Problem 36 - Precalculus Mathematics for Calculus

Question

Use the graphical method to find all solutions of the system of equations, rounded to two decimal places. $$\left\\{\begin{aligned} x^{2}+y^{2} &=17 \\ x^{2}-2 x+y^{2} &=13 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are approximately \((2, 3.61)\) and \((2, -3.61)\).

Step 1: Identify the equations

We have two equations in the system:1. \(x^2 + y^2 = 17\)2. \(x^2 - 2x + y^2 = 13\)

Step 2: Simplify the second equation

Rearrange the second equation for clarity. By subtracting 13 from both sides, we get:\[x^2 - 2x + y^2 - 13 = 0\], which can be written as \[(x^2 + y^2) - 2x = 13\].

Step 3: Subtract the first equation from the second

Subtract the first equation \(x^2 + y^2 = 17\) from the second equation \(x^2 - 2x + y^2 = 13\). This yields:\[-2x = 13 - 17\], which simplifies to \[-2x = -4\].

Step 4: Solve for x

From the equation \(-2x = -4\), solve for \(x\):\[x = \frac{-4}{-2} = 2\].

Step 5: Substitute x back into the first equation

Substitute \(x = 2\) back into the first equation \(x^2 + y^2 = 17\):\[(2)^2 + y^2 = 17\]. This simplifies to \[4 + y^2 = 17\].

Step 6: Solve for y

Rearrange the equation \[4 + y^2 = 17\] to find \(y^2\):\[y^2 = 17 - 4\], which simplifies to \[y^2 = 13\].Take the square root of both sides to get two solutions for \(y\):\[y = \pm \sqrt{13}\].

Step 7: Write the solution set

The solutions of the system are the points where these values of \(x\) and \(y\) meet:- For \(x = 2\), \(y = \sqrt{13}\) and \(y = -\sqrt{13}\).Rounding \(\sqrt{13}\) to two decimal places gives approximately \(3.61\) and \(-3.61\). Therefore, the solutions are \((2, 3.61)\) and \((2, -3.61)\).

Key Concepts

Understanding Systems of Equations

When we talk about a system of equations, we're dealing with more than one equation at the same time. Essentially, a system of equations is a set that contains two or more equations with the same variables. In our exercise, we have two equations, both involving the variables \(x\) and \(y\).

• The first equation is a circle equation: \(x^2 + y^2 = 17\).
• The second one is a slightly different circle equation: \(x^2 - 2x + y^2 = 13\).

The purpose of solving a system of equations is to find the values of the variables that make all equations true at the same time. This means that we’re looking for points (or a point) where the equations intersect if they are graphed. In other words, where do our circle equations meet in the graph? Solving this system will lead us to these solution points.

Finding the Solution Set

The solution set of a system of equations is basically all possible sets of values that satisfy all equations within the system. For the system we're interested in, finding the solution set means determining if there are particular values of \(x\) and \(y\) where both equations are true.
Think of the graphs of the equations. One graph is a circle with radius \(\sqrt{17}\), and the other is a slightly shifted circle with respect to the origin. Where they overlap on the graph (intersecting points), provides us with the solution set.
After finding \(x = 2\), we plugged it back into one of the equations and solved for \(y\). The results were the points \((2, \sqrt{13})\) and \((2, -\sqrt{13})\). However, as the exercise requested rounding to two decimal places, we get approximately \((2, 3.61)\) and \((2, -3.61)\). This rounded result is the solution set that tells us where exactly these circle equations intersect in decimal form.

Simplifying Equations Step-by-Step

Simplifying equations is an important step in solving systems of equations, as it makes them easier to work with. In the original problem, the second equation was simplified by subtracting the constant 13 from both sides. This refinement gave us a clearer view by rearranging it into

• \(x^2 + y^2 - 2x = 13\)

To further simplify, we subtract the first equation from the second one. This allows us to eliminate common terms (like \(x^2 + y^2\)), isolating \(x\) for easier solving.
This subtraction revealed \(-2x = -4\), which simplifies to \(x = 2\) after dividing both sides by \(-2\). Simplification broke down a seemingly complicated problem into more manageable pieces, eventually allowing us to find our solutions by substitution and basic algebraic manipulation. Always remember to keep equations as clean and simple as possible for efficient solving!