Problem 9 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{aligned} &x+2 y=5\\\ &x^{2}+y^{2}=10 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions to the given system of equations are: \((x, y) = (3, 1)\) and \((x, y) = (-1, 3)\).

Step 1: Solving the linear equation for one variable

From equation (1), we can write one variable in terms of the other. Let's solve the equation for x: $$x = 5-2y$$

Step 2: Substituting into the quadratic equation

Now, substitute the expression for x that we found in step 1 into the equation (2): \[(5-2y)^2 + y^2 = 10\]

Step 3: Simplifying and rearranging the equation

Expand and simplify the equation obtained in step 2: \begin{align*} (5-2y)^2 + y^2 &= 10 \\ (25 - 20y + 4y^2) + y^2 &= 10 \\ 5y^2 - 20y + 15 &= 0 \end{align*}

Step 4: Solve the quadratic equation for y

Divide the equation by 5 to make it easier to solve: \[y^2 - 4y + 3= 0\] Now, factor the quadratic equation: \[(y-3)(y-1) = 0\] From this, we find two possible values for y: \[y = 1 \text{ or } y = 3\]

Step 5: Solve for x using the values of y

For each value of y, find the corresponding value of x using the expression found in step 1: 1. When \(y = 1\), we have: \[x = 5-2(1) = 3\] So one solution is \((x, y) = (3, 1)\). 2. When \(y = 3\), we have: \[x = 5-2(3) = -1\] So another solution is \((x, y) = (-1, 3)\).

Step 6: Write down the final solution

The solutions to the given system of equations are: \[(x, y) = (3, 1) \text{ and } (x, y) = (-1, 3)\]

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation. They are typically written in the standard form:

• \( ax^2 + bx + c = 0 \)

The graph of a quadratic equation is a curve known as a parabola. It can open upwards or downwards depending on the sign of the coefficient of the \( x^2 \) term (\( a \)). A quadratic equation can have:

• Two real solutions,
• One real solution, or
• No real solutions (if complex solutions are involved)

These solutions represent the x-values where the parabola intersects the x-axis.
To solve quadratic equations, there are several methods one could use:

• Factoring (as done in the original exercise)
• Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

In the exercise, we focused on factoring, which involves expressing the quadratic as a product of two binomials.This method is efficient and straightforward when the quadratic is factorable.

Linear Equations

Linear equations describe straight lines when graphed. They follow the general form:

• \( ax + by = c \)

In these equations, both x and y are raised only to the first power, hence the name 'linear'.
We can solve linear equations using a variety of approaches:

• Graphical method - where lines are plotted, and intersections represent solutions.
• Substitution method - which involves substituting one equation into another.
• Elimination method - which involves eliminating one variable to solve for another.

In our problem, we specifically utilize the substitution method. We rearrange the linear equation to solve for one variable in terms of the other, making it easier to use in conjunction with another type of equation - in this case, a quadratic one. The beauty of linear equations lies in their simplicity and directness, ensuring a straightforward pathway to finding a solution.

Substitution Method

The substitution method is a popular technique to solve systems of equations, especially when they consist of different types of equations. In this method, one equation is solved to express one variable in terms of another variable.

• For example, if the system consists of a linear and a quadratic equation, you start by solving the linear equation for one variable.
• You then substitute this expression into the other equation.

This method is particularly handy when dealing with systems where one equation is easier to manipulate.
Our exercise demonstrated this by first rewriting the linear equation \( x + 2y = 5 \) to express \( x \) in terms of \( y \) as \( x = 5 - 2y \). This expression was substituted into the quadratic equation. This led us to create a single variable equation, which is much simpler to solve. It is important to:

• Solve the substitution equation carefully to avoid mistakes.
• Check each solution in the context of the original system of equations to ensure they both satisfy all the constraints given initially.

This method is valuable for developing a step-by-step approach to finding precise solutions.