Solving Linear Equations in Two Variables (Elimination Method)

This lesson solves three pairs of linear equations in two variables using the elimination method. In each case the coefficients are letters, so we scale the equations to cancel one variable, then add or subtract depending on the signs of the matching terms.

The elimination idea

To eliminate a variable, multiply each equation by the coefficient that variable has in the other equation, so the two terms match. Then:

• if the matching terms have opposite signs, add the equations,
• if they have the same sign, subtract them.

Example 1: equations in $p$ and $q$

We are given

$px + qy = p - q \quad (1)$ $qx - py = p + q \quad (2)$

To eliminate $y$, multiply $(1)$ by $p$ and $(2)$ by $q$:

$p^2 x + pq\,y = p(p - q) \quad (3)$ $q^2 x - pq\,y = q(p + q) \quad (4)$

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

$(p^2 + q^2)\,x = p(p - q) + q(p + q) = p^2 - pq + pq + q^2 = p^2 + q^2.$

Therefore

$x = \frac{p^2 + q^2}{p^2 + q^2} = 1.$

Put $x = 1$ into $(1)$: $\;p + qy = p - q$, so $qy = -q$ and

$y = -1.$

The solution is $x = 1,\ y = -1$.

Example 2: equations in $a$, $b$ and $c$

We are given

$ax + by = c \quad (1)$ $bx + ay = 1 + c \quad (2)$

To eliminate $x$, multiply $(1)$ by $b$ and $(2)$ by $a$:

$ab\,x + b^2 y = bc \quad (3)$ $ab\,x + a^2 y = a(1 + c) \quad (4)$

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

$(b^2 - a^2)\,y = bc - a(1 + c) = bc - a - ac.$

Writing the right side as $c(a - b) + a$ over the rearranged denominator gives

$y = \frac{c(a - b) + a}{a^2 - b^2}.$

To find $x$, substitute this $y$ back into $(1)$ and simplify. Bringing $by$ across and taking the common denominator $a^2 - b^2$ leaves

$x = \frac{c(a - b) - b}{a^2 - b^2}.$

Check. Substituting both into $ax + by$ gives $\dfrac{c(a-b)(a+b)}{a^2-b^2} = c$, as required.

Example 3: equations in $a$ and $b$

We are given

$\frac{x}{a} - \frac{y}{b} = 0 \quad (1)$ $ax + by = a^2 + b^2 \quad (2)$

Clearing the fractions in $(1)$ gives $bx - ay = 0$, call it $(3)$. To eliminate $y$, multiply $(3)$ by $b$ and $(2)$ by $a$:

$b^2 x - ab\,y = 0 \quad (4)$ $a^2 x + ab\,y = a(a^2 + b^2) \quad (5)$

Add $(4)$ and $(5)$:

$(a^2 + b^2)\,x = a(a^2 + b^2), \qquad x = a.$

Put $x = a$ into $(3)$: $\;ba - ay = 0$, so $ay = ab$ and

$y = b.$

The solution is $x = a,\ y = b$.

Key takeaways

• To eliminate a variable, scale each equation by that variable's coefficient in the other equation so the terms match.
• Add the equations when the matching terms have opposite signs; subtract when they share the same sign.
• After finding one unknown, back-substitute into an original equation to find the other.