SOLUTION: The length of a rectangle is 2cm less than 7 times the width. The perimeter is 60cm. Find the width and length.
The second problem is:
The first side of a triangle is 8m shor
Question 15536: The length of a rectangle is 2cm less than 7 times the width. The perimeter is 60cm. Find the width and length.
The second problem is:
The first side of a triangle is 8m shorter than the second side. The third side is 4times as long as the first side. The perimeter is 26m. Find the length of each side.
Thank you so much for your help. I am in desperate need with college algebra.
You can put this solution on YOUR website! The length of a rectangle is 2cm less than 7 times the width. The perimeter is 60cm. Find the width and length.
let length be =L
let width be = W
7 times width =7W
2 cm. less than above = 7W-2 = L
perimeter =2L+2W = 60
2*(7W-2)+2W = 60
14W+2W=60+2*2=64
16W = 64
W =64/16 = 4
L=7*4-2 = 26
SECOND PROBLEM
The first side of a triangle is 8m shorter than the second side. The third side is 4times as long as the first side. The perimeter is 26m. Find the length of each side.
Let the first side =x
let the second side =y
let the third side =z
8m shorte than second side = y-8 = x OR y = x+8
4 times the first side =4*x= z
perimeter =x+y+z = 26
x+(x+8)+4*x = 26
x+x+4x=26-8 = 18
6x = 18
x=18/6 = 3
y= 3+8 = 11
z=4*3 = 12
You can put this solution on YOUR website! OK Problem 1)
We have to derive an equation:Perimeter of rectangle is
2 x width+2 x length
If we have width = x
The length will be 7x-2
So our equation will be:
2x+2(7x-2)=60 Multiply expression in bracket by 2
2x+14x-4=60 Rearrange to get:
16x=64 Divide both sides by 16
x=4cm which is the width
Length is (7x4)-2=26
Problem 2)
Again need to derive an equation:
Let second side=x
so first side=x-8 and
third=4(x-8).So our equation will be:
x-8+x+4(x-8)=26 get rid of brackets by multiplying by the 4
x-8+x+4x-32=26
6x-40=26 Add 40 to both sides
6x=66
x=11 So 1st side=3m,2nd side=11m and 3rd=12m
hope this helps
Pete
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