3x + 2y = 64x + y = 1Solve the system of equations.no solution(8/11, -21/11)(-4/5, 21/5)NEXT QUESTION

1. Break Down the Problem

We need to solve the system of equations given by:

1. $3x + 2y = 6$
2. $4x + y = 1$

2. Relevant Concepts

We can solve the system using the substitution method or the elimination method. For this solution, we will use the substitution method.

3. Analysis and Detail

1. From the second equation, solve for $y$: $y = 1 - 4x$

2. Substitute $y$ in the first equation: $3x + 2(1 - 4x) = 6$ Simplifying this: $3x + 2 - 8x = 6$ $-5x + 2 = 6$ $-5x = 4$ $x = -\frac{4}{5}$

3. Substitute $x$ back into the expression for $y$: $y = 1 - 4(-\frac{4}{5})$ $y = 1 + \frac{16}{5} = \frac{5}{5} + \frac{16}{5} = \frac{21}{5}$

4. Verify and Summarize

We found $x = -\frac{4}{5}$ and $y = \frac{21}{5}$. Let's verify by substituting back into the original equations:

• For $3x + 2y = 6$: $3(-\frac{4}{5}) + 2(\frac{21}{5}) = -\frac{12}{5} + \frac{42}{5} = \frac{30}{5} = 6$

• For $4x + y = 1$: $4(-\frac{4}{5}) + \frac{21}{5} = -\frac{16}{5} + \frac{21}{5} = \frac{5}{5} = 1$

Both equations are satisfied.

Final Answer

The solution to the system of equations is: $\left( -\frac{4}{5}, \frac{21}{5} \right)$