Solve the system of equations 3x + 2y = 12 and x - y = 1.

Step-by-step solution

1. 1

   Pick the equation that is cheapest to rearrange. The second equation has a coefficient of $1$ on $x$, so isolating a variable there costs no fractions:

   $x - y = 1 \quad \Longrightarrow \quad x = y + 1$

   Choosing the other equation would force you to divide by $3$ immediately and carry thirds through the whole calculation.

2. 2

   Substitute into the first equation. Replacing $x$ by $y + 1$ turns a two-variable problem into a one-variable one:

   $3(y + 1) + 2y = 12$

3. 3

   Expand and collect like terms.

   $3y + 3 + 2y = 12$

   $5y + 3 = 12$

   $5y = 9 \quad \Longrightarrow \quad y = \frac{9}{5}$

   The answer is not an integer, which is normal: $5$ does not divide $9$, and nothing in the problem promised whole-number solutions.

4. 4

   Back-substitute to recover x. Using $x = y + 1$ with a common denominator of $5$:

   $x = \frac{9}{5} + 1 = \frac{9}{5} + \frac{5}{5} = \frac{14}{5}$

5. 5

   Verify in both original equations. A solution must satisfy the system, not just the equation you last touched:

   $3\left(\frac{14}{5}\right) + 2\left(\frac{9}{5}\right) = \frac{42}{5} + \frac{18}{5} = \frac{60}{5} = 12 \ \checkmark$

   $\frac{14}{5} - \frac{9}{5} = \frac{5}{5} = 1 \ \checkmark$

   So the two lines meet at the single point $\left(\tfrac{14}{5}, \tfrac95\right) = (2.8,\, 1.8)$.