Solving Pairs of Linear Equations by Elimination

This lesson covers the elimination method for solving a pair of linear equations in two variables. The key idea is to combine the two equations so that one variable cancels out, leaving a single equation in one variable.

The rule for adding or subtracting

First check whether either variable has the same coefficient in both equations. To eliminate that variable, look at the signs of its terms:

• If the signs are opposite, add the equations.
• If the signs are the same, subtract one equation from the other.

Example 1: opposite signs, so add

Solve the pair

$x + y = 10 \qquad (1)$ $x - y = 2 \qquad (2)$

The $y$ terms have opposite signs, so add $(1)$ and $(2)$:

$2x = 12 \implies x = \tfrac{12}{2} = 6$

Substitute $x = 6$ into $(1)$:

$6 + y = 10 \implies y = 4$

So the solution is $x = 6,\ y = 4$.

Example 2: same signs, so subtract

Solve the pair

$2x + 3y = 2 \qquad (1)$ $2x + 5y = 2 \qquad (2)$

The $x$ terms are identical with the same sign, so subtract $(2)$ from $(1)$:

$-2y = 0 \implies y = 0$

Substitute $y = 0$ into $(1)$:

$2x = 2 \implies x = 1$

So the solution is $x = 1,\ y = 0$.

Example 3: different coefficients, so scale first

Solve the pair

$10x + 3y = 75 \qquad (1)$ $6x - 5y = 11 \qquad (2)$

Here neither variable has matching coefficients, so first make one match.

Eliminating $y$

Multiply $(1)$ by $5$ (the coefficient of $y$ in equation 2) and $(2)$ by $3$ (the coefficient of $y$ in equation 1):

$50x + 15y = 375 \qquad (3)$ $18x - 15y = 33 \qquad (4)$

The $y$ terms now have opposite signs, so add $(3)$ and $(4)$:

$68x = 408 \implies x = 6$

Substitute $x = 6$ into $(1)$:

$60 + 3y = 75 \implies 3y = 15 \implies y = 5$

Eliminating $x$ instead

The same pair can be solved by removing $x$. Multiply $(1)$ by $6$ and $(2)$ by $10$:

$60x + 18y = 450 \qquad (3)$ $60x - 50y = 110 \qquad (4)$

The $x$ terms match with the same sign, so subtract $(4)$ from $(3)$:

$68y = 340 \implies y = 5$

Substitute $y = 5$ into $(2)$:

$6x - 25 = 11 \implies 6x = 36 \implies x = 6$

Either route gives the solution $x = 6,\ y = 5$.

Example 4: decimal coefficients

Solve the pair

$0.4x - 1.5y = 6.5 \qquad (1)$ $0.3x + 0.2y = 0.9 \qquad (2)$

Clear the decimals by multiplying each equation by $10$:

$4x - 15y = 65 \qquad (3)$ $3x + 2y = 9 \qquad (4)$

Eliminate $y$. Multiply $(3)$ by $2$ and $(4)$ by $15$:

$8x - 30y = 130 \qquad (5)$ $45x + 30y = 135 \qquad (6)$

The $y$ terms have opposite signs, so add $(5)$ and $(6)$:

$53x = 265 \implies x = 5$

Substitute $x = 5$ into $(3)$:

$20 - 15y = 65 \implies -15y = 45 \implies y = -3$

So the solution is $x = 5,\ y = -3$.

Key takeaways

• To eliminate a variable, make its coefficients equal in both equations, scaling each equation if needed.
• Add the equations when the matching terms have opposite signs; subtract when they have the same sign.
• Once one variable is found, substitute it back into the simpler equation to find the other, and clear decimals to whole numbers first to keep the working easy.