Solve the system 3x + 2y = 5 and 2x - 3y = -1.

Step-by-step solution

1. 1

   Choose elimination and pick the variable to remove. Neither equation has a coefficient of $1$, so substitution would introduce fractions. Elimination is cleaner. The $y$ coefficients are $2$ and $-3$, with least common multiple $6$ — and they already carry opposite signs, so scaling alone will make them cancel on addition.

2. 2

   Scale each equation. Multiply the first by $3$ and the second by $2$:

   $3(3x+2y=5)\;\Longrightarrow\;9x+6y=15,$

   $2(2x-3y=-1)\;\Longrightarrow\;4x-6y=-2.$

   Every term must be multiplied, right-hand sides included — the $15$ and the $-2$ are where this step usually goes wrong.

3. 3

   Add the equations to eliminate $y$. The $+6y$ and $-6y$ cancel:

   $(9x+4x)+(6y-6y)=15+(-2)\;\Longrightarrow\;13x=13\;\Longrightarrow\;x=1.$

4. 4

   Back-substitute to find $y$. Using the original first equation:

   $3(1)+2y=5\;\Longrightarrow\;2y=2\;\Longrightarrow\;y=1.$

   Substituting into the original equation rather than a scaled one means an error in step 2 cannot propagate silently into $y$.

5. 5

   Check in both original equations. First: $3(1)+2(1)=5$ ✓. Second: $2(1)-3(1)=-1$ ✓. Geometrically the two lines have slopes $-\tfrac32$ and $\tfrac23$, which are negative reciprocals, so they meet perpendicularly at the single point $(1,1)$ — a unique solution, as the nonzero determinant $3(-3)-2(2)=-13$ guarantees.