Algrebra Question!! Fastest correct answer gets points!!!1?

Question

The length of one of the equal legs of an isosceles triangle is 3 inches more than 2 times the length of the base. If the perimeter is 31 in, find the length of one of the equal legs.

I am helping my son with his home work and need help with a few questions. Fastest correct answer wins 10 points.

Answer 1

Let base = x
31 = 5x +6
x = 5
base = 5inches
equal sides are 13 inches each

Answer 2

If you represent (algebraically) the equal leg of the triangle it should be written as 2x+3 (two times the length of the base which is x plus 3 inches). Since there are two sides that equal this and to find the perimeter (a known sum of 31) you just need to add all of the sides together to get your base equation:
2x+3 (side 1) +2x+3 (side 2)+ x (base)= 31 or 2x+3+2x+3+x=31
when added together you get : 5x+6=31
5x=25
x=5

Plugging the value for x back into the equation for the length of the side: 2x+3
2(5)+3= 13 inches

Answer 3

If x the length of one of the equal legs in inches and y the length of the base then:
x+x+y=perimeter=31
We also know that the length of one of the equal legs of an isosceles triangle is 3 inches more than 2 times the length of the base:
x=3+2y
Then:
2x+y=31
x=3+2y
=>2(3+2y)=31
6+4y=31
4y=31-4
y=27/4
y=6.75inches
and x=3+2y => x=3+2*6.75 => x=16.5inches

Answer 4

Let the base be x:

Therefore your equation is:

x + 2(2x +3) = 31

> x + 4x + 6 = 31

> 5x = 25

> x = 5

Therefore equal leg = 13

Answer 5

The length of each leg = 13 inches.

Answer 6

x = base
leg = 2x+3
So perimeter = x+2x+3+2x+3 = 31
5x+6 = 31
5x = 25
x=5
2x+3 = 13 = length of leg

Answer 7

5" base, 13" legs