Problem 3 - Algebra and Trigonometry

Question

Use the substitution method to find all solutions of the system of equations. \(\left\\{\begin{array}{l}{y=x^{2}} \\ {y=x+12}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions are \((4, 16)\) and \((-3, 9)\).

Step 1: Set Equations Equal

Since both equations are set to equal to \(y\), we can set them equal to each other: \(x^2 = x + 12\). This gives us a single equation to solve for \(x\).

Step 2: Rearrange the Equation

Rearrange the equation from Step 1 to form a standard quadratic equation: \(x^2 - x - 12 = 0\).

Step 3: Factor the Quadratic Equation

Factor the quadratic equation \(x^2 - x - 12\). We need two numbers that multiply to \(-12\) and add to \(-1\). The numbers \(-4\) and \(3\) satisfy this condition, so the factored equation is \((x - 4)(x + 3) = 0\).

Step 4: Solve for x

Set each factor equal to zero to solve for \(x\). This gives us: \(x - 4 = 0\) and \(x + 3 = 0\). Solving these, we get \(x = 4\) and \(x = -3\).

Step 5: Substitute x-values Back to Find y

Substitute \(x = 4\) and \(x = -3\) back into one of the original equations to find the corresponding \(y\) values. Using \(y = x + 12\):- For \(x = 4\), \(y = 4 + 12 = 16\).- For \(x = -3\), \(y = -3 + 12 = 9\).

Step 6: Write the Solution as Points

The solutions to the system of equations are the points \((4, 16)\) and \((-3, 9)\).

Key Concepts

System of Equations

A system of equations is a set of two or more equations that have common variables. In our exercise, we looked at two equations: \( y = x^2 \) and \( y = x + 12 \). These equations share the variables \( x \) and \( y \), and they describe relationships between these variables. The goal is to find values for \( x \) and \( y \) that satisfy both equations at the same time.

There are several methods to solve systems of equations, such as substitution, elimination, and graphing. The substitution method is particularly useful when one equation is already solved for one variable. Here, since both equations are set equal to \( y \), we can substitute one equation into the other without changing what they are equal to.

Substituting \( y = x + 12 \) into \( y = x^2 \) gives us a single equation to solve for \( x \), making it easier to find the solution. After finding the values of \( x \), we plug them back into one of the original equations to find the corresponding \( y \) values, thus finding the solution to the system of equations.

Quadratic Equations

A quadratic equation is a polynomial equation of degree two. It can be written in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. In our exercise, we formed a quadratic equation by rearranging the equation \( x^2 = x + 12 \) into \( x^2 - x - 12 = 0 \).

Quadratic equations can have two solutions, one solution, or no real solutions, depending on the values of \( a \), \( b \), and \( c \). The two solutions in our case were found by factoring the quadratic equation. Factoring is a method of breaking down a quadratic into simpler expressions that can multiply to give the original polynomial. The factored form \((x - 4)(x + 3) = 0\) gives us two potential solutions for \( x \). Each factor set equal to zero gives us a straightforward equation to solve, yielding the \( x \) values that satisfy both the original equations.

Factoring Polynomials

Factoring polynomials involves expressing a polynomial as the product of its factors. This is a key step in solving many quadratic equations. For a quadratic like \( x^2 - x - 12 \), we found two numbers that both multiply to \(-12\) and add to \(-1\).

These numbers, \(-4\) and \(3\), become the factors \((x - 4)\) and \((x + 3)\). This process not only simplifies solving equations but also helps in understanding the nature of the polynomial better.

Factoring is often straightforward, but sometimes it may involve techniques like grouping or using the quadratic formula if simpler methods don’t work. Successful factoring leads to simple equations \((x - 4) = 0\) and \((x + 3) = 0\), which are easy to solve for the variable \( x \). Thus, factoring is a powerful tool in algebra, particularly useful for simplifying and solving quadratic equations.