Help with Math

Question

The length of a rectangle is 5 feet less than three times its width, and the area of the rectangle is 50 ft^2. Find the dimensions of the rectangle

Answer 1

|w| improves the precision of the answer.

Answer 2

The good news is you already know how to do this! I'm sure you know that the area of any rectangle is length*width, so 
A=l*w

 

You know that the area is 50 ft^2, so 50=l*w

 

For this proble, you know that the length is "5 less than 3 times the width", or l=3w-5. Let's plug that in to 50=l*w, so we only have 1 variable to deal with. 

 

50=l*w

50=(3w-5)*w        //Plug in 

50=3w^2-5w        //Distribute

0=3w^2-5w-50     //Subtract the 50 over

0=(3w+10)(w-5)   //Factor to solve the quadratic

w=-10/5 or w=5    //Solve for w

 

Now, the width could either be -10/5 or 5. But having a negative width doesn't make any sense, so the width must be 5. Now just plug in 5 for w and solve for l ! 

 

50=l*w

50=l*5    //Plug in 

10=l       //Divide by 5 to solve for l

 

So, the length is 10. Now you know the answer: The dimensions are 5 by 10 feet. (You know it's feet because the area was feet^2)

Answer 3

w=x then l=3x-5

 

and x( 3x-5)=50

3x^2-5x-50=0 factoring gives

(x-5)(3x+10)=0

 

x=5 since x can not be negative so width=5 and length= 3(5)-5 gives length=10

Answer 4

Area =  length*width

Area = (3*width – 5)*width           (length is 5 feet less than 3*width)
Area =  (3W – 5)*W = 50       (let W = width)
         3W^2 – 5*W -50 = 0       (distribute W; subtract 50 from both sides)
         W=5   or   W=-3.3333      (either factor or use quadratic formula; disregard negative result)

Remember:  L = 3W – 5 =  15- 5 = 10

Check (very important):
   Is L = 3W – 5   ?
       10 = 15 – 5 ?   yes.
   Is area 50 ft^2  ?
        10 * 5 = 50     yes.