Solve the system of equations: 4x=-7+3y 8x+4y=-1

To solve the system of equations:

\[
\begin{aligned}
1. & \quad 4x = -7 + 3y \\
2. & \quad 8x + 4y = -1
\end{aligned}
\]

### Step 1: Simplify the first equation

First, isolate \(y\) in the first equation. 

\[
4x = 3y - 7
\]

Now, solve for \(y\):

\[
3y = 4x + 7 \quad \Rightarrow \quad y = \frac{4x + 7}{3}
\]

### Step 2: Substitute \(y\) into the second equation

Substitute \(y = \frac{4x + 7}{3}\) into the second equation:

\[
8x + 4\left(\frac{4x + 7}{3}\right) = -1
\]

### Step 3: Solve for \(x\)

Multiply everything out to eliminate the fraction:

\[
8x + \frac{16x + 28}{3} = -1
\]

Multiply the entire equation by 3 to eliminate the denominator:

\[
3(8x) + 16x + 28 = -3
\]

Simplify:

\[
24x + 16x + 28 = -3
\]

Combine like terms:

\[
40x + 28 = -3
\]

Now, isolate \(x\):

\[
40x = -3 - 28
\]

\[
40x = -31
\]

\[
x = -\frac{31}{40}
\]

### Step 4: Solve for \(y\)

Now that we have \(x = -\frac{31}{40}\), substitute this value back into the equation for \(y\):

\[
y = \frac{4\left(-\frac{31}{40}\right) + 7}{3}
\]

Simplify:

\[
y = \frac{-\frac{124}{40} + \frac{280}{40}}{3}
\]

\[
y = \frac{\frac{156}{40}}{3}
\]

\[
y = \frac{156}{120} = \frac{39}{30} = \frac{13}{10}
\]

### Final Answer:
\[
x = -\frac{31}{40}, \quad y = \frac{13}{10}
\]

Would you like further details on any step? Here are some related questions you might find interesting:

1. How would the solution change if the coefficients in the first equation were different?
2. Can this system be solved using matrix methods?
3. What if the equations had been dependent or inconsistent?
4. How can you graphically interpret the solution of this system?
5. What does the solution tell us about the intersection of the two lines?

**Tip:** Always double-check your substitution to avoid simple arithmetic errors, especially when working with fractions.

Learn how to solve the system of equations 4x = -7 + 3y and 8x + 4y = -1 with a detailed step-by-step solution. This problem involves systems of linear equations and uses the substitution method to find the values of x and y. Suitable for high school students (Grades 9-12).

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