Problem 18 - Algebra and Trigonometry

Question

Solve each system by the substitution method. \(\left\\{\begin{array}{l}{y=-\frac{1}{2} x+2} \\ {y=\frac{3}{4} x+7}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions for the set of equations are (0, 4) and (1, 2).

Step 1: Express one variable in terms of the other from the linear equation

From the first equation \(2x + y = 4\), express \(y\) in terms of \(x\). This yields \(y = 4 - 2x\).

Step 2: Substitute \(y\) in the circle equation

Substitute \(y = 4 - 2x\) in \((x + 1)^2 + (y - 2)^2 = 4\). This gives \((x + 1)^2 + [(4 - 2x) - 2]^2 = 4\). Simplify to obtain the quadratic equation \((x + 1)^2 + (2x - 2)^2 = 4\). Expanding and simplifying gives \(4x^2 - 4x = 0\).

Step 3: Solve the quadratic equation

Solving \(4x^2 - 4x = 0\) for \(x\), we get \(x=0\) or \(x=1\).

Step 4: Substitute \(x\) into the linear equation to find \(y\)

Substitute \(x = 0\) into \(y = 4 - 2x\), to get \(y = 4\). Substitute \(x = 1\) into the same equation to get \(y = 2\). Therefore, the solutions are \((x, y) = (0, 4)\) and \((x, y) = (1, 2)\).

Key Concepts

Understanding Systems of Equations

A system of equations consists of two or more equations that are considered simultaneously. The primary goal is to find a common solution that satisfies all equations in the system.

There are several methods to solve these systems, including graphing, substitution, and elimination. For this exercise, we focus on the **substitution method**, which involves solving one equation for one variable, and then substituting this expression into another equation in the system.

This method is particularly useful when one of the equations is easily solvable for one variable, as it allows you to reduce the multiple equations into a single equation problem.

Linear Equations and Their Role

Linear equations are equations of the first degree, meaning they have no exponents higher than one. They graph as straight lines in a coordinate system. The general form of a linear equation in two variables is expressed as:

\[ ax + by = c \]

Here, \(a\), \(b\), and \(c\) are constants, while \(x\) and \(y\) are variables.

In our example, the equation \(2x + y = 4\) is a linear equation. By rearranging it, we can express it as \(y = 4 - 2x\). This equation represents a line whose slope is \(-2\) and y-intercept is \(4\). Linear equations are critical in substitution because they allow you to easily express one variable in terms of the other, facilitating the substitution process.

Working with Quadratic Equations

Quadratic equations involve terms up to the second degree. They generally take the form:

\[ ax^2 + bx + c = 0 \]

Quadratic equations graph as parabolas, which may open upwards or downwards.

In our problem, the transformed equation from the circle equation is a quadratic: \(4x^2 - 4x = 0\). Solving quadratic equations can be done using various methods like factoring, completing the square, or using the quadratic formula.

Here, the equation is factored to find \(x\):

• Factoring the equation: \(4x(x - 1) = 0\).
• Setting each factor to zero gives: \(x = 0\) or \(x = 1\).

Quadratic equations thus allow for finding multiple potential solutions within the context of the system.