Problem 27 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{c} 9 x^{2}+y^{2}=9 \\ x^{2}+y^{2}=5 \end{array}$$

Step-by-Step Solution

Answer

The solutions to the given system of equations are: 1. \(\left(\sqrt{\frac{1}{2}},\sqrt{\frac{9}{2}}\right)\) 2. \(\left(\sqrt{\frac{1}{2}},-\sqrt{\frac{9}{2}}\right)\) 3. \(\left(-\sqrt{\frac{1}{2}},\sqrt{\frac{9}{2}}\right)\) 4. \(\left(-\sqrt{\frac{1}{2}},-\sqrt{\frac{9}{2}}\right)\)

Step 1: Analyze the Equations

We have the following system of equations: $$ \begin{array}{c} 9 x^{2}+y^{2}=9 \\\ x^{2}+y^{2}=5 \end{array} $$ Notice that both equations are in the form \(x^2+y^2=k\), where \(k\) is a constant. This represents a circle with its center at the origin \((0,0)\). We can use this fact to help us find the intersections of the two circles, which represent the solutions to the system.

Step 2: Subtract the Second Equation from the First Equation

In order to eliminate one of the variables, we will subtract the second equation from the first equation: $$ (9x^2+y^2)-(x^2+y^2)=9-5 $$ This simplifies to: $$ 8x^2=4 $$

Step 3: Solve for x

Now, we will solve for the variable \(x\): $$ x^2=\frac{4}{8}=\frac{1}{2} $$ Taking the square root of both sides, we obtain the two possible values for \(x\): $$ x=\pm\sqrt{\frac{1}{2}} $$

Step 4: Substitute the Values of x into the Second Equation

Now that we have the values of \(x\), we will substitute each value into the second equation to find the corresponding values of \(y\). For \(x=\sqrt{\frac{1}{2}}\): $$ \left(\sqrt{\frac{1}{2}}\right)^2+y^2=5 \\ \frac{1}{2}+y^2=5 \\ y^2=\frac{9}{2} $$ Taking the square root of both sides, we obtain the two possible values for \(y\): $$ y=\pm\sqrt{\frac{9}{2}} $$ For \(x=-\sqrt{\frac{1}{2}}\): $$ \left(-\sqrt{\frac{1}{2}}\right)^2+y^2=5 \\ \frac{1}{2}+y^2=5 \\ y^2=\frac{9}{2} $$ Taking the square root of both sides, we obtain the same two possible values for \(y\): $$ y=\pm\sqrt{\frac{9}{2}} $$

Step 5: Write the Solutions

Since we have two possible values for \(x\) and two possible values for \(y\), we have four possible solutions for the system of equations. These solutions are given by the ordered pairs: 1. \(\left(\sqrt{\frac{1}{2}},\sqrt{\frac{9}{2}}\right)\) 2. \(\left(\sqrt{\frac{1}{2}},-\sqrt{\frac{9}{2}}\right)\) 3. \(\left(-\sqrt{\frac{1}{2}},\sqrt{\frac{9}{2}}\right)\) 4. \(\left(-\sqrt{\frac{1}{2}},-\sqrt{\frac{9}{2}}\right)\) Therefore, the solutions to the given system of equations are the four ordered pairs listed above.

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). These equations play a pivotal role in algebra due to their parabolic shapes when graphed. In our exercise, the quadratic expressions are found in the form of circles, specifically \(x^2\) and \(y^2\).
This specific form represents the equation of a circle centered at the origin. In our problem, we have a system where the expression appears as \(x^2 + y^2 = k\), indicating circles.

• Each circle's radius can be found from the expression \(r^2 = k\).
• These equations can hence be transformed to resemble quadratic equations by isolating one variable or both.
• The solutions to these equations reflect where these curves (or shapes) intersect with the axes or each other.

In this context, the solutions to such systems give rich insights into the nature of intersections and point geometry.

Circle Equations

A circle equation in the standard form is \(x^2 + y^2 = r^2\), where \(r\) is the circle's radius. When two such equations are solved together, they form a system representing two circles.
The key here is that both circles have their centers at the origin, and their radii determine their positioning and potential interaction.

• The circle described by \(9x^2 + y^2 = 9\) changes form to \(x^2 + y^2 = 1\) upon dividing by 9, having a radius of 1.
• The second circle given by \(x^2 + y^2 = 5\) has a radius of \(\sqrt{5}\).
• Understanding how these circles overlap or intersect involves calculating the points meeting both equations.

Recognizing these equations and their geometrical meanings makes solving and visualizing their intersections much easier.

Intersection of Curves

When we talk about the intersection of curves, we are discussing points that lie on both curves. For circles, this translates into finding solutions that meet in the same points for both equations.
In this exercise, the quadratic forms of circle equations determine where these two circles intersect in a coordinate plane.

• Subtraction of one circle's equation from another allows us to solve for one variable at a time.
• The intersection results become coordinates solving both equations together.
• Once you solve for \(x\), you substitute back to find \(y\).

This method is useful when comparing shapes that share identical axes or space within a coordinate geometry framework.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to isolate variables and solve them. In systems of equations involving curves or quadratics, this is crucial.
In our exercise, algebraic manipulation involved simplifying the circle equations to find common solutions efficiently.

• Starting by subtracting equations makes a complex problem simpler.
• Dividing equations by a common factor can simplify them to an easier comparable form.
• Squaring or taking roots as needed allows solving for variables without additional unknowns.

Mastering algebraic manipulation opens up a range of problem-solving techniques, especially when dealing with polynomials, offering clarity to often confusing systems of equations.