Solving a special type of linear equations

Some pairs of linear equations have a handy symmetry: the coefficient of $x$ in the first equation equals the coefficient of $y$ in the second, and the coefficient of $y$ in the first equals the coefficient of $x$ in the second. When this happens you do not need full elimination or substitution. Adding and subtracting the equations, then dividing by the common coefficient, leads straight to the answer.

The idea

If the pair has the swapped-coefficient form, then:

• Adding the two equations makes the coefficients of $x$ and $y$ equal, so dividing through gives a simple equation in $x + y$.
• Subtracting them does the same for $x - y$.

The two simple equations are then quick to solve.

Example 1

Given the pair:

$71x + 37y = 253 \quad (1)$ $37x + 71y = 287 \quad (2)$

Add (1) and (2). The coefficients combine to be equal:

$108x + 108y = 540$

Divide every term by $108$:

$x + y = 5 \quad (3)$

Subtract (2) from (1).

$34x - 34y = -34$

Divide every term by $34$:

$x - y = -1 \quad (4)$

Solve (3) and (4). Adding them eliminates $y$:

$2x = 4 \implies x = 2$

Substituting $x = 2$ into (3):

$2 + y = 5 \implies y = 3$

So the solution is $(2, 3)$. Check: $71(2) + 37(3) = 142 + 111 = 253$ and $37(2) + 71(3) = 74 + 213 = 287$.

Example 2

Given the pair:

$41x - 17y = 99 \quad (1)$ $17x - 41y = 75 \quad (2)$

Add (1) and (2). Both $y$ terms are negative:

$58x - 58y = 174$

Divide by $58$:

$x - y = 3 \quad (3)$

Subtract (2) from (1).

$24x + 24y = 24$

Divide by $24$:

$x + y = 1 \quad (4)$

Solve (3) and (4). Adding them eliminates $y$:

$2x = 4 \implies x = 2$

Substituting $x = 2$ into (3):

$2 - y = 3 \implies y = -1$

So the solution is $(2, -1)$. Check: $41(2) - 17(-1) = 82 + 17 = 99$ and $17(2) - 41(-1) = 34 + 41 = 75$.

Key takeaways

• The shortcut applies only when the x and y coefficients are swapped between the two equations.
• Add the equations and divide by the common coefficient to get an equation in $x + y$; subtract and divide to get one in $x - y$.
• Solve the simplified pair by adding to eliminate a variable, then back-substitute, and always check the answer in the original equations.