Applications of Linear Equations
Example 108.
The perimeter of a rectangle is \(44\). The length is \(5\) less than double the width. Find the dimensions.
| Length \(x\) | We will make the length \(x\) |
| Width \(2x − 5\) | Width is five less than two times the length |
| \(P =2L + 2W\) | The formula for perimeter of a rectangle |
| \((44) =2(x)+ 2(2x − 5)\) | Replace \(P\), \(L\), and \(W\) with labeled values |
| \(44 = 2x +4x − 10\) | Distribute through parenthesis |
| \( 44 =6x − 10\) | Combine like terms \(2x +4x\) |
| \(\underline {+ 10 + 10}\) | Add \(10\) to both sides |
| \( \underline {54 =6x}\) | The variable is multiplied by \(6\) |
| \(6 \quad \quad 6\) | Divide both sides by \(6\) |
| \(9= x\) | Our solution for \(x\) |
| Length \(9\) | Replace \(x\) with \(9\) in the original list of sides |
| Width \(2(9) − 5 = 13\) | The dimensions of the rectangle are \(9\) by \(13\). |
We have seen that it is important to start by clearly labeling the variables in a short list before we begin to solve the problem. This is important in all word problems involving variables, not just consecutive numbers or geometry problems. This is shown in the following example.