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Chapter 2: Problem 2

Sketch the following graphs: a \(y=3\) b \(x=3\) c \(2 x+2 y=6\) d \(x+2 y=-3\) e \(x-y-3=0\) f \(2 x+7 y-5=0\)

Short Answer

Expert verified
a. y=3: horizontal line at y=3; b. x=3: vertical line at x=3; c. y=3-x; d. y=-\frac{1}{2}x -\frac{3}{2}; e. y=x-3; f. y= -\frac{2}{7}x + \frac{5}{7}

Step by step solution

01

Graph of y=3

This is a horizontal line where the y-value is always 3. Draw a horizontal line across the graph at y = 3.
02

Graph of x=3

This is a vertical line where the x-value is always 3. Draw a vertical line across the graph at x = 3.
03

Graph of 2x + 2y = 6

Rewrite the equation in slope-intercept form. First, divide by 2: \[x + y = 3\] Then, solve for y: \[y = 3 - x\] Plot the y-intercept (3,0) and the x-intercept (0,3) and draw the line.
04

Graph of x + 2y = -3

Rewrite in slope-intercept form: \[2y = -x - 3\] Solve for y: \[y = -\frac{1}{2}x - \frac{3}{2}\] Plot the y-intercept (-3/2, 0) and the x-intercept (0, -3/2) and draw the line.
05

Graph of x - y - 3 = 0

Rewrite in slope-intercept form: \[x - y = 3\] Solve for y: \[y = x - 3\] Plot the y-intercept (3,0) and the x-intercept (0, -3) and draw the line.
06

Graph of 2x + 7y - 5 = 0

Rewrite in slope-intercept form: \[7y = -2x + 5\] Solve for y: \[y = -\frac{2}{7}x + \frac{5}{7}\]. Plot the y-intercept (5/7,0) and the x-intercept (0, -5/2) and draw the line.

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Key Concepts

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

graphing linear equations
When graphing linear equations, our goal is to represent the equation on a coordinate plane. Linear equations form straight lines, and each point on the line is a solution to the equation. The general form of a linear equation in two variables is \(Ax + By = C\). It means the same relationship exists between x and y everywhere on the line.
slope-intercept form
The slope-intercept form of a linear equation is \(y = mx + b\) where \(m\) is the slope and \(b\) is the y-intercept. This form is especially handy for graphing because it directly shows the starting point (y-intercept) and the direction and steepness of the line (slope).
intercepts in graphing
Intercepts are the points where the graph crosses the x-axis and y-axis. To find the y-intercept, set \(x = 0\) and solve for \(y\). To find the x-intercept, set \(y = 0\) and solve for \(x\). These points are crucial for graphing since they provide key reference points for drawing the line.

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Most popular questions from this chapter

[electrical principles] The voltage, \(V\), across a variable load resistor of resistance \(R_{\mathrm{L}}\), is given by $$ V=\frac{E R_{\mathrm{L}}}{R+R_{\mathrm{L}}} $$ where \(E\) is the source e.m.f. and \(R\) is the source resistance. Plot the graphs (on different axes) of \(V\) versus \(R_{\mathrm{L}}\) for the corresponding values: a \(E=60\) volts, \(R=10 \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 20\) b \(E=15\) volts, \(R=3 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 6 \times 10^{3}\) c \(E=10\) volts, \(R=15 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 30 \times 10^{3}\) From each of your graphs, determine the value of \(V\) at \(R_{\mathrm{L}}=R\). Do you notice any relationship between \(V\) and \(E\) at \(R_{\mathrm{L}}=R\) ? Show algebraically that if \(R_{\mathrm{L}}=R\) then \(V=\frac{E}{2}\).

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