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Chapter 2: Problem 7

The following problems represent calculations of the slopes of different lines. Solve for the variable in each equation. a. \(\frac{150-75}{20-10}=m\) c. \(\frac{182-150}{28-x}=4\) b. \(\frac{70-y}{0-8}=0.5\) d. \(\frac{6-0}{x-10}=0.6\)

Short Answer

Expert verified
a) m = 7.5, b) x = 20, c) y = 74, d) x = 20.

Step by step solution

01

Title - Solve for m in equation a

Given the equation \( \frac{150-75}{20-10} = m \), simplify the numerator and the denominator: \( 150 - 75 = 75 \) and \( 20 - 10 = 10 \). So, \( m = \frac{75}{10} = 7.5 \).
02

Title - Solve for x in equation b

Given the equation \( \frac{182-150}{28-x}=4 \), solve for \( x \). Cross-multiply to get: \( 182 - 150 = 4(28 - x) \). Simplify: \( 32 = 112 - 4x \). Move all terms involving \( x \) to one side and solve: \( 4x = 112 - 32 \) which simplifies to \( 4x = 80 \). So, \( x = 20 \).
03

Title - Solve for y in equation c

Given the equation \( \frac{70-y}{0-8}=0.5 \), solve for \( y \). Cross-multiply to get: \( 70 - y = 0.5 (-8) \). Simplify: \( 70 - y = -4 \). Move all terms involving \( y \) to one side and solve: \( 70 + 4 = y \). So, \( y = 74 \).
04

Title - Solve for x in equation d

Given the equation \( \frac{6-0}{x-10}=0.6 \), solve for \( x \). Cross-multiply to get: \( 6 = 0.6(x - 10) \). Simplify: \( 6 = 0.6x - 6 \). Move all terms involving \( x \) to one side and solve: \( 6 + 6 = 0.6x \). This simplifies to \( 12 = 0.6x \). So, \( x = 20 \).

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Key Concepts

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

Algebra
Algebra is a branch of mathematics that involves variables and the relationships between them. By using algebra, you can form equations and solve for unknown variables. In the provided exercise, we have multiple algebraic equations where the unknown variables need to be solved by manipulating the equations to isolate the variable on one side.
For example, in equation a: \( m = \frac{150-75}{20-10} \), the goal is to simplify the numerator and the denominator to find the value of m.
Here鈥檚 the detailed process:
  • Simplify the numerator: 150 - 75 equals 75.
  • Simplify the denominator: 20 - 10 equals 10.
  • Now, division: 75 / 10 equals 7.5.
So, m equals 7.5. These steps involve basic arithmetic, but algebra is the tool we use to structure and manipulate these operations.
Linear Equations
Linear equations are equations of the first degree, meaning they involve variables raised only to the power of one. They represent straight lines when plotted on a graph. In the exercise, each equation is a form of a linear equation.
For instance, equation b: \( \frac{182-150}{28-x}=4 \) can be solved by cross-multiplying to isolate the variable x.
Here鈥檚 how to solve it:
  • First, cross-multiply: \( 182 - 150 = 4 \times (28 - x) \).
  • Simplify: \( 32 = 112 - 4x \).
  • Move all x terms to one side: \( 4x = 112 - 32 \), which simplifies to \( 4x = 80 \).
  • Finally, solve for x: \( x = \frac{80}{4} \), resulting in \( x = 20 \).
Linear equations can be solved by performing operations such as addition, subtraction, and cross-multiplication.
Solving for Variables
Solving for variables involves isolating the unknown variable on one side of the equation. This is typically done through various arithmetic operations until the variable is by itself. Let's see how this concept applies to equation c:
Equation c: \( \frac{70-y}{0-8}=0.5 \)
The steps to solve for y are:
  • Cross-multiply: \( 70 - y = 0.5 \times (-8) \).
  • Simplify the multiplication: \( 70 - y = -4 \).
  • Isolate y by moving all terms involving y to one side: \( 70 + 4 = y \).
  • So, \( y = 74 \).
For equation d: \( \frac{6-0}{x-10}=0.6 \)
  • Cross-multiply: \( 6 = 0.6 \times (x - 10) \).
  • Simplify: \( 6 = 0.6 x - 6 \).
  • Move terms involving x: \( 6 + 6 = 0.6 x \), which simplifies to \( 12 = 0.6 x \).
  • Finally, solve for x: \( x = \frac{12}{0.6} \), resulting in \( x = 20 \).
This process requires carefully following the steps and performing basic arithmetic operations.

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Most popular questions from this chapter

Describe the graphs of the following equations. a. \(y=-2\) c. \(x=\frac{2}{3}\) e. \(y=324\) b. \(x=-2\) d. \(y=\frac{x}{4}\) \(\mathbf{f} . y=\frac{2}{3}\)

You read in the newspaper that the river is polluted with 285 parts per million (ppm) of a toxic substance, and local officials estimate they can reduce the pollution by \(15 \mathrm{ppm}\) each year. a. Derive an equation that represents the amount of pollution, \(P\), as a function of time, \(t\). b. The article states the river will not be safe for swimming until pollution is reduced to \(40 \mathrm{ppm} .\) If the cleanup proceeds as estimated, in how many years will it be safe to swim in the river?

According to the U.S. Burcau of the Census, the percentage of persons 25 years old and over completing 4 or more years of college was 4.6 in 1940 and 27.6 in 2005 . a. Plot the data, labeling both axes and the coordinates of the points. b. Calculate the average rate of change in percentage points per year. c. Write a topic sentence summarizing what you think is the central idea to be drawn from these data.

Using the general formula \(y=m x\) that describes direct proportionality, find the value of \(m\) if: a. \(y\) is directly proportional to \(x\) and \(y=2\) when \(x=10\). b. \(y\) is directly proportional to \(x\) and \(y=0.1\) when \(x=0.2\). c. \(y\) is directly proportional to \(x\) and \(y=1\) when \(x=\frac{1}{4}\).

The consumption of margarine (in pounds per person) decreased from 9.4 in 1960 to 7.5 in 2000 . What was the annual average rate of change? (Source: wwweensus.gov)
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