Math Problem Statement
Solve the system of equations: x^2 + y^2 = 41 and x^2 - y^2 = -9
Solution
To solve the system of equations:
x^2 + y^2 = 41 \\ x^2 - y^2 = -9 \end{cases}$$ we can approach it by using the method of addition and subtraction to simplify the expressions. ### Step 1: Adding the Equations Add the two equations: $$(x^2 + y^2) + (x^2 - y^2) = 41 + (-9)$$ This simplifies to: $$2x^2 = 32$$ Divide by 2: $$x^2 = 16$$ Taking the square root of both sides: $$x = \pm 4$$ ### Step 2: Substituting $$x$$ back into the Equations Now, substitute $$x = \pm 4$$ into one of the original equations to find $$y$$. Let’s use $$x^2 + y^2 = 41$$. 1. **For $$x = 4$$:** $$4^2 + y^2 = 41$$ $$16 + y^2 = 41$$ $$y^2 = 25$$ Taking the square root of both sides: $$y = \pm 5$$ So, for $$x = 4$$, we have two solutions: $$(4, 5)$$ and $$(4, -5)$$. 2. **For $$x = -4$$:** $$(-4)^2 + y^2 = 41$$ $$16 + y^2 = 41$$ $$y^2 = 25$$ Taking the square root of both sides: $$y = \pm 5$$ So, for $$x = -4$$, we have two solutions: $$(-4, 5)$$ and $$(-4, -5)$$. ### Final Solution The solutions to the system are: $$(4, 5), (4, -5), (-4, 5), (-4, -5)$$ Would you like more details or have any questions? Here are some additional questions you might find interesting: 1. How would the solution change if the equations involved $$x^3$$ and $$y^3$$? 2. What if the constant terms in the equations were different? 3. Can this method be used to solve systems with three variables? 4. How would graphing these equations help verify the solution? 5. What if we change the signs in the second equation? **Tip:** In systems of equations involving squares, adding and subtracting equations can often help eliminate variables effectively.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Systems of Equations
Addition and Subtraction of Equations
Formulas
Addition of equations
Square root extraction
Theorems
Properties of equality
Suitable Grade Level
Grades 10-12
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