = 18

. Adding this to

-3x + 12y = -18

gives

0 = 0

. Both variables vanished and the leftover statement is true, so no single point is being pinned down; instead, every pair that satisfies the first equation also satisfies the second. In fact the second equation is the first multiplied by

-3

, so the two equations graph as the same line. A complete answer says "infinitely many solutions" and names the line, rather than writing something like

x = 0

A student solves

4x + y = 9

and

4x + 3y = 19

by subtracting the first equation from the second and writes

2y = 10

, so

y = 5

, and then reports the solution as

y = 5

. Identify every error and give the correct solution.

Answer: The subtraction is correct ( $2y = 10$ , $y = 5$ ), but the answer is incomplete: the student never found $x$ . Substituting $y = 5$ into $4x + y = 9$ gives $4x = 4$ , so $x = 1$ , and the solution is the ordered pair $(1, 5)$ .

Subtracting the equations gives

(4x - 4x) + (3y - y) = 19 - 9

, which is

2y = 10

, so the elimination step is fine. The trouble is stopping early. A system of two equations in two variables asks where two lines meet, and a meeting place is a point, not a number. Always back-substitute to find the second coordinate, then check the pair in the equation you did not use:

4(1) + 3(5) = 4 + 15 = 19

FAQ

When should I use elimination instead of substitution?: Use elimination when both equations are in standard form $Ax + By = C$ and no variable already has a coefficient of $1$ . Substitution shines when a variable is alone, like $y = 3x - 5$ , because you can plug it straight in. If one equation has a variable with coefficient $1$ , either method is quick; if all four coefficients are 2 or larger, elimination usually avoids messy fractions.
Do I have to add the equations, or can I subtract?: Both are valid. Adding works when the matching coefficients are opposites; subtracting works when they are equal. Many students choose to always add, and get there by multiplying one equation by a negative constant. Subtraction requires distributing the minus sign across every term, including the constant, and that is where sign errors creep in.
What does it mean if I end up with $0 = 5$ ?: It means the system has no solution. Both variables canceled and what remains is false for every possible $x$ and $y$ , so no ordered pair satisfies both equations. Graphically the lines are parallel with different $y$ -intercepts. Write "no solution" as your answer — do not write $0$ or leave it blank.
Can elimination handle a system with fractions or decimals?: Yes, and the first move is to clear them. Multiply each equation by the least common denominator, or by a power of ten for decimals, to get whole-number coefficients. For example, $\frac{1}{2}x + \frac{2}{3}y = 5$ becomes $3x + 4y = 30$ after multiplying every term by $6$ . Then eliminate as usual.

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