Problem 27
Question
Without solving them, say whether the equations have a positive solution, a negative solution, a zero solution, or no solution. Give a reason for your answer. $$ 2-11 c=8-3 c $$
Step-by-Step Solution
Verified Answer
Based on the step by step solution provided, the given linear equation has a negative solution as the value of c is \(-\frac{3}{4}\).
1Step 1: Isolate the variable c
In order to do this, we will perform algebraic operations to make c the subject of the equation. Start by adding \(3c\) to both sides of the equation:
$$
2 - 11c + 3c = 8 - 3c + 3c
$$
This simplifies to:
$$
2 - 8c = 8
$$
Now, add \(8c\) to both sides of the equation:
$$
2 = 8 + 8c
$$
Finally, subtract \(8\) from both sides of the equation:
$$
-6 = 8c
$$
Divide both sides by \(8\) to isolate \(c\):
$$
c = \frac{-6}{8}
$$
Which simplifies to:
$$
c = -\frac{3}{4}
$$
2Step 2: Analyze the solution
Now that we have isolated \(c\) and found its value, we can determine the type of solution this equation has. Since \(c = -\frac{3}{4}\), it is a negative number, which means the equation has a negative solution.
Key Concepts
Solving EquationsNegative SolutionsAlgebraic Operations
Solving Equations
Solving equations is like solving a puzzle. The goal is to find the value of the unknown variable, in this case, "c". You do this by rearranging the equation until the variable is isolated. To start, apply basic algebraic operations such as addition, subtraction, multiplication, or division to balance the equation. Remember, whatever you do on one side of the equation, you must do on the other side to keep it balanced.
- Identify the variable you want to solve for.
- Use inverse operations to simplify the equation, step by step.
- Check your work by substituting your solution back into the original equation.
Negative Solutions
A negative solution occurs when the variable, once solved, results in a negative number. In the example provided, after performing the algebraic operations, we found that the variable \(c\) equals \(-\frac{3}{4}\). This is clearly a negative number.
Negative solutions are just as valid as positive solutions and occur frequently in algebra. They often arise from equations where the sum or the differences of variables lead to numbers less than zero.
Negative solutions are just as valid as positive solutions and occur frequently in algebra. They often arise from equations where the sum or the differences of variables lead to numbers less than zero.
Understanding Negative Values
When interpreting a negative solution, it's important to consider the context. In real-world applications, negative solutions might represent loss, decrease, direction, or position. Here are some examples to put it into context:- Loss of height (-2 meters could indicate depth below sea level).
- A decrease in temperature (-5°C means 5 degrees below zero).
- Debt (owes -\(100 represents owing \)100).
Algebraic Operations
Algebraic operations are the tools you use to manipulate the equations in order to solve for the variable. They include addition, subtraction, multiplication, and division. Let's see how each was applied in the example:
- Addition: Adding \(3c\) to both sides to simplify the equation further.
- Subtraction: Subtracting numbers from both sides to help isolate the variable.
- Multiplication/Division: Dividing or multiplying when necessary to solve for the variable itself.
Key Points to Remember
- Perform operations on both sides of the equation to maintain balance.
- Use opposite operations: if a term is subtracted, add it to both sides and vice versa.
- Simplify step by step, checking each step for errors.
Other exercises in this chapter
Problem 27
Solve the system of equations graphically. $$ \left\\{\begin{array}{r} 2 x+5 y=7 \\ -3 x+2 y=1 \end{array}\right. $$
View solution Problem 27
Put the equation in standard form. $$ x+4=3(y-1) $$
View solution Problem 27
Identify the slope and \(y\) -intercept and graph the function. $$ f(x)=3 x-2 $$
View solution Problem 27
Find a possible formula for the linear function \(h(x)\) if \(h(-30)=80\) and \(h(40)=-60\)
View solution