Solve the system: 2x + y = 3 and 4x + 3y = 1.

Step-by-step solution

1. 1

   Choose the method by looking at the coefficients. In the first equation $y$ has coefficient $1$, so solving for it introduces no fractions. That makes substitution the least error-prone route here (elimination would work too, by doubling the first equation).

2. 2

   Isolate y in the first equation.

   $y=3-2x$

3. 3

   Substitute into the second equation. Replace $y$ everywhere it appears:

   $4x+3(3-2x)=1$

   Distributing the $3$ across both terms — including the $-2x$ — is essential:

   $4x+9-6x=1$

4. 4

   Solve the resulting single-variable equation. Combine the $x$ terms:

   $-2x+9=1\qquad\Longrightarrow\qquad -2x=-8\qquad\Longrightarrow\qquad x=4$

   Dividing two negatives gives a positive, so $x=4$, not $-4$.

5. 5

   Back-substitute to find y. Use the simpler original equation:

   $2(4)+y=3\qquad\Longrightarrow\qquad 8+y=3\qquad\Longrightarrow\qquad y=-5$

6. 6

   Verify in both original equations. First: $2(4)+(-5)=8-5=3$ ✓. Second: $4(4)+3(-5)=16-15=1$ ✓. A solution must satisfy both equations — checking only one is the classic way to miss an arithmetic slip.