Problem 8 - Beginning and Intermediate Algebra

Question

Solve each system. $$\begin{array}{c} x^{2}+y=1 \\ -x+y=-5 \end{array}$$

Step-by-Step Solution

Answer

The short version of the answer is: The solutions for the given system of equations are \( (x, y) = (-3, -8) \) and \( (x, y) = (2, -3) \).

Step 1: Solve the linear equation for y

We can rearrange the linear equation to isolate y: $$-x + y = -5$$ Adding x to both sides, we get: $$y = x - 5$$

Step 2: Substitute the value of y in the quadratic equation

Replace y in the quadratic equation with the expression we found in Step 1: $$x^2 + (x - 5) = 1$$

Step 3: Solve the quadratic equation for x

Now, we simplify the equation and solve for x: $$x^2 + x - 5 - 1 = 0$$ $$x^2 + x - 6 = 0$$ Factoring the quadratic equation, we get: $$(x + 3)(x - 2) = 0$$ Setting each factor to zero, we find the solutions for x: $$x + 3 = 0 \Rightarrow x = -3$$ $$x - 2 = 0 \Rightarrow x = 2$$

Step 4: Solve for y using the values of x

Now that we have the values for x, we can substitute them into the equation we found in Step 1 to find the corresponding values for y: When \(x = -3\): $$y = -3 - 5 \Rightarrow y = -8$$ When \(x = 2\): $$y = 2 - 5 \Rightarrow y = -3$$

Step 5: State the solutions

The solutions for the given system are (x, y) pairs: $$(x, y) = (-3, -8)$$ and $$(x, y) = (2, -3)$$

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They have the general form:

• \(ax^2 + bx + c = 0\)

where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
These equations can be visualized as parabolas when graphed on a coordinate plane. The solutions, or roots, of a quadratic equation are where the parabola intersects the x-axis.
There are various methods to solve quadratic equations including factoring, using the quadratic formula, or completing the square.
In this context, we have solved the quadratic equation by factoring, which is a straightforward method if the equation can be rewritten as a product of binomials.

Linear Equations

Linear equations are first-degree polynomial equations and take the form:

• \(ax + by = c\)

where \(a\), \(b\), and \(c\) are constants.
These equations graph as straight lines. Finding the solution to a linear equation means finding points on the plane that satisfy the equation (usually expressed as \(x, y\) pairs).
In the exercise provided, the linear equation \(-x + y = -5\) was rearranged to express \(y\) in terms of \(x\), transforming our system of equations into a form more suitable for substitution.

Factoring Quadratics

Factoring quadratics involves rewriting a polynomial as a product of smaller, simpler polynomials. This is a crucial step that can simplify solving equations.
For a quadratic like \(x^2 + x - 6 = 0\), the method includes looking for two numbers that multiply to \(c\) (the constant term) and add to \(b\) (the coefficient of \(x\)).
In this case, those numbers are +3 and -2, allowing us to express the quadratic as:

• \((x + 3)(x - 2) = 0\)

Once factored, we apply the Zero Product Property which states if a product is zero, at least one of the multiplicands must be zero. Thus, we find the solutions \(x = -3\) and \(x = 2\).

Substitution Method

The substitution method is a technique for solving systems of equations. It involves solving one equation for one variable and then substituting this into another equation.
Here, we isolated \(y\) from the linear equation \(-x + y = -5\), getting \(y = x - 5\).
This expression was substituted into the quadratic equation, transforming it from a two-variable equation into a single-variable equation:

• \(x^2 + (x-5) = 1\)

After substitution, we ended with a simpler quadratic equation \(x^2 + x - 6 = 0\). Solving this quadratic gives us the values of \(x\), and consequently, by back-substitution, we find the values of \(y\). This method is particularly powerful as it reduces complexity by breaking down the problem into more manageable parts.