Math Problem Statement
\begin{cases}
\frac{1}{2}x-3=\frac{5}{4}y\
3x+5y+2=0
\end{cases}
Solution
We are given the system of linear equations:
\frac{1}{2}x - 3 = \frac{5}{4}y \\ 3x + 5y + 2 = 0 \end{cases}$$ ### Step 1: Solve the first equation for $$x$$ or $$y$$. Start by solving the first equation for $$x$$: $$\frac{1}{2}x - 3 = \frac{5}{4}y$$ Add $$3$$ to both sides: $$\frac{1}{2}x = \frac{5}{4}y + 3$$ Multiply through by 2 to eliminate the fraction: $$x = 2 \left( \frac{5}{4}y + 3 \right)$$ Simplify: $$x = \frac{5}{2}y + 6$$ ### Step 2: Substitute into the second equation. Now, substitute $$x = \frac{5}{2}y + 6$$ into the second equation $$3x + 5y + 2 = 0$$: $$3\left( \frac{5}{2}y + 6 \right) + 5y + 2 = 0$$ Distribute the 3: $$\frac{15}{2}y + 18 + 5y + 2 = 0$$ Simplify the equation: $$\frac{15}{2}y + 5y + 20 = 0$$ Convert $$5y$$ to have a denominator of 2: $$\frac{15}{2}y + \frac{10}{2}y + 20 = 0$$ Combine like terms: $$\frac{25}{2}y + 20 = 0$$ ### Step 3: Solve for $$y$$. Subtract 20 from both sides: $$\frac{25}{2}y = -20$$ Multiply both sides by 2 to eliminate the denominator: $$25y = -40$$ Solve for $$y$$: $$y = -\frac{40}{25} = -\frac{8}{5}$$ ### Step 4: Solve for $$x$$. Substitute $$y = -\frac{8}{5}$$ into $$x = \frac{5}{2}y + 6$$: $$x = \frac{5}{2} \left( -\frac{8}{5} \right) + 6$$ Simplify: $$x = -4 + 6 = 2$$ ### Final Answer: The solution to the system is: $$x = 2, \quad y = -\frac{8}{5}$$ Would you like a more detailed explanation or have any questions? Here are 5 related questions: 1. How do you solve a system of equations using substitution? 2. What is the elimination method in solving systems of equations? 3. How do you handle equations involving fractions? 4. How can you check if a solution is correct for a system of equations? 5. What are the applications of solving systems of linear equations in real life? **Tip**: Always double-check your substitution in the second equation to avoid errors in solving systems of equations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Linear Equations
System of Equations
Substitution Method
Formulas
Solving for x or y in terms of the other variable
Substitution into the second equation
Combining like terms
Solving for variables
Theorems
Substitution Method for solving systems of linear equations
Suitable Grade Level
Grades 9-10
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