Problem 1 - Algebra and Trigonometry

Question

Solve each system by the substitution method. $$\left\\{\begin{array}{l}{x+y=2} \\\\{y=x^{2}-4}\end{array}\right.$$

Step-by-Step Solution

Answer

The solutions to the system of equations are (1,-3) and (-4,12).

Step 1: Substitute y from the second equation into the first one

Using the second equation \(y=x^{2}-4\), substitute \(y\) in the first equation \(x+y=2\). This gives \(x+(x^{2}-4)=2\).

Step 2: Simplify and solve for x

Simplify the equation to: \(x^{2}+x-4=0\). This is a quadratic equation in the form \(ax^{2}+bx+c=0\). Use the quadratic formula \(x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) to find the two possible values for \(x\), which are \(x=1\) and \(x=-4\).

Step 3: Find the corresponding y values

Substitute \(x=1\) and \(x=-4\) back into the second equation. For \(x=1\), we get \(y=1^{2}-4=-3\) and for \(x=-4\), we get \(y=(-4)^{2}-4=12\). Thus, the solutions to the system are (1,-3) and (-4,12).

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. In this problem, the system consists of two equations: \(x + y = 2\) and \(y = x^2 - 4\). These equations share the variables \(x\) and \(y\). When solving a system, the goal is to find values for the variables that satisfy all equations simultaneously. By using methods like substitution or elimination, intersections of these equations can be found, which gives the solution for the system. In this exercise, since one equation is linear and the other is quadratic, the solutions will reveal points where a line intersects a parabola.

Quadratic Equation

A quadratic equation is any equation that can be rearranged in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). In our step-by-step solution, we derived a quadratic equation: \(x^2 + x - 4 = 0\) by substituting \(y = x^2 - 4\) into the linear equation \(x + y = 2\). This rearrangement helps set us up to find the values of \(x\) using methods like factorization, completing the square, or the quadratic formula, which is often the most reliable method when the equation does not easily factorize.

Quadratic Formula

The quadratic formula is a powerful tool that allows us to find solutions to any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In using the quadratic formula, we follow these simple steps:

• Identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation.
• Substitute these values into the formula.
• Solve for \(x\) to find the two solutions.

In our exercise, we identified \(a = 1\), \(b = 1\), and \(c = -4\), and applying the formula gave us the solutions \(x = 1\) and \(x = -4\). The results give the possible \(x\)-values where the two equations intersect.

Solving Equations

Solving equations involves finding the value(s) of the variable(s) that make the equation true. In the case of a system of equations, solving can entail finding where the equations intersect. Methods like substitution, elimination, and graphing help achieve this. In our exercise, we utilized the substitution method, which involves replacing one variable with an equivalent expression from another equation. This method is particularly useful when one equation is already solved for a variable, as was the case with \(y = x^2 - 4\). By substituting this expression into the other equation, we simplify the system into a single quadratic equation, allowing us to solve for one variable at a time. This method effectively breaks down complex problems into more manageable steps.