Given the system 2x + 3y = 14 and 3x + 2y = 15, find x + y without solving the system.

Step-by-step solution

1. 1

   Look at the symmetry of the coefficients. The coefficient pairs are $(2,3)$ and $(3,2)$ — the same two numbers in swapped roles. Whenever a system has that symmetry, adding the equations produces equal coefficients on $x$ and $y$, which is exactly the combination $x+y$ that is wanted.

2. 2

   Add the two equations.

   $(2x+3y)+(3x+2y)=14+15$

   $5x+5y=29$

3. 3

   Factor out the common 5.

   $5(x+y)=29\quad\Longrightarrow\quad x+y=\frac{29}{5}$

   $\boxed{x+y=\dfrac{29}{5}=5.8}$

4. 4

   Note the companion shortcut. Subtracting the equations instead gives $x-y=1$, so the two symmetric combinations come out for free. Together they even solve the system: $x=\tfrac12\left(\tfrac{29}{5}+1\right)=\tfrac{17}{5}$ and $y=\tfrac{12}{5}$.

5. 5

   Check. With $x=\tfrac{17}{5}$ and $y=\tfrac{12}{5}$: $2\left(\tfrac{17}{5}\right)+3\left(\tfrac{12}{5}\right)=\tfrac{34+36}{5}=\tfrac{70}{5}=14$ ✓ and $3\left(\tfrac{17}{5}\right)+2\left(\tfrac{12}{5}\right)=\tfrac{51+24}{5}=\tfrac{75}{5}=15$ ✓, and indeed $x+y=\tfrac{29}{5}$.