Solving Simultaneous Linear Equations Algebraically

Theory

A pair of simultaneous linear equations in two unknowns is solved by finding the values of $x$ and $y$ that satisfy both equations. The two standard algebraic methods are substitution (rearrange one equation, sub into the other) and elimination (add or subtract the equations to remove a variable). Always check by substituting back.

A pair of simultaneous equations is two linear equations in the same two unknowns. The solution $(x, y)$ is the pair of values that makes both equations true at once.

Geometrically, each linear equation represents a straight line. The solution is the point where the two lines meet. A pair of simultaneous equations has either exactly one solution (lines cross), infinitely many (the same line written two ways), or none (parallel and distinct). The standard exam questions have exactly one solution.

There are two standard algebraic methods: substitution and elimination. Pick whichever suits the form of the equations.

The first diagram shows the geometric meaning of a unique solution: two lines cross at exactly one point, and its coordinates are the values of $x$ and $y$ that satisfy both equations. The second is a flowchart of the substitution method for a typical example.

The lines

y = x + 1

and

y = - x + 5

meet at

(2, 3)

— the unique solution.

Substitution flow: replace

y

in equation 2 with

x - 3

, solve, then back-substitute.

There are no new formulas — just two procedures. Pick the one that fits the form of the equations.

Substitution method

Best when one equation already has a variable isolated ( $y = \dots$ or $x = \dots$ ) or is easy to rearrange. Replace that variable in the other equation with its expression.

Elimination method

Best when both equations are in the form $a x + b y = c$ . Scale one or both equations so the coefficients of one variable match in size, then add or subtract.

Signs of the variable to eliminate	What to do
Same (both $+ 3 y$ or both $- 3 y$ )	Subtract the equations
Opposite ( $+ 3 y$ and $- 3 y$ )	Add the equations

Always check. Substitute the values you found back into both original equations. If both are satisfied, the solution is correct.

Substitution method — steps

Rearrange one equation to make a single variable the subject.
Substitute that expression into the other equation, giving one equation in one unknown.
Solve for that unknown.
Back-substitute into the rearranged equation to find the other unknown.

Elimination method — steps

Multiply one or both equations by suitable numbers so the coefficients of one variable become equal (or equal magnitude with opposite signs).
Add (if signs are opposite) or subtract (if signs are the same) to eliminate that variable.
Solve the resulting one-variable equation.
Back-substitute into one of the original equations to find the other unknown.

EXAMPLE 1 — SUBSTITUTION

Solve

y = x - 3

and

2 x + 3 y = 26

.

SOLUTION

The first equation already has $y$ by itself. Substitute $y = x - 3$ into the second.

$2 x + 3 (x - 3)$	$=$	$26$
$2 x + 3 x - 9$	$=$	$26$
$5 x$	$=$	$35$
$x$	$=$	$7$

Back-substitute: $y = 7 - 3 = 4$ . Solution: $(7, 4)$ .

x = 7, y = 4

EXAMPLE 2 — ELIMINATION (SUBTRACT)

Solve

3 x + 2 y = 7

and

x + 3 y = 7

.

SOLUTION

Multiply the second equation by $3$ so the $x$ coefficients match.

$3 x + 9 y$	$=$	$21$
$3 x + 2 y$	$=$	$7$

Both have $+ 3 x$ , so subtract to eliminate $x$ .

$7 y$	$=$	$14$
$y$	$=$	$2$

Back-substitute into $x + 3 y = 7$ : $x + 6 = 7$ , so $x = 1$ . Solution: $(1, 2)$ .

x = 1, y = 2

EXAMPLE 3 — ELIMINATION (ADD)

Solve

2 x + 3 y = 10

and

5 x - 3 y = 4

.

SOLUTION

The $y$ terms are $+ 3 y$ and $- 3 y$ — opposite signs. Add the equations.

$7 x$	$=$	$14$
$x$	$=$	$2$

Back-substitute into $2 x + 3 y = 10$ : $4 + 3 y = 10$ .

$3 y$	$=$	$6$
$y$	$=$	$2$

Solution: $(2, 2)$ .

x = 2, y = 2

EXAMPLE 4 — MULTIPLY BOTH EQUATIONS

Solve

2 x + 3 y = 12

and

3 x - 5 y = - 1

.

SOLUTION

Eliminate $y$ . Multiply the first equation by $5$ and the second by $3$ so the $y$ terms become $+ 15 y$ and $- 15 y$ .

$10 x + 15 y$	$=$	$60$
$9 x - 15 y$	$=$	$- 3$

Opposite signs — add.

$19 x$	$=$	$57$
$x$	$=$	$3$

Back-substitute into $2 x + 3 y = 12$ : $6 + 3 y = 12$ , so $y = 2$ . Solution: $(3, 2)$ .

x = 3, y = 2

Common pitfalls

Watch the signs when eliminating. If both equations have

+ 3 y

, subtract to remove it. If one has

+ 3 y

and the other has

- 3 y

, add. Mixing this up is the most common elimination error.

Multiply every term. When you multiply an equation by a number to line up coefficients, that number multiplies every term — including the right-hand side. Forgetting one term gives a wrong system.

Don't stop after one variable. Finding

x = 7

is not a complete answer. Back-substitute to get

y

too, and state the solution as the ordered pair

(x, y)

.

Check with the other equation. Substituting back into the equation you already used will always work — even if you made an error earlier. Use the equation you haven't used yet to verify.

Frequently asked questions

What are simultaneous equations?

Simultaneous equations are two or more equations that share the same unknowns. A solution is a set of values that makes every equation true at the same time. For a pair of linear equations in two unknowns, the solution is an ordered pair (x, y).

What is the substitution method?

Substitution works by rearranging one equation to make a single variable the subject (for example y equals ...) and substituting that expression into the other equation. This gives one equation in one unknown, which you can solve, then substitute back to find the other unknown.

What is the elimination method?

Elimination works by multiplying one or both equations so that the coefficients of one variable match (or are opposite). You then add the equations if the signs are opposite, or subtract if the signs are the same. This eliminates one variable and leaves a single equation in one unknown.

Should I add or subtract when eliminating?

Add when the signs of the variable you want to eliminate are opposite (for example plus 3y in one and minus 3y in the other). Subtract when the signs are the same (for example plus 3y in both).

How do I check my answer to simultaneous equations?

Substitute both values back into the equation you did not use for the final substitution. If both sides match, the solution is correct. Checking in the equation you already used will always work — even if you made an error — so use the other one.

When should I use substitution and when should I use elimination?

Use substitution when one equation already has a variable by itself, or is easy to rearrange. Use elimination when both equations are in the form a x plus b y equals c, especially when the coefficients line up nicely after multiplying by a small integer.

Video Lessons

Practice Questions

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Practice Questions

Solving Simultaneous Linear Equations Algebraically

📖 Theory

Substitution method

Elimination method

Substitution method — steps

Elimination method — steps

Common pitfalls

Frequently asked questions

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✏️ Practice Questions

Theory

Video Lessons

Practice Questions