Solve using a system of two equations in two unknowns.
the length of a rectangle is 9 inches more than twice its
width. if the perimeter of the rectangle is 48 inches, find
its length and width.

Hello!

To solve this, first write two equations. We are given two facts about the situation, so we can write the equations accordingly.

Say the length of the rectangle is l, and the width is w.

The length of a rectangle is 9 inches more than twice its width: 2w + 9 = l, as you're adding 9 to two times the width.

The perimeter of the rectangle is 48 inches: The equation for perimeter is 2l + 2w, so we can just use that in this case to make the equation - 2l + 2w = 48

Now, set up the system of equations.

Now, we can already use substitution to solve. We get from one of the equations that l = 2w + 9, so we can substitute 2w + 9 for l in the other equation, and then solve for w.

2l + 2w = 48

2 (2w + 9) + 2w = 48

4w + 18 + 2w = 48

6w = 30

w = 5

We know one of our variables now. Now, all that's left to do is substitute 5 for w in one of the original equations to solve for l.

2w + 9 = l

2 (5) + 9 = l

10 + 9 = l

19 = l

Therefore, we now have our dimensions. The length of the rectangle is 19 inches, and the width is 5.

Hope this helps!