*The length of a rectangle is 5m more than three times the width. The area is 232m2. Find the length and width of the rectangle.*

**Step 1** - What are the unknowns you're asked to find?

*Find the length and width of the rectangle.*

Let L= the length of the rectangle

Let W = its width

**Step 2** - We have 2 unknowns (L and W), so we need two equations that relate the unknowns

Equation 1. *the length of a rectangle is 5m more than three times the width*: L = 5 + 3W

Equation 2. *The area is 232m2*: Area = L*W = 232

**Step 3** Substitution

Let's substitute 5+3W (from equation 1) in place of L in equation 2:

**L***W = 232 [Equation 2]

(**5+3W**)*W = 232 [Substituted 5+3W from equation 1 in place of L]

5W + 3W^{2} = 232 [We now have an equation with only variable, W, to solve for]

3W^{2} + 5W - 232 = 0 [A quadratic equation!]

**Step 4** - Solve for W. Use the quadratic formula.

W = -(b/2a) ± (1/2a)√(b^{2}-4ac) where a=3, b=5, c= -232 from our quadratic equation

W = -(5/6) ± (1/6) √(5^{2}-(4)(3)(-232))

W = -(5/6) ± (1/6) √(2809)

W = -(5/6) ± (53/6)

W = 8, -58/6

We can't have a negative width (-58/6), so **W = 8 meters**

**Step 5** - Solve for L

L = 5 + 3W [Equation 1]

L = 5 + 3(8) [Put in W=8]

L = 5 + 24

**L = 29 meters**

**Step 6** Check

L*W = 232 [Equation 2]

(29)(8) = 232 [Put in L=29, W=8]

232=232 CHECK!