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Chapter 2: Problem 7

Put the following equations in \(y=m x+b\) form, then identify the slope and the vertical intercept. a. \(2 x-3 y=6\) b. \(3 x+2 y=6\) c. \(\frac{1}{3} x+\frac{1}{2} y=6\) d. \(2 y-3 x=0\) e. \(6 y-9 x=0\) f. \(\frac{1}{2} x-\frac{2}{3} y=-\frac{1}{6}\)

Short Answer

Expert verified
a) Slope: \( \frac{2}{3} \), Intercept: \( -2 \) b) Slope: \( -\frac{3}{2} \), Intercept: \( 3 \) c) Slope: \( -\frac{2}{3} \), Intercept: \( 12 \) d) Slope: \( \frac{3}{2} \), Intercept: \( 0 \) e) Slope: \( \frac{3}{2} \), Intercept: \( 0 \) f) Slope: \( \frac{3}{4} \), Intercept: \( \frac{1}{4} \)

Step by step solution

01

Transform equation (a)

Start with the equation: \[2x - 3y = 6\] To put it in the form \( y = mx + b \), first isolate the \( y \)-term by subtracting \( 2x \) from both sides: \[-3y = -2x + 6\] Then, divide by \( -3 \) to solve for \( y \): \[y = \frac{2}{3}x - 2\]
02

Identify slope and intercept for (a)

The slope \( m \) is \( \frac{2}{3} \) and the vertical intercept \( b \) is \( -2 \).
03

Transform equation (b)

Start with the equation: \[3x + 2y = 6\] Subtract \( 3x \) from both sides: \[2y = -3x + 6\] Divide by \( 2 \) to solve for \( y \): \[y = -\frac{3}{2} x + 3\]
04

Identify slope and intercept for (b)

The slope \( m \) is \( -\frac{3}{2} \) and the vertical intercept \( b \) is \( 3 \).
05

Transform equation (c)

Start with the equation: \[\frac{1}{3}x + \frac{1}{2}y = 6\] Subtract \( \frac{1}{3}x \) from both sides: \[\frac{1}{2}y = -\frac{1}{3}x + 6\] Multiply by \(2\) to isolate \( y \): \[y = -\frac{2}{3} x + 12\]
06

Identify slope and intercept for (c)

The slope \( m \) is \( -\frac{2}{3} \) and the vertical intercept \( b \) is \( 12 \).
07

Transform equation (d)

Start with the equation: \[2y - 3x = 0\] Add \( 3x \) to both sides: \[2y = 3x\] Divide by \( 2 \) to solve for \( y \): \[y = \frac{3}{2} x\]
08

Identify slope and intercept for (d)

The slope \( m \) is \( \frac{3}{2} \) and the vertical intercept \( b \) is \( 0 \).
09

Transform equation (e)

Start with the equation: \[6y - 9x = 0\] Add \( 9x \) to both sides: \[6y = 9x\] Divide by \( 6 \) to solve for \( y \): \[y = \frac{3}{2} x\]
10

Identify slope and intercept for (e)

The slope \( m \) is \( \frac{3}{2} \) and the vertical intercept \( b \) is \( 0 \).
11

Transform equation (f)

Start with the equation: \[\frac{1}{2}x - \frac{2}{3}y = -\frac{1}{6}\] Add \( \frac{2}{3}y \) to both sides: \[\frac{1}{2}x = \frac{2}{3}y - \frac{1}{6}\] Multiply everything by \( 6 \) to clear the fractions: \[3x - 4y = -1\] Add \( 4y \) to both sides: \[3x + 1 = 4y\] Divide by \( 4 \) to solve for \( y \): \[y = \frac{3}{4}x + \frac{1}{4}\]
12

Identify slope and intercept for (f)

The slope \( m \) is \( \frac{3}{4} \) and the vertical intercept \( b \) is \( \frac{1}{4} \).

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Key Concepts

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

solving linear equations
Solving linear equations is essential in algebra as it forms the basis for many other concepts. A linear equation typically has the form \(Ax + By = C\). To solve, you isolate one variable. Here's the process:

For instance, from the equation \(3x + 2y = 6\), subtract \(3x\) from both sides to isolate \(y\):
\[2y = -3x + 6\]
Then, divide by \(2\) to solve for \(y\):
\[y = -\frac{3}{2} x + 3\]
This skill will be useful in transforming equations into different forms, such as the slope-intercept form. Using simple steps and breaking down complex equations makes learning approachable.
slope and intercept
Understanding the slope and intercept in the slope-intercept form \(y = mx + b\) is crucial. The slope (\(m\)) represents the steepness of the line, and the intercept (\(b\)) is where the line crosses the y-axis.

The formula helps easily identify these components. For example, in \(y = \frac{2}{3}x - 2\), the slope \(m\) is \(\frac{2}{3}\), signalling a rise of 2 for every 3 units run. The intercept \(b\) is \(-2\), meaning the line crosses the y-axis at -2.

Visualizing this makes graphing more straightforward, aiding in comprehensive understanding and application to various problems.
transforming equations
Transforming equations involves changing their form while keeping their relationships intact. Often, this means converting to the slope-intercept form \(y = mx + b\).

Starting from an equation like \(2x - 3y = 6\), you isolate \(y\) with steps:
  • Subtract \2x\: \(-3y = -2x + 6\)
  • Divide by \(-3\): \(y = \frac{2}{3} x - 2\)
This systematic approach enables clear visualization of the slope and intercept.

Resolving complex fractions is similar. For \(\frac{1}{3} x + \frac{1}{2} y = 6\), isolate \(y\):
  • Subtract \(\frac{1}{3} x\): \(\frac{1}{2} y = -\frac{1}{3} x + 6\)
  • Multiply by \2\: \(y = -\frac{2}{3} x + 12\)
Practicing this boosts familiarity, making any equation transformation logical and intuitive.

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Most popular questions from this chapter

On the same axes, graph (and label with the correct equation) three lines that go through the point (0,2) and have the following slopes: a. \(m=\frac{1}{2}\) b. \(m=2\) c. \(m=\frac{5}{6}\)

The following table shows U.S. first-class stamp prices (per ounce) over time. $$ \begin{array}{lc} \hline \text { Year } & \text { Price for First-Class Stamp } \\ \hline 2001 & 34 \text { cents } \\ 2002 & 37 \text { cents } \\ 2006 & 39 \text { cents } \\ \hline \end{array} $$ a. Construct a step function describing stamp prices for \(2001-2006 .\) b. Graph the function. Be sure to specify whether each of the endpoints is included or excluded. c. In 2007 the price of a first-class stamp was raised to 41 cents. How would the function domain and the graph change?

Which equation has the steepest slope? a. \(y=2-7 x\) b. \(y=2 x+7\) c. \(y=-2+7 x\)

Determine which of the following tables represents a linear function. If it is linear, write the equation for the linear function. a. $$ \begin{array}{rr} \hline x & y \\ \hline 0 & 3 \\ 1 & 8 \\ 2 & 13 \\ 3 & 18 \\ 4 & 23 \\ \hline \end{array} $$ b.$$ \begin{array}{rr} \hline q & R \\ \hline 0 & 0.0 \\ 1 & 2.5 \\ 2 & 5.0 \\ 3 & 7.5 \\ 4 & 10.0 \\ \hline \end{array} $$ c. $$ \begin{array}{rr} \hline x & g(x) \\ \hline 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ 4 & 16 \\ \hline \end{array} $$ d.$$ \begin{array}{cc} \hline t & r \\ \hline 10 & 5.00 \\ 20 & 2.50 \\ 30 & 1.67 \\ 40 & 1.25 \\ 50 & 1.00 \\ \hline \end{array} $$ e. $$ \begin{array}{rr} \hline x & \multicolumn{1}{c} {h(x)} \\ \hline 20 & 20 \\ 40 & -60 \\ 60 & -140 \\ 80 & -220 \\ 100 & -300 \\ \hline \end{array} $$ f. $$ \begin{array}{cc} \hline p & T \\ \hline 5 & 0.25 \\ 10 & 0.50 \\ 15 & 0.75 \\ 20 & 1.00 \\ 25 & 1.25 \\ \hline \end{array} $$

On the scale of a map 1 inch represents a distance of 35 miles. a. What is the distance between two places that are 4.5 inches apart on the map? b. Construct an equation that converts inches on the map to miles in the real world.
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