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Chapter 1: Problem 15

The value of the variable a solution to the equation? $$ 4(r-3)=4 r-3, r=1 $$

Short Answer

Expert verified
Answer: -5

Step by step solution

01

Substitute the given value of r

Since r=1, replace r with 1 in the equation: $$ 4(1-3)=4\cdot1-3 $$
02

Simplify the equation

Perform the operations inside the parentheses and the multiplication: $$ 4(-2)=4-3 $$
03

Solve for a

Now that we have simplified the equation, we can solve for a: $$ -8=a-3 $$ Add 3 to both sides of the equation: $$ -5=a $$ The value of the variable a solution to the equation is -5.

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Key Concepts

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

Substitution Method
The substitution method is a crucial technique in mathematics for solving equations, where you replace a variable with its given value. It is especially helpful when you are solving for a specific variable within an equation that includes multiple variables.
To use the substitution method, follow these steps:
  • Identify the variable you need to substitute. In the example exercise, this is given as \( r = 1 \).
  • Replace the variable everywhere it appears in the equation with its given value. In our example, each instance of \( r \) in the equation \( 4(r-3)=4r-3 \) was replaced by \( 1 \).
This method helps convert an equation with variables into a simpler arithmetic expression. The substitution method is effective in linear equations but is also applicable in other types of equations, providing flexibility and accuracy.
Simplification
Simplification is a process of reducing complexity in a mathematical expression or equation. After substituting the values, the next step is simplifying any complex parts, such as parentheses and multiplications.
In our exercise example, simplification involved:
  • Performing operations inside parentheses: \( 4(1-3) \equiv 4(-2) \).
  • Carrying out the multiplication: you multiply 4 by -2 to get -8 on the left side of the equation.
The goal is to make the equation easier to solve by turning a potentially complicated equation into one that is straightforward and manageable. Simplifying an equation allows you to focus on understanding and solving the crucial parts of the math problem.
Linear Equations
Linear equations are algebraic expressions that represent relationships of straight lines when graphed. They typically appear in the form \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable you are solving for. Linear equations have consistent solutions and do not include exponents higher than one.
In the exercise, after simplification, the equation becomes \( -8 = a - 3 \). This is a linear equation where the unknown variable \( a \) can be isolated by algebraic manipulation.
Some key points about linear equations:
  • They generally have one solution that can be found through basic algebraic operations like addition, subtraction, multiplication, or division.
  • The solution to a linear equation is the value of the variable that makes the equation true.
  • Solving a linear equation involves rearranging the equation until the variable stands alone on one side, making it possible to easily identify its value, as we did by adding 3 to both sides to solve for \( a \).

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Most popular questions from this chapter

In Problems \(52-55,\) decide whether the expressions can be put in the form $$ \frac{a x}{a+x} . $$ For those that are of this form, identify \(a\) and \(x\). $$ \frac{b^{2} \theta^{2}}{b^{2}+\theta^{2}} $$

Evaluate and simplify the expressions in Problems 44-53. $$ h(2 x-b-3) \text { given } h(x)=2 x-b-2 \text { . } $$

In Problems \(46-49\), say which of the given pair of expressions is larger. $$ \frac{12}{z^{2}+4} \text { and } \frac{12}{z^{2}+3} $$

Oil well number 1 produces \(r_{1}\) barrels per day, and oil well number 2 produces \(r_{2}\) barrels per day. Each expression in Problems \(35-40\) describes the production at another well. What does the expression tell you about the well? $$ r_{1}-80 $$

(a) Write an expression for the total cost of buying 8 apples at \(\$ a\) each and 5 pears at \(\$ p\) each. (b) Find the total cost if apples cost \(\$ 0.40\) each and pears cost \(\$ 0.75\) each.
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