Math Problem Statement
Solve the system of equations: (1/4)x + y = 1 and (3/2)x - y = 4/3
Solution
Let's solve this system of linear equations:
\frac{1}{4}x + y = 1 \\ \frac{3}{2}x - y = \frac{4}{3} \end{cases}$$ ### Step 1: Eliminate $$y$$ To eliminate $$y$$, let's add the two equations together. But first, we’ll rewrite each equation for clarity: 1. $$\frac{1}{4}x + y = 1$$ 2. $$\frac{3}{2}x - y = \frac{4}{3}$$ By adding these two equations together, the $$y$$ terms will cancel each other out. $$\frac{1}{4}x + y + \frac{3}{2}x - y = 1 + \frac{4}{3}$$ Simplifying, we get: $$\left(\frac{1}{4} + \frac{3}{2}\right)x = 1 + \frac{4}{3}$$ ### Step 2: Simplify the Coefficients Convert $$\frac{1}{4}$$ and $$\frac{3}{2}$$ to have a common denominator: $$\frac{1}{4} = \frac{1}{4}, \quad \frac{3}{2} = \frac{6}{4}$$ Adding these fractions: $$\frac{1}{4} + \frac{6}{4} = \frac{7}{4}$$ So we have: $$\frac{7}{4}x = 1 + \frac{4}{3}$$ ### Step 3: Simplify the Right Side Convert $$1$$ to $$\frac{3}{3}$$ so that both terms on the right side have a common denominator: $$1 = \frac{3}{3}$$ Now, $$1 + \frac{4}{3} = \frac{3}{3} + \frac{4}{3} = \frac{7}{3}$$ So our equation is now: $$\frac{7}{4}x = \frac{7}{3}$$ ### Step 4: Solve for $$x$$ To isolate $$x$$, multiply both sides by the reciprocal of $$\frac{7}{4}$$, which is $$\frac{4}{7}$$: $$x = \frac{7}{3} \times \frac{4}{7} = \frac{28}{21} = \frac{4}{3}$$ So, $$x = \frac{4}{3}$$. ### Step 5: Substitute to Find $$y$$ Now, substitute $$x = \frac{4}{3}$$ into the first equation to find $$y$$: $$\frac{1}{4} \left(\frac{4}{3}\right) + y = 1$$ Simplifying, $$\frac{4}{12} + y = 1$$ $$\frac{1}{3} + y = 1$$ Subtract $$\frac{1}{3}$$ from both sides: $$y = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$$ ### Solution The solution to the system is: $$x = \frac{4}{3}, \quad y = \frac{2}{3}$$ Would you like further details or have any questions? --- 1. How can systems of equations be solved by substitution? 2. What are the benefits of elimination versus substitution methods? 3. How can fractions in systems of equations be managed effectively? 4. What are other methods to solve linear equations apart from elimination and substitution? 5. How can these techniques apply to real-world problems? **Tip:** Always check your solution by substituting back into the original equations to verify the accuracy of the answer.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Linear Equations
System of Equations
Fraction Operations
Formulas
Elimination method in systems of equations
Theorems
Properties of linear equations
Basic fraction addition and subtraction
Suitable Grade Level
Grades 8-10