Problem 42 - College Algebra

Question

Solve each system by the method of your choice. $$\begin{aligned} &x-3 y=-5\\\ &x^{2}+y^{2}-25=0 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions to the system of equations are \((x, y) = (-5, 0)\) and \((x, y) = (4, 3)\).

Step 1: Rewrite the first equation

Rearrange the first equation such that we obtain \(x = 3y - 5\). This is done by adding \(3y\) to both sides.

Step 2: Substitute into the second equation

Substitute \(x = 3y - 5\) into the second equation. We get \((3y - 5)^2 + y^2 - 25 = 0\). This is a quadratic equation in \(y\).

Step 3: Solve the quadratic equation

Simplify and solve the quadratic equation. This gives us a quadratic equation \(10y^2 -30y = 0\). Solve this equation to get \(y = 0\) and \(y = 3\).

Step 4: Find the corresponding x values

Substitute \(y = 0\) and \(y = 3\) back into the first equation individually to get the corresponding \(x\) value for each solution. This will give us solutions \((x, y) = (-5, 0)\) and \((x, y) = (4, 3)\).

Key Concepts

Quadratic Equations

Quadratic equations are a cornerstone of algebra and appear in many math problems. A quadratic equation is any equation that can be rearranged to the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. The hallmark of a quadratic equation is the \(x^2\) term—this makes it a second-degree polynomial equation.

Quadratic equations can have two, one, or no real solutions. They are usually solved by factoring, completing the square, or using the quadratic formula, which is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Understanding how to solve these equations is crucial because they model many real-world situations, such as projectile motion, area calculations, and economic models. In the context of the problem, we ended up with a quadratic form after substituting one equation into another, requiring us to use algebraic methods to find the value of \(y\).

Substitution Method

The substitution method is a technique for solving systems of equations. It involves isolating one variable in one of the equations and substituting that expression into the other equation(s). This is particularly useful when one equation is linear, making it easy to solve for one variable directly.

Here's how it works in the original problem:

• We started with the linear equation \(x - 3y = -5\) and solved for \(x\), getting \(x = 3y - 5\).
• This expression was then substituted into the second equation \(x^2 + y^2 - 25 = 0\). When we substituted \(3y - 5\) for \(x\), the second equation became a quadratic equation in terms of \(y\).

This approach simplifies the process because it reduces the number of variables in the second equation, making it easier to solve it. Once you find the numerical values of one variable, you just plug them back into the linear equation to find the values of the other variable.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to make them easier to work with. It is a fundamental skill that underpins solving any algebraic equation or system of equations.

In the original problem, we used algebraic manipulation in several steps:

• We first rearranged the linear equation \(x - 3y = -5\) into \(x = 3y - 5\) by adding \(3y\) to both sides.
• We then substituted \(x = 3y - 5\) into the quadratic equation, which required careful expansion and simplification: \((3y - 5)^2 + y^2 - 25 = 0\).
• This led us to the simplified quadratic form \(10y^2 -30y = 0\), which could then be solved using factoring to find the values of \(y\).

These steps show the power of algebraic manipulation. It enables the transformation of complex or seemingly intractable problems into simpler forms that can be easily solved. Mastering these techniques is key to excelling in algebra and ultimately understanding more complex mathematical concepts.