Solve $\frac{4}{7}(c+1)=-\frac{2}{7} c-2$.
To solve the equation \(\frac{4}{7}(c+1) = -\frac{2}{7} c - 2\), we first look at the structure of the equation. We have a variable \(c\) on both sides and fractions with a common denominator of 7. To simplify the equation, we can start by eliminating the fractions. We multiply every term on both sides by 7:
\[7 \times \left( \frac{4}{7}(c+1) \right) = 7 \times \left( -\frac{2}{7} c - 2 \right)\]
This simplifies to:
\[4(c+1) = -2c - 14\]
Next, we distribute the constant \(4\) on the left side of the equation:
\[4c + 4 = -2c - 14\]
Now, we want to collect all the terms containing the variable \(c\) on one side and the constant terms on the other side. We add \(2c\) to both sides of the equation:
\[4c + 2c + 4 = -14\]
Combining the \(c\) terms gives us:
\[6c + 4 = -14\]
Now, we isolate the \(c\) term by subtracting 4 from both sides:
\[6c = -14 - 4\]
\[6c = -18\]
Finally, we solve for \(c\) by dividing both sides by 6:
\[c = \frac{-18}{6}\]
\[c = -3\]
Thus, the solution is \(c = -3\). This matches the fourth option provided.
This type of linear equation solving is essential in various real-life scenarios, such as financial planning or engineering design. For instance, if you are budgeting for a construction project, you might have fixed costs (like material fees) and variable costs (that depend on the amount of material, represented by a variable like \(c\)). Being able to isolate variables in an equation allows you to calculate the exact amount of resources needed to stay within your budget. By simplifying complex expressions into manageable linear steps, you can accurately determine the value that satisfies your requirements.