Problem 10 - College Algebra with Corequisite Support

Question

For each of the following exercises, find the \(x\) -intercept and the \(y\) -intercept without graphing. Write the coordinates of each intercept. $$ 2 x-\frac{2}{3}=\frac{3}{4} y+3 $$

Step-by-Step Solution

Answer

x-intercept: \((\frac{11}{6}, 0)\); y-intercept: \((0, -\frac{44}{9})\).

Step 1: Simplifying the Equation

The given equation is \(2x - \frac{2}{3} = \frac{3}{4}y + 3\). First, let's rearrange it to isolate terms involving \(x\) and \(y\) to make it easier to find the intercepts. We start by adding \(\frac{2}{3}\) to both sides to get: \[2x = \frac{3}{4}y + 3 + \frac{2}{3}\] Next, to combine the constants, find a common denominator and compute: \[2x = \frac{3}{4}y + \frac{11}{3}\]

Step 2: Finding the x-intercept

To find the \(x\)-intercept, set \(y = 0\) and solve for \(x\). Substitute \(y = 0\) into the equation obtained: \[2x = \frac{3}{4}(0) + \frac{11}{3}\] Simplify to find: \[2x = \frac{11}{3}\] Now, solve for \(x\) by dividing both sides by 2: \[x = \frac{11}{6}\] Thus, the \(x\)-intercept is \(\left(\frac{11}{6}, 0\right)\).

Step 3: Finding the y-intercept

To find the \(y\)-intercept, set \(x = 0\) and solve for \(y\). Substitute \(x = 0\) into the equation: \[2(0) = \frac{3}{4}y + \frac{11}{3}\] This simplifies to: \[0 = \frac{3}{4}y + \frac{11}{3}\] Rearrange to solve for \(y\): \[\frac{3}{4}y = -\frac{11}{3}\] Multiply both sides by \(\frac{4}{3}\) to solve for \(y\): \[y = -\frac{44}{9}\] Thus, the \(y\)-intercept is \(\left(0, -\frac{44}{9}\right)\).

Key Concepts

Equation Simplification

Simplifying an equation can make it much easier to solve and understand. In mathematics, the ability to reformulate an equation often plays a crucial role in finding solutions efficiently. Let's look at the given equation: \(2x - \frac{2}{3} = \frac{3}{4}y + 3\). The goal is to isolate the terms involving \(x\) and \(y\). This often involves simple algebraic operations like addition, subtraction, and gathering like terms together.

Our first step was to add \(\frac{2}{3}\) to both sides. This gives you a clean start where terms on one side are simplified to just \(2x\). This maneuver is vital because it helps in focusing on each term without additional clutter.

After rearranging correctly and combining constants, you'll have \(2x = \frac{3}{4}y + \frac{11}{3}\). Employing a common denominator is crucial when combining numbers to ensure accuracy and simplicity. Remember, simplified equations aren't just easier on the eyes, they are also the pathway to straightforward solutions.

X-intercept

Finding the \(x\)-intercept is all about understanding where the graph of an equation crosses the x-axis. At this point, the \(y\) value will always be zero because you're essentially identifying where the curve or line touches or crosses the x-axis.

To find the \(x\)-intercept for our equation, set \(y = 0\). Plug this into our simplified equation: \[2x = \frac{3}{4}(0) + \frac{11}{3}\]. Simplifying gives \(2x = \frac{11}{3}\). Solve for \(x\) by dividing both sides by 2, giving us \(x = \frac{11}{6}\).

So, the coordinates of the \(x\)-intercept are \(\left(\frac{11}{6}, 0\right)\). This means at approximately 1.833 along the x-axis, the line will intersect or touch. Understanding this provides insights into the graph's behavior regarding its inclination or trajectory.

Y-intercept

The \(y\)-intercept highlights where the graph of an equation crosses the y-axis. This is crucial because it shows how the line behaves when the x-value is zero, providing an essential anchor point for the entire graph.

To determine the \(y\)-intercept, set \(x = 0\) in our simplified equation: \[2(0) = \frac{3}{4}y + \frac{11}{3}\]. This simplifies to \(0 = \frac{3}{4}y + \frac{11}{3}\). To isolate \(y\), subtract \(\frac{11}{3}\) from both sides and then multiply by \(\frac{4}{3}\): \[y = -\frac{44}{9}\].

Thus, the \(y\)-intercept coordinates are \((0, -\frac{44}{9})\), precisely where the graph touches the y-axis. This gives you an idea of the initial point of contact or the position it crosses through when projected against the vertical axis.