Question

the perimeter of a rectangle is 56m. The length is 1m more than twice the width. Find the dimensions.

L=
W=

Answer 1

Hi there! Let's solve a linear system! Call the length L and the width W. First, can you write an equation that relates L and W to the perimeter?

Answer 2

2L+2W=56

Answer 3

Yep, you got it. Now, can you write another equation, also using L and W, that expresses " The length is 1m more than twice the width"?

Answer 4

L=2W+1

Answer 5

Perfect!! Now solve the system using the substitution method. You know L=2W+1, so substitute "2W+1" in every place L appears in the first equation.

Answer 6

2(2W+1)+2W=56

Answer 7

Perfect. Now expand that first part out by distributing the 2. Then collect like terms.

Answer 8

4W+2+2W=56

Answer 9

right on. Now collect all the W terms on the left and the constant terms on the right.

Answer 10

2W=56

Answer 11

Oops, not quite. You had 4W + 2 + 2W = 56, that was right. What do you get when you collect the W terms together? i.e. 4W + 2W?

Answer 12

4W=56

Answer 13

Let me write it again in a way that makes it look a little easier:
4W + 2 + 2W = 56
is the same as
4W + 2W + 2 = 56

Can you combine the 4W and 2W together?

Answer 14

Pretend W is actually the word "marble". Add 4 marbles and 2 marbles.

Answer 15

ok now you just lost me

Answer 16

You understand how we got to 4W + 2W + 2 = 56, right?

Answer 17

yes

Answer 18

W=9

Answer 19

OK, so now we combine the "W" terms together -- that is, the terms that have a W in them.

Answer 20

Oops, yep, you skipped ahead - that's great! w = 9. Now substitute w=9 into either of the two equations you started with in order to find L.

Answer 21

L=2(9)+1

Answer 22

Now just solve for L.

Answer 23

L=19

Answer 24

you got it. you have the solution!