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Chapter 20: Problem 598

Solve the system \(\mathrm{x}^{2}+\mathrm{y}^{2}=25\) \(x-y=1\)

Short Answer

Expert verified
The solutions to the system \(x^2 + y^2 = 25\) and \(x - y = 1\) are \((-3, -4)\) and \((4, 3)\).

Step by step solution

01

Solve the linear equation for one variable

Solve the linear equation \(x - y = 1\) for one of the variables, let's say \(x\): \(x = y + 1\)
02

Substitute the expression for x in the circle equation

Replace \(x\) in \(x^2 + y^2 = 25\) with the expression we found in Step 1 (\(y + 1\)): \((y + 1)^2 + y^2 = 25\)
03

Simplify the equation

Expand and simplify the new equation to solve for \(y\): \((y^2 + 2y + 1) + y^2 = 25 \\ 2y^2 + 2y + 1 - 25 = 0 \\ 2y^2 + 2y - 24 = 0\)
04

Solve the quadratic equation

Solve the quadratic equation \(2y^2 + 2y - 24 = 0\) for \(y\). First, we can divide the entire equation by 2 to simplify: \(y^2 + y - 12 = 0\) Now, try factoring the equation: \((y + 4)(y - 3) = 0\) The two possible values for \(y\) are: \(y = -4\) and \(y = 3\)
05

Back-substitute to find x values

Using the expression we found in Step 1 (\(x = y + 1\)), substitute the two values of \(y\) to find the corresponding \(x\) values. For \(y = -4\): \(x = -4 + 1 = -3\) For \(y = 3\): \(x = 3 + 1 = 4\)
06

Write the solutions

The solutions for the system of equations are the two sets of \(x\) and \(y\) values we found: 1. \( (x, y) = (-3, -4) \) 2. \( (x, y) = (4, 3) \) Therefore, the solutions to the system \(x^2 + y^2 = 25\) and \(x - y = 1\) are \((-3, -4)\) and \((4, 3)\).

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