Problem 28 - College Algebra Essentials

Question

In Exercises \(19-28,\) solve each system by the addition method. $$ \left\\{\begin{array}{l} x^{2}-2 y=8 \\ x^{2}+y^{2}=16 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions to the system of equations are \((2, -4), (-2, -4), (2, 2), (-2, 2)\)

Step 1: Write down the system of equations

First thing to do is to write out the system of equation given: \( \left\{\begin{array}{l} x^{2}-2 y=8 \ x^{2}+y^{2}=16 \end{array}\right. \)

Step 2: Addition method to eliminate a variable

To eliminate one of the variables, the equations can be added together. The good thing about this system is that when the second equation is subtracted from the first, x will be eliminated: \((x^{2}-2 y) - (x^{2}+y^{2}) = 8 - 16\) which simplifies to: \(-2y - y^{2} = -8\)

Step 3: Rearranging the equation

Rewrite the equation in ascending powers of y by adding 8 to both sides of the equation: \(y^{2} + 2y - 8= 0\)

Step 4: Solving the equation

Solving for y using the quadratic formula (since the equation is a quadratic equation): \(y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) where \(a=1, b=2, c=-8\) Substituting a, b and c, gives two possible solutions for y, namely \(y = -4, 2\)

Step 5: Substitute y values to find x values

Substitute the values of y in one of the original equations to find the corresponding x values. Let's use \(x^{2} - 2y = 8\) (\(x^{2} - 2*(-4) = 8\) gives \(x = \pm2\)) and (\(x^{2} - 2*2 = 8\) gives \(x = \pm2\)).

Step 6: The solution to the system of equations

Therefore, the solutions to the system are \((2, -4), (-2, -4), (2, 2), (-2, 2)\)

Key Concepts

Systems of Equations

A system of equations involves finding values for variables that satisfy all equations in the system simultaneously. In this exercise, we are dealing with two equations that involve both subtraction and quadratic terms. These are:

• \(x^2 - 2y = 8\)
• \(x^2 + y^2 = 16\)

These two equations form a system because they involve the same variables, \(x\) and \(y\), and we are looking for the set of values that will satisfy both equations. Solving systems of equations can involve various methods including substitution and elimination.

Addition Method

The addition method, often called elimination method, helps in solving systems by adding or subtracting equations to eliminate one variable. It's particularly useful when you have coefficients set up for easy elimination. In our exercise, the system:

• \(x^2 - 2y = 8\)
• \(x^2 + y^2 = 16\)

allows us to subtract one equation from the other, eliminating \(x^2\) easily:\[(x^2 - 2y) - (x^2 + y^2) = 8 - 16\]This simplifies to:\[-y^2 - 2y = -8\]By eliminating \(x^2\), we're left with an equation just in terms of \(y\), simplifying the problem.

Quadratic Formula

Upon simplifying the system using the elimination method, we end up with a quadratic equation in terms of \(y\):\[y^2 + 2y - 8 = 0\]The quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) helps solve this, where \(a = 1\), \(b = 2\), and \(c = -8\). Plugging these into the formula provides the solutions:

• \(y = -4\)
• \(y = 2\)

Quadratic equations like these often have two solutions owing to their parabolic nature, meaning \(y\) could be at either of these values for the vertices of the parabola (or solutions in this context).

Solution Sets

Now we substitute our \(y\) values back into the original equations to find the corresponding \(x\) values. Using \(x^2 - 2y = 8\), if \(y = -4\), we get:\[x^2 - 2(-4) = 8 \Rightarrow x = \pm 2\]Similarly, if \(y = 2\):\[x^2 - 2(2) = 8 \Rightarrow x = \pm 2\]The complete solution set for this system is:

• \((2, -4)\)
• \((-2, -4)\)
• \((2, 2)\)
• \((-2, 2)\)

These solutions form the set of coordinates that make both equations true, thereby solving the system of equations.