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Chapter 4: Problem 6

Solve each equation for the indicated variable. a. \(y=3(x+8)\) solve for \(x\) b. \(\frac{y-3}{x-4}=10\) solve for \(y\) c. \(4(2 y-5)-12=x\) solve for \(y\)

Short Answer

Expert verified
a. \(x = \frac{y-24}{3}\); b. \(y = 10x-37\); c. \(y = \frac{x+32}{8}\).

Step by step solution

01

Distribute and Simplify the Equation for 'a'

The equation given is \(y = 3(x + 8)\). Start by distributing the 3 across the terms inside the parentheses: \(y = 3x + 24\).
02

Isolate 'x' in 'a'

Subtract 24 from both sides to isolate the term with \(x\): \(y - 24 = 3x\). Now divide both sides by 3 to solve for \(x\): \(x = \frac{y - 24}{3}\).
03

Simplify the Equation for 'b'

The original equation is \(\frac{y-3}{x-4} = 10\). Eliminate the fraction by multiplying both sides by \(x-4\): \(y - 3 = 10(x - 4)\).
04

Distribute and Isolate 'y' in 'b'

Distribute 10 on the right-hand side: \(y - 3 = 10x - 40\). Add 3 to both sides to isolate \(y\): \(y = 10x - 37\).
05

Simplify and Rearrange for 'c'

Simplify the equation: \(4(2y - 5) - 12 = x\). First, distribute the 4: \(8y - 20 - 12 = x\). Combine like terms: \(8y - 32 = x\).
06

Solve for 'y' in 'c'

Add 32 to both sides: \(8y = x + 32\). Divide each side by 8 to solve for \(y\): \(y = \frac{x + 32}{8}\).

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Key Concepts

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

Solving for Variables
Solving for variables involves finding the value of one unknown variable in an equation. This process often requires understanding and implementing different algebraic techniques to isolate the variable of interest.

Consider an equation like: \( y = 3(x + 8) \). If we need to solve for \( x \), we start by controlling other numbers surrounding it. We perform operations such as addition, subtraction, multiplication, or division on both sides of the equation to maintain balance.

  • Identify the variable to solve for (e.g., in this case, \( x \)).
  • Perform inverse operations to unravel the variable from other terms or constants.
  • Always check your solution by substituting it back into the original equation to ensure accuracy.
Solving for variables is crucial, as it allows us to understand relationships between different quantities in mathematics and real-life problems.
Distributive Property
The distributive property is an essential algebraic tool used to simplify expressions, especially when solving equations. It states that a term multiplied by terms in parentheses can be distributed across the addition or subtraction within the parentheses.

For example: \( 3(x + 8) \) becomes \( 3 \times x + 3 \times 8 \) which simplifies to \( 3x + 24 \).

This property helps to:
  • Easily break down complex expressions into simpler ones.
  • Simplify the process of solving equations by removing parentheses.
Using the distributive property early in the process can make it easier to isolate and solve for specific variables later on in the equation.
Isolating Variables
Isolating variables is the technique used to get the variable you need by itself on one side of the equation. Often it involves several steps and the use of different algebraic operations.

Take the equation \( y - 24 = 3x \) from the prior example. To isolate \( x \), we adjust the equation so that \( x \) stands alone on one side:
  • Subtract 24 from both sides as needed.
  • Divide or multiply throughout to eliminate coefficients next to the variable.
After following these steps, you will have an expression where that variable is isolated, setting you up to find its exact value.
Linear Equations
Linear equations are equations where variables are raised only to the power of one. These equations form a straight line when plotted on a graph and include no intricate curves or powers higher than one.

The general form of a linear equation is often kept as \( y = mx + b \), where \( m \) denotes the slope and \( b \) denotes the y-intercept.

Solving linear equations requires:
  • Understanding how they represent relationships between variables.
  • Performing operations to isolate and solve for what's needed.
Linear equations are foundational in algebra, used widely in various fields to predict trends, understand data, and model different scenarios in both academic and practical situations.

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Most popular questions from this chapter

A sample labeled "50 grains" weighs \(3.24\) grams on a balance. What is the conversion factor for grams to grains?

Explain how to find the equation of a line when you know a. The slope and the \(y\)-intercept. b. Two points on that line.

Rewrite each equation in intercept form. Show your steps. Check your answer by using a calculator graph or table. a. \(y=14+3(x-5)\) b. \(y=-5-2(x+5)\) (a) c. \(6 x+2 y=24\)

APPLICATION The volume of a gas is \(3.50 \mathrm{~L}\) at \(280 \mathrm{~K}\). The volume of any gas is directly proportional to its temperature on the Kelvin scale \((\mathrm{K})\). a. Find the volume of this gas when the temperature is \(330 \mathrm{~K}\). b. Find the temperature when the volume is \(2.25 \mathrm{~L}\).

For each of these tables of \(x\) - and \(y\)-values, decide if the values indicate a direct variation, an inverse variation, or neither. Explain how you made your decision. If the values represent a direct or inverse variation, write an equation. $$ \begin{aligned} &\begin{array}{|c|c|} \hline \text { a. } & \boldsymbol{x} \end{array}\\\ &\text { b. } \begin{array}{|c|c|} \hline \text { (a) } & \boldsymbol{y} \\ \hline-20 & -5 \\ \hline-8 & -12.5 \\ \hline 2 & 50 \\ \hline 10 & 10 \\ \hline 25 & 4 \\ \hline \end{array}\\\ &\text { c. } \begin{array}{|r|r|} \hline {}{}{\boldsymbol{x}} & {}{}{\boldsymbol{y}} \\ \hline 0 & 0 \\ \hline-6 & 15 \\ \hline 8 & -20 \\ \hline-12 & 30 \\ \hline 4 & -10 \\ \hline \end{array}\\\ &\text { d. }\\\ &\begin{array}{|r|c|} \hline {}{}{\boldsymbol{x}} & \boldsymbol{y} \\ \hline 78 & 6 \\ \hline 31.2 & 2.4 \\ \hline-145.6 & -11.2 \\ \hline 14.3 & 1.1 \\ \hline-44.2 & -3.4 \\ \hline \end{array} \end{aligned} $$
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