Problem 15 - Elementary Algebra

Question

For the following problems, solve the linear equations in two variables. $$ -x-y=0, \text { if } y=\frac{14}{3} $$

Step-by-Step Solution

Answer

Answer: The values of x and y that satisfy the given equation and condition are x = 14/3 and y = 14/3.

Step 1: Write down the given equation and condition

We are given the equation: $$ -x-y=0 $$ And the condition: $$ y = \frac{14}{3} $$

Step 2: Substitute the condition into the equation

We know the value of y, so we can substitute it into the equation to find the value of x: $$ -x - \left(\frac{14}{3}\right) = 0 $$

Step 3: Solve for x

Now we need to solve the equation for x: $$ -x=-\frac{14}{3} $$ To solve for x, multiply both sides by -1: $$ x = \frac{14}{3} $$

Step 4: Write the solution as a coordinate pair

The solution to the system of equations is the values of x and y that satisfy both the equation and the condition. These are: $$ x = \frac{14}{3} \text{ and } y = \frac{14}{3} $$ The solution can be written as a coordinate pair: $$ \left(\frac{14}{3}, \frac{14}{3}\right) $$

Key Concepts

Substitution Method

When solving linear equations with two variables, the substitution method is a powerful tool that allows for solving systems efficiently. It involves replacing one variable with another to reduce the system into a single equation with one variable. In the exercise provided, you are asked to solve the equation \( -x - y = 0 \) with the condition \( y = \frac{14}{3} \).

To apply the substitution method, first, acknowledge the condition given for \( y \) and replace \( y \) in the original equation with \( \frac{14}{3} \). This step simplifies the equation to one with a single variable x, making it easier to solve. The substitution method is particularly useful when you can easily isolate one of the variables in one of the equations—preferably resulting in less complex algebraic manipulation.

Algebraic Manipulation

Algebraic manipulation encompasses a variety of techniques required to rearrange and solve equations. In the context of the substitution method, after substituting \( y \) with \( \frac{14}{3} \) you get to the simplified equation \(-x - \frac{14}{3} = 0\). To solve for \( x \), algebraic manipulation comes into play.

You need to keep \( x \) on one side of the equation and move everything else to the other side. In this case, you add \( \frac{14}{3} \) to both sides to isolate the \( x \) term. Remember to perform the same operation on both sides of the equation to maintain equality. After this step, you'll find that \( x = \frac{14}{3} \). Mastering the art of algebraic manipulation is crucial and includes operations such as adding, subtracting, dividing, multiplying, and factoring.

Coordinate Pairs

Coordinate pairs are a fundamental concept in algebra and geometry, allowing us to specify the exact location of points on a plane. A coordinate pair is written in the form \( (x, y) \), where \( x \) and \( y \) correspond to positions on the horizontal and vertical axes, respectively.

In the solution to the linear equation, after finding the value of \( x \) using substitution and algebraic manipulation, the exercise asks us to express the solution as a coordinate pair. Consequently, we combine the values of \( x \) and \( y \) as \( (\frac{14}{3}, \frac{14}{3}) \). Such coordinate pairs serve as critical points of reference in graphing linear equations and can represent solutions to systems of equations in two variables. Understanding how to formulate and interpret these pairs is essential for analyzing mathematical relationships graphically.