Question

1. (6 pts) the length of a rectangle is 5 less than three - times its width. if the perimeter of the rectangle is 30, find the length and width of the rectangle.

length = ________
width = ________

1. (3 pts) find three consecutive odd integers with a sum of 123.
2. (3 pts) solve the equation.

7(p + 3)+9 = 5(p - 2)-3p

1. (3 pts) what is the solution to the inequality shown?

-2(x + 2)<8 - x

Explanation:

1.

Step1: Set up equations

Let the width of the rectangle be $w$ and the length be $l$. Given $l = 3w - 5$, and the perimeter $P=2(l + w)=240$. Substitute $l$ into the perimeter - formula: $2((3w - 5)+w)=240$.

Step2: Simplify the equation

First, simplify the left - hand side of the equation: $2(3w - 5+w)=2(4w - 5)=8w-10$. So, $8w - 10 = 240$.

Step3: Solve for $w$

Add 10 to both sides: $8w=240 + 10=250$. Then, divide both sides by 8: $w=\frac{250}{8}=31.25$.

Step4: Solve for $l$

Substitute $w = 31.25$ into $l = 3w - 5$: $l=3\times31.25-5=93.75 - 5=88.75$.

Step1: Let the first odd integer be $x$

The next consecutive odd integers are $x + 2$ and $x+4$. The sum of these three integers is $x+(x + 2)+(x + 4)=123$.

Step2: Simplify the left - hand side

Combine like terms: $3x+6 = 123$.

Step3: Solve for $x$

Subtract 6 from both sides: $3x=123 - 6 = 117$. Divide both sides by 3: $x = 39$.

Step4: Find the other two odd integers

$x+2=41$ and $x + 4=43$.

Step1: Expand both sides

Expand $7(p + 3)+9$ to get $7p+21 + 9=7p+30$. Expand $5(p - 2)-3p$ to get $5p-10-3p=2p-10$. So, the equation becomes $7p+30=2p - 10$.

Step2: Move the terms with $p$ to one side

Subtract $2p$ from both sides: $7p-2p+30=2p-2p - 10$, which simplifies to $5p+30=-10$.

Step3: Solve for $p$

Subtract 30 from both sides: $5p=-10 - 30=-40$. Divide both sides by 5: $p=-8$.

Answer:

Length = $88.75$, Width = $31.25$

2.