| $3(x+2)-4=-7$ | |
Distribute (to get rid of parentheses) | |
Combine any Like Terms | |
Undo addition or subtraction first | |
Undo multiplication or division next | |
We are given the linear equation \(3(x+2)-4=-7\). Our goal is to solve for \(x\) by following a systematic approach: distributing the constant across the parentheses, combining like terms, isolating the variable term, and finally solving for \(x\). This process ensures that we follow the algebraic properties of equality to reach the correct value.
The first step is to distribute the constant \(3\) into the parentheses \((x+2)\). Distributing means multiplying \(3\) by each term inside the parentheses:
\[3 \times x + 3 \times 2 = 3x + 6.\]
Substituting this back into the original equation, we get:
\[3x + 6 - 4 = -7.\]
Now, we look for terms on the same side of the equation that can be combined. On the left side, we have the constants \(6\) and \(-4\). Combining these gives:
\[6 - 4 = 2.\]
The equation now simplifies to:
\[3x + 2 = -7.\]
To isolate the term with \(x\), we need to undo the addition of \(2\) on the left side. We do this by subtracting \(2\) from both sides of the equation:
\[3x + 2 - 2 = -7 - 2,\]
which results in:
\[3x = -9.\]
Finally, we need to isolate \(x\) by undoing the multiplication by \(3\). We do this by dividing both sides of the equation by \(3\):
\[\frac{3x}{3} = \frac{-9}{3},\]
which gives us:
\[x = -3.\]
This method of solving linear equations is foundational for balancing budgets. For instance, if you have a base budget for a project and need to account for recurring costs that are multiplied by the number of units purchased, you can represent this as a linear equation. Solving for the unknown number of units allows you to calculate exactly how many you can afford while staying within your total budget limit. If you have \(\$100\) and need to pay a \(\$10\) fixed fee plus \(\$5\) for every item, the equation \(10 + 5x = 100\) helps you determine that \(x = 18\) items can be purchased.