Question

(20) the perimeter of a rectangle is 48 inches. if the length is 4 inches more than twice the width, find the dimensions.

1. the sum of two consecutive integers add to 12 is equal to -11 find both numbers.

Explanation:

Step1: Set up variables for problem 20

Let the width of the rectangle be \(w\) inches. Then the length \(l=(2w + 4)\) inches. The formula for the perimeter of a rectangle is \(P = 2(l + w)\). Given \(P=48\) inches.
Substitute \(l\) and \(P\) into the formula: \(48=2((2w + 4)+w)\)

Step2: Simplify the equation for problem 20

First, simplify the right - hand side: \(48=2(3w + 4)\). Then \(48 = 6w+8\).
Subtract 8 from both sides: \(48−8=6w\), so \(40 = 6w\). Divide both sides by 6: \(w=\frac{40}{6}=\frac{20}{3}\approx6.67\) inches.
Then \(l=2w + 4=2\times\frac{20}{3}+4=\frac{40}{3}+4=\frac{40 + 12}{3}=\frac{52}{3}\approx17.33\) inches.

Step1: Set up variables for problem 21

Let the first consecutive integer be \(n\), then the second consecutive integer is \(n + 1\). The equation is \(n+(n + 1)+12=-11\)

Step2: Simplify the equation for problem 21

Combine like terms: \(2n+13=-11\). Subtract 13 from both sides: \(2n=-11 - 13=-24\). Divide both sides by 2: \(n=-12\). The second integer is \(n + 1=-12 + 1=-11\)

Answer:

For problem 20: The width is \(\frac{20}{3}\) inches and the length is \(\frac{52}{3}\) inches.
For problem 21: The two consecutive integers are \(-12\) and \(-11\)