Question

this question has two parts. first, answer part a. then, answer part b. part a rectangle wxyz has a length that is 5 more than three times its width. a. draw and label a figure for rectangle wxyz on a separate sheet of paper. b. write an algebraic expression for the perimeter of the rectangle. an expression for the perimeter, where x is the width, would be ___. part b c. find the width if the perimeter is 58 millimeters. explain how you can check that your answer is correct. solving 58 = 8x+ for x, the width is found to be mm. to check that this answer is correct, use the value of the width to

Explanation:

Step1: Define length in terms of width

Let the width of the rectangle be $x$. The length is 5 more than three - times its width, so the length $l = 3x + 5$.

Step2: Recall perimeter formula

The perimeter formula of a rectangle is $P=2(l + w)$. Substitute $l = 3x + 5$ and $w=x$ into the formula:
$P = 2((3x + 5)+x)$.

Step3: Simplify the expression

First, combine like - terms inside the parentheses: $(3x + 5)+x=4x + 5$. Then, multiply by 2: $P = 2(4x + 5)=8x+10$.

Step4: Solve for width when perimeter is 58

Set $P = 58$ in the equation $P = 8x+10$. So, $58=8x + 10$.
Subtract 10 from both sides: $58−10=8x+10−10$, which gives $48 = 8x$.
Divide both sides by 8: $\frac{48}{8}=x$, so $x = 6$.

Step5: Check the answer

Substitute $x = 6$ into the length formula $l = 3x+5$, we get $l=3\times6 + 5=18 + 5=23$.
Then use the perimeter formula $P = 2(l + w)$. Substitute $l = 23$ and $w = 6$: $P=2(23 + 6)=2\times29 = 58$.

Answer:

b. $8x + 10$
c. Solving $58=8x + 10$ for $x$, the width is found to be $6$ mm. To check that this answer is correct, use the value of the width to find the length ($l=3\times6 + 5 = 23$), then use the perimeter formula $P = 2(l + w)$ and see if $P=2(23 + 6)=58$.