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

By Image

Drop file here or Click Here to upload

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