MATH SOLVE

2 months ago

Q:
# Use the three steps to solve the problem. The length of a rectangle is 3 times the width, and the perimeter is 22. Find the dimensions of the rectangle. {Width is 8.25 a0, Length is 2.75 a1}

Accepted Solution

A:

If you recall, the perimeter of a rectangle is equal to 2 times the length, plus 2 times the width.

P=2l + 2w

It would be hard to solve for two variables in this problem, but you are given that the length is equal to 3 times the width.

l=3w

We can substitute this into the original equation so that we’re only looking for “w”.

P=2l + 2w

P=2(3w) + 2w substitution

P=6w + 2w

P=8w simplification

We’re also given that P=22, so we can substitute that into the final equation and solve for “w”.

P=8w

22=8w substitution

2.75=w solving for w

We now know the width, but we’re not done! Use the fact that we defined length in terms of width to calculate the value of the length.

w = 2.75

l=3w what we defined length

l=3(2.75) substitution

l=8.25

So, width is equal to 2.75, and length is equal to 8.25.

P=2l + 2w

It would be hard to solve for two variables in this problem, but you are given that the length is equal to 3 times the width.

l=3w

We can substitute this into the original equation so that we’re only looking for “w”.

P=2l + 2w

P=2(3w) + 2w substitution

P=6w + 2w

P=8w simplification

We’re also given that P=22, so we can substitute that into the final equation and solve for “w”.

P=8w

22=8w substitution

2.75=w solving for w

We now know the width, but we’re not done! Use the fact that we defined length in terms of width to calculate the value of the length.

w = 2.75

l=3w what we defined length

l=3(2.75) substitution

l=8.25

So, width is equal to 2.75, and length is equal to 8.25.