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Chapter 5: Problem 25

Find the slope and the \(y\) -intercept of the line with the given equation and sketch the graph using the slope and the \(y\) -intercept. A calculator can be used to check your graph. $$5 x-2 y=40$$

Short Answer

Expert verified
The slope is \(\frac{5}{2}\) and the \(y\)-intercept is \(-20\).

Step by step solution

01

Rewrite to Slope-Intercept Form

The standard form of the line is given as \(5x - 2y = 40\). Our first task is to rewrite it in the slope-intercept form, which is \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. Solve for \(y\) by isolating it on one side of the equation: Start with the equation: \[5x - 2y = 40\]Subtract \(5x\) from both sides: \[-2y = -5x + 40\]Divide each term by \(-2\) to solve for \(y\): \[y = \frac{5}{2}x - 20\]Now, the equation is in the form \(y = mx + b\) with \(m = \frac{5}{2}\) and \(b = -20\).
02

Identify Slope and y-intercept

Using the equation \(y = \frac{5}{2}x - 20\), identify the slope \(m\) and the \(y\)-intercept \(b\). The slope \(m\) is the coefficient of \(x\), which is \(\frac{5}{2}\). The \(y\)-intercept \(b\) is the constant term, which is \(-20\).
03

Plot the y-intercept

On a Cartesian plane, start by plotting the \(y\)-intercept. This is the point where the line crosses the \(y\)-axis. Given \(b = -20\), locate the point \((0, -20)\) on the graph.
04

Use Slope to Find Another Point

The slope \(m = \frac{5}{2}\) means that for every increase of 2 units in \(x\), \(y\) increases by 5 units. Start from the \(y\)-intercept \((0, -20)\) and move 2 units to the right (to \(x = 2\)) and 5 units up (to \(y = -15\)). Plot the second point \((2, -15)\).
05

Draw the Line

With the two points \((0, -20)\) and \((2, -15)\) plotted, draw a straight line through these points. This line represents the graph of the equation \(5x - 2y = 40\).
06

Verify with a Calculator

Optionally, use a graphing calculator to confirm the graph by inputting the equation \(y = \frac{5}{2}x - 20\). Ensure that the slope appears consistent and that the line correctly crosses the \(y\)-axis at \(-20\).

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Key Concepts

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

Linear Equations
Linear equations represent relationships between two variables, usually in the form of a straight line when graphed. These equations have a standard form: \(Ax + By = C\). In this standard form, \(A\), \(B\), and \(C\) are constants. The equation given in the original exercise is an example of such a form: \(5x - 2y = 40\).

To analyze it easily, you can rewrite it into the slope-intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. This form makes it simpler to understand how the line behaves on a graph. The slope shows how much \(y\) changes with \(x\), and the y-intercept tells you the starting point on the y-axis.

Here鈥檚 why this is useful:
  • It's simple to identify the slope and y-intercept.
  • Easy to interpret the graph鈥檚 line.
  • Makes solving for one variable straightforward.
Graphing Lines
Graphing lines from a linear equation is a step-by-step process. Once the equation is in the slope-intercept form \(y = mx + b\), you can graph it using these steps:

  • Start by identifying the y-intercept \(b\). This is where the line will cross the y-axis. For our exercise, \(b\) is \(-20\), so you plot the point \((0, -20)\).
  • The slope \(m\) tells you the angle of the line. In this exercise, it is \(\frac{5}{2}\), meaning for every 2 units you move horizontally, you move 5 units vertically. This rise-over-run relationship helps you find another point on the line.
  • After plotting the y-intercept, use the slope to find another point by moving 2 units right and 5 units up from \((0, -20)\) to get \((2, -15)\).
  • Draw a straight line through both points to complete the graph.
Visualizing the line helps you understand its relation to the axes and how changes in either direction (up or right) affect the values of y.
Mathematical Solutions
Solving linear equations and understanding graphs involves finding precise mathematical solutions. Here are the steps demonstrated in the solution:

  • Convert the equation into a workable form, such as slope-intercept, for clarity and ease of graphing.
  • Identify the key elements like slope and intercept that allow for practical graphing and understanding of linear relationships.
  • Verify your graph manually, or optionally, with a calculator, to ensure accuracy.
To convert the given equation \(5x - 2y = 40\) to slope-intercept form, rearrange to isolate \(y\). This step ensures you can read the slope \(m = \frac{5}{2}\) and the y-intercept \(b = -20\) directly. Such solutions provide a clear pathway from an abstract equation to a visual graph, making mathematical ideas tangible and logical.

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Most popular questions from this chapter

Solve the given systems of equations by use of determinants. (Exercises \(17-26\) are the same as Exercises \(3-12\) of Section \(5.6 .)\) $$\begin{aligned} &3.0 x+4.5 y-7.5 z=10.5\\\ &4.8 x-3.6 y-2.4 z=1.2\\\ &4.0 x-0.5 y+2.0 z=1.5 \end{aligned}$$

Solve the given systems of equations by either method of this section. $$\begin{aligned} &66 x+66 y=-77\\\ &33 x-132 y=143 \end{aligned}$$

Solve the indicated or given systems of equations by an appropriate algebraic method. Solve the following system of equations by (a) solving the first equation for \(x\) and substituting, and (b) solving the first equation for \(y\) and substituting. $$\begin{aligned} &2 x+y=4\\\ &3 x-4 y=-5 \end{aligned}$$

Solve the given problems by determinants. In Exerciser-46,\( set up appropriate systems of equations. All numbers are accurate to at least two significant digits. A company budgets \)\$ 750,000$ in salaries, hardware, and computer time for the design of a new product. The salaries are as much as the others combined, and the hardware budget is twice the computer budget. How much is budgeted for each?

Set up appropriate systems of two linear equations and solve the systems algebraically. All data are accurate to at least two significant digits. Regarding the forces on a truss, a report stated that force \(F_{1}\) is twice force \(F_{2}\) and that twice the sum off the two forces less 6 times \(F_{2}\) is 6 N. Explain your conclusion about the magnitudes of the forces found from this support.
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