Math Problem Statement
Solution
The image contains two sets of equations to solve. Let's solve each set step by step.
Problem 1: Solve the system of equations
1.\quad & x^2 + y^2 = 29 \\ 2.\quad & x - y = 3 \end{aligned}$$ #### Step 1: Express $$x$$ from the second equation From the second equation: $$x = y + 3$$ #### Step 2: Substitute $$x = y + 3$$ into the first equation Substituting into the first equation: $$(y + 3)^2 + y^2 = 29$$ Expand and simplify: $$y^2 + 6y + 9 + y^2 = 29$$ Combine like terms: $$2y^2 + 6y + 9 = 29$$ Subtract 29 from both sides: $$2y^2 + 6y - 20 = 0$$ Divide the entire equation by 2 to simplify: $$y^2 + 3y - 10 = 0$$ #### Step 3: Solve the quadratic equation The quadratic equation can be solved using the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For $$a = 1$$, $$b = 3$$, and $$c = -10$$: $$y = \frac{-3 \pm \sqrt{3^2 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}$$ So: $$y = \frac{4}{2} = 2 \quad \text{or} \quad y = \frac{-10}{2} = -5$$ #### Step 4: Find $$x$$ Using $$x = y + 3$$: - If $$y = 2$$, then $$x = 2 + 3 = 5$$. - If $$y = -5$$, then $$x = -5 + 3 = -2$$. #### Final Solutions The solutions are: $$(x, y) = (5, 2) \quad \text{or} \quad (-2, -5)$$ ### Problem 2: Solve the system of equations $$\begin{aligned} 1.\quad & \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = \frac{5}{2} \\ 2.\quad & x + y = 10 \end{aligned}$$ #### Step 1: Set up substitutions Let: $$\sqrt{\frac{x}{y}} = a \quad \text{and} \quad \sqrt{\frac{y}{x}} = \frac{1}{a}$$ Thus, the first equation becomes: $$a + \frac{1}{a} = \frac{5}{2}$$ #### Step 2: Multiply both sides by $$a$$ This gives: $$a^2 + 1 = \frac{5a}{2}$$ Multiply by 2 to clear the fraction: $$2a^2 - 5a + 2 = 0$$ #### Step 3: Solve the quadratic equation Using the quadratic formula: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For $$a = 2$$, $$b = -5$$, and $$c = 2$$: $$a = \frac{5 \pm \sqrt{(-5)^2 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}$$ So: $$a = 2 \quad \text{or} \quad a = \frac{1}{2}$$ #### Step 4: Relate to $$x$$ and $$y$$ For $$a = 2$$: $$\sqrt{\frac{x}{y}} = 2 \quad \Rightarrow \quad \frac{x}{y} = 4 \quad \Rightarrow \quad x = 4y$$ For $$a = \frac{1}{2}$$: $$\sqrt{\frac{x}{y}} = \frac{1}{2} \quad \Rightarrow \quad \frac{x}{y} = \frac{1}{4} \quad \Rightarrow \quad x = \frac{y}{4}$$ #### Step 5: Use $$x + y = 10$$ For $$x = 4y$$: $$4y + y = 10 \quad \Rightarrow \quad 5y = 10 \quad \Rightarrow \quad y = 2, \, x = 8$$ For $$x = \frac{y}{4}$$: $$\frac{y}{4} + y = 10 \quad \Rightarrow \quad \frac{5y}{4} = 10 \quad \Rightarrow \quad y = 8, \, x = 2$$ #### Final Solutions The solutions are: $$(x, y) = (8, 2) \quad \text{or} \quad (2, 8)$$ --- Would you like more details or have any other questions? Here are some related questions: 1. How does solving quadratic equations help in systems of equations? 2. What is the significance of substituting in solving equations with radicals? 3. Can systems with radicals have extraneous solutions? 4. What is the importance of checking all possible solutions in non-linear equations? 5. How can graphical methods be used to solve systems of equations? **Tip:** When dealing with radicals in equations, always check if the solutions satisfy the original equation to avoid including extraneous solutions.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Systems of Equations
Quadratic Equations
Radicals
Formulas
Quadratic formula
Formula for solving equations with radicals
Theorems
-
Suitable Grade Level
Grades 10-12