Problem 29 - Introductory and Intermediate Algebra for College Students 4th

Question

Use intercepts and a checkpoint to graph each equation. $$3 x=5 y-15$$

Step-by-Step Solution

Answer

The graph of the equation \(3x = 5y - 15\) should be a straight line passing through the points (-5,0), (0,3), and (5,6)

Step 1: Finding the y-intercept

To find the y-intercept, set x equal to 0 in the equation \(3x = 5y - 15\). Solving this gives \(0 = 5y -15 \), and y is 3, meaning that the y-intercept (where the line crosses the y axis) is at the point (0,3).

Step 2: Finding the x-intercept

To find the x-intercept, set y equal to 0 in the equation \(3x = 5y - 15\). Solving this for x gives \(3x = -15\), and x is -5, meaning that the x-intercept (where the line crosses the x axis) is at the point (-5,0).

Step 3: Choosing a checkpoint

A good checkpoint might be when x=5 and y=6, as these values satisfy the equation \(5(6)=3*5+15\). This confirms that the point (5,6) is on the line.

Step 4: Graphing the equation

Now, to graph the equation, first plot the y-intercept (0,3), then the x-intercept (-5,0), and then the checkpoint (5,6). Finally, draw a line connecting these three points, giving the graph of the linear equation \(3x = 5y - 15\)

Key Concepts

Finding Y-Intercept

The y-intercept is a crucial starting point for graphing linear equations. It represents the location where the line crosses the y-axis. To find the y-intercept, we set the x-variable to 0 and solve the equation for y. Let's take a closer look at our example:

In the equation \(3x = 5y - 15\), when x is 0, we get \(0 = 5y - 15\). By adding 15 to both sides and then dividing by 5, we find that y equals 3. So, our y-intercept is the point \(0, 3\). Placing this point on the graph gives us a crucial anchor around which the rest of the line will be built.

Finding X-Intercept

Parallel to finding the y-intercept, the x-intercept is where the line meets the x-axis. To compute the x-intercept, the y-variable is set to 0. Following this method in our example equation \(3x = 5y - 15\), substituting y with 0 leaves us with \(3x = -15\). Solving for x, we divide both sides by 3 and discover that x equals -5. This means our x-intercept is located at point \( -5, 0\). With both intercepts found, we can begin to visualize the path our line will take across the graph.

Linear Equation Checkpoints

After locating the intercepts, it's beneficial to identify at least one more point to ensure our line is graphed accurately. This is known as a checkpoint. A good checkpoint is a set of x and y values that satisfy the original equation when substituted. In the given example, \(3x = 5y - 15\), we chose \(x = 5\) and found that \(y = 6\) is a suitable checkpoint because it satisfies the equation:\[5(6) = 3(5) + 15\]. Verifying this point guarantees that we aren't merely connecting the intercepts but accurately representing the entire line.

Plotting Linear Equations

With our y-intercept \(0, 3\), x-intercept \( -5, 0\), and checkpoint \(5, 6\), we are ready to plot the linear equation. Start by marking the y-intercept on the vertical axis and the x-intercept on the horizontal axis of the graph. Then, plot the checkpoint. Ensure that each point is accurately placed according to its coordinates. Once all points are plotted, draw a straight line through them with the help of a ruler. This line represents the graph of the given linear equation \(3x = 5y - 15\). By connecting these points, the visual representation of the equation emerges, providing a clear and precise graph that captures the behavior of the mathematical relationship defined by the equation.