Math Problem Statement

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

Solution

To solve the system of equations:

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.

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Math Problem Analysis

Mathematical Concepts

Systems of Linear Equations
Substitution Method
Fraction Arithmetic

Formulas

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Theorems

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Suitable Grade Level

Grades 9-12