Question

the width of a rectangle 8in. greater than the length. the perimeter is 102in. find the dimensions of the rectangle.
linear equations

Answer 1

@mathstudent55 @misty1212

Answer 2

HI!!

Answer 3

HI. :)

Answer 4

lets add up to find the perimeter

Answer 5

\[x+x+8+x+x+8=4x+16\] right?

Answer 6

right

Answer 7

and you know that is (102\) so your job is to solve
\[4x+16=102\] for \(x\)

Answer 8

can you do that?

Answer 9

yes

Answer 10

ok let me know what you get

Answer 11

kk one sec

Answer 12

-26

Answer 13

@misty1212

Answer 14

oh no

Answer 15

\[4x+16=102\] first subtract \(16\) from both sides

Answer 16

-86

Answer 17

@misty1212

Answer 18

You did 16 - 102.
You need to do 102 - 16

Answer 19

ohhhhh

Answer 20

86-

Answer 21

@misty1212 @mathstudent55

Answer 22

Good,

Now you have

4x = 86

Answer 23

21.5

Answer 24

lol what does \(86-\) mean ??

Answer 25

Correct.

Answer 26

thanks. :)

Answer 27

That is x.
Now you need x + 8 = 21.5 + 8
What is 21.5 + 8 ?

Answer 28

one sec

Answer 29

29.5

Answer 30

Correct.

Answer 31

thanks.