Solve the system: x - 3y = 2 and 2x + 2y = 10.

Step-by-step solution

1. 1

   Simplify before eliminating. Every term of the second equation is divisible by $2$:

   $2x+2y=10\qquad\Longrightarrow\qquad x+y=5$

   Now both equations have $x$ with coefficient $1$, which sets up a one-step elimination. Skipping this reduction makes the arithmetic needlessly heavier.

2. 2

   Write the reduced system.

   $x-3y=2,\qquad x+y=5$

3. 3

   Eliminate x by subtracting. Take the second minus the first so the $x$ terms cancel:

   $(x+y)-(x-3y)=5-2$

   Distributing the subtraction across the parenthesis — $-(-3y)=+3y$ — gives

   $4y=3$

4. 4

   Solve for y.

   $y=\frac{3}{4}$

   A fractional answer is perfectly normal; the system is not required to have integer solutions.

5. 5

   Back-substitute to find x. Using $x+y=5$:

   $x=5-\frac34=\frac{20}{4}-\frac34=\frac{17}{4}$

6. 6

   Verify in both original equations. First: $\tfrac{17}{4}-3\cdot\tfrac34=\tfrac{17}{4}-\tfrac94=\tfrac84=2$ ✓. Second: $2\cdot\tfrac{17}{4}+2\cdot\tfrac34=\tfrac{34}{4}+\tfrac64=\tfrac{40}{4}=10$ ✓. Both check exactly in fraction arithmetic, with no rounding involved.