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Chapter 1: Problem 70

Find the \(x\) and \(y\)-intercepts of the line, and draw its graph. \(\frac{1}{3} x-\frac{1}{5} y-2=0\)

Short Answer

Expert verified
The x-intercept is (6, 0) and the y-intercept is (0, -10).

Step by step solution

01

Find the x-intercept

To find the x-intercept, set \( y = 0 \) in the equation. The equation becomes \( \frac{1}{3}x - 2 = 0 \). Solve this equation for \( x \): \( \frac{1}{3}x = 2 \), which simplifies to \( x = 6 \). Hence, the x-intercept is \((6, 0)\).
02

Find the y-intercept

To find the y-intercept, set \( x = 0 \) in the equation. The equation becomes \( -\frac{1}{5}y - 2 = 0 \). Solve this equation for \( y \): \( -\frac{1}{5}y = 2 \), which simplifies to \( y = -10 \). Therefore, the y-intercept is \((0, -10)\).
03

Plot the intercepts

With the intercepts \((6, 0)\) and \((0, -10)\), plot these points on a coordinate plane. The x-intercept is on the x-axis at 6, and the y-intercept is on the y-axis at -10.
04

Draw the line

Draw a straight line through the points \((6, 0)\) and \((0, -10)\). This line represents the graph of the given equation \(\frac{1}{3}x - \frac{1}{5}y - 2 = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the X-Intercept
The x-intercept of a line is the point where the line crosses the x-axis. At this point, the value of y is always 0. Finding the x-intercept is crucial when graphing a linear equation, as it provides one of the key locations through which the line passes.
To find the x-intercept for an equation like \(\frac{1}{3}x - \frac{1}{5}y - 2 = 0\), you simply set \(y = 0\) and solve for \(x\).
  • Replace \(y\) with 0 in the equation: \(\frac{1}{3}x - 2 = 0\).
  • Solve for \(x\), which results in \(x = 6\).
This means the x-intercept is at the point \((6, 0)\), which you can easily plot on the x-axis when drawing the line.
Understanding the Y-Intercept
The y-intercept is the point where the line crosses the y-axis. Here, the value of x is always 0. Like the x-intercept, finding the y-intercept is vital for sketching the graph of a line, as it marks another key point.
To find the y-intercept in a linear equation such as \(\frac{1}{3}x - \frac{1}{5}y - 2 = 0\), follow these steps:
  • Substitute \(x = 0\) into the equation: \(-\frac{1}{5}y - 2 = 0\).
  • Solve for \(y\), leading to \(y = -10\).
Thus, the y-intercept is at \((0, -10)\). This point can be plotted on the y-axis, aiding in the accurate drawing of the line.
Grasping the Linear Equation
A linear equation is a mathematical expression that creates a straight line when graphed. Such equations typically involve two variables, \(x\) and \(y\), and are represented in standard form, slope-intercept form, or other similar forms.
The standard form of a linear equation is \(Ax + By = C\). The given equation \(\frac{1}{3}x - \frac{1}{5}y - 2 = 0\) is a variation of this form, showing the constant terms and coefficients.
When rearranged, you can better understand the relationship:
  • The coefficients of \(x\) and \(y\) (in this case, \(\frac{1}{3}\) and \(-\frac{1}{5}\)) indicate the steepness and orientation of the line.
  • Consistent solutions for \(x\) and \(y\), such as intercepts, help solve and graph the equation.
Comprehending these expressions is beneficial for predicting how different changes will affect the line鈥檚 appearance on a graph.
Visualizing the Graph of a Line
Creating the graph of a line involves plotting points and drawing the connecting line, primarily using the x- and y-intercepts for guidance. In our scenario, we have the intercepts \((6, 0)\) and \((0, -10)\), which serve as anchor points.
To construct a graph:
  • Start by plotting the x-intercept \((6, 0)\) on the x-axis.
  • Next, plot the y-intercept \((0, -10)\) on the y-axis.
  • Draw a straight line between these two points, ensuring it extends beyond each intercept across the plane.
This line accurately represents the relationship described by the equation \(\frac{1}{3}x - \frac{1}{5}y - 2 = 0\). Understanding this concept allows you to visualize the equation鈥檚 solution as a straight line on a graph.

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Most popular questions from this chapter

Solving an Equation in Different Ways We have learned several different ways to solve an equation in this section. Some equations can be tackled by more than one method. For example, the equation \(x-\sqrt{x}-2=0\) is of quadratic type. We can solve it by letting \(\sqrt{x}=u\) and \(x=u^{2},\) and factoring. Or we could solve for \(\sqrt{x},\) square each side, and then solve the resulting quadratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers. (a) \(x-\sqrt{x}-2=0 \quad\) quadratic type; solve for the radical, and square (b) \(\frac{12}{(x-3)^{2}}+\frac{10}{x-3}+1=0 \quad \begin{array}{l}\text { quadratic type; multiply } \\ \text { by } \mathrm{LCD}\end{array}\)

Test the equation for symmetry. $$y=x^{4}+x^{2}$$

Simplify the expression. (a) \(\left(8 y^{3}\right)^{-2 / 3}\) (b) \(\left(u^{4} v^{6}\right)^{-1 / 3}\)

The Lens Equation If \(F\) is the focal length of a convex lens and an object is placed at a distance \(x\) from the lens, then its image will be at a distance \(y\) from the lens, where \(F, x,\) and \(y\) are related by the lens equation $$\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$$ Suppose that a lens has a focal length of \(4.8 \mathrm{cm}\) and that the image of an object is \(4 \mathrm{cm}\) closer to the lens than the object itself. How far from the lens is the object?

Rationalize Put each fractional expression into standard form by rationalizing the denominator. (a) \(\frac{1}{\sqrt{5 x}}\) (b) \(\sqrt{\frac{x}{5}}\) (c) \(\sqrt[5]{\frac{1}{x^{3}}}\)
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