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Chapter 3: Problem 288

In the following exercises, determine whether each number is a solution of the given equation. $$ \begin{array}{l}{\text { 288. } 7 c+42=-56} \\ {\text { (a) } c=2 \quad \text { (b) } c=-2} \\ {\text { (c) } c=-14}\end{array} $$

Short Answer

Expert verified
Only \(c = -14\) is a solution.

Step by step solution

01

- Understand the equation

The given equation is \(7c + 42 = -56\). You need to determine if each value of \(c\): 2, -2, and -14 is a solution to this equation.
02

- Substitute each value into the equation

For each value of \(c\), substitute it into the equation and check if the left-hand side equals the right-hand side.
03

- Test for \(c = 2\)

Substitute \(c = 2\) into the equation: \(7(2) + 42 = 14 + 42 = 56 eq -56\). Thus, \(c = 2\) is not a solution.
04

- Test for \(c = -2\)

Substitute \(c = -2\) into the equation: \(7(-2) + 42 = -14 + 42 = 28 eq -56\). Thus, \(c = -2\) is not a solution.
05

- Test for \(c = -14\)

Substitute \(c = -14\) into the equation: \(7(-14) + 42 = -98 + 42 = -56 = -56\). Thus, \(c = -14\) is a solution.

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Key Concepts

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

equation solving
Solving equations is a fundamental skill in algebra. You use it to find out which numbers make an equation true. An equation is simply a statement that two expressions are equal, with an '=' sign in between.
For example, in the equation: \(7c + 42 = -56\), we want to determine the value of \(c\) that makes this statement true.
To solve an equation:
  • Start by isolating the variable (usually represented by letters like \(c\), \(x\), or \(y\)).
  • Use inverse operations to simplify the equation step-by-step.
  • Check if the final value satisfies the original equation.
In our case, for each given value of \(c\) (2, -2, and -14), we will substitute them back into the equation and verify if the equation holds true.
substitution method
The substitution method is used to determine if a particular value is a solution to an equation. This method requires you to replace the variable in the equation with a given number and then perform the necessary calculations.
Here's how it works with our equation \(7c + 42 = -56\):
  • For \(c = 2\), substitute \(2\) into the equation: \(7(2) + 42 = 14 + 42 = 56\), which is not equal to -56. So, \(c = 2\) is not a solution.
  • For \(c = -2\), substitute \(-2\) into the equation: \(7(-2) + 42 = -14 + 42 = 28\), which is not equal to -56. So, \(c = -2\) is not a solution.
  • For \(c = -14\), substitute \(-14\) into the equation: \(7(-14) + 42 = -98 + 42 = -56\). This matches -56 exactly, meaning \(c = -14\) is a solution.
By substituting and calculating, we can see which values make the equation true.
algebraic solutions
An algebraic solution involves finding a value for a variable that makes the equation true. The steps are straightforward and rely on algebraic principles. Let's break it down:
  • Identify the given values and the equation.
  • Substitute each given value into the equation.
  • Perform arithmetic operations to see if both sides of the equation match.
In this exercise, we saw that out of the three given potential solutions, only \(c = -14\) satisfies the equation \(7c + 42 = -56\).
This means -14 is the algebraic solution to the equation.
Solving algebraic equations helps build a solid foundation in algebra. It strengthens problem-solving skills and prepares students for more complex math.

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