math problem...?

Question

The 1st side of a triangle is 3 less than twice the 2nd side. The 3rd side of the triangle is 3 times the 1st side. The perimeter of the triangle is 78 yards. Find the length of each side of the triangle.

Answer 1

1st leg = 2x -3
2nd leg = x
3rd leg = 3(2x-3)

(2x-3) + x + [3(2x-3)]=78

2x - 3 + x + 6x -9 = 78
9x -12 = 78
9x = 90
x = 10

leg 1 = 17
leg 2 = 10
leg 3 = 51

Answer 2

Ok you need to first figure out each side in terms of the given sides.

so Side 1 = 2*side2-3
Side 2 = side2
Side 3 = 3*side1

so the equation becomes

(2*side2 - 3) + (side2) + (3*side1) = 78

since there are two variables we need to change the one different variable to show in terms of the common variable.

so 3*side1 = 3(2*side2 - 3)

the equation then becomes

(2*side2 - 3) + (side2) + 3(2*side2 - 3) = 78

then combine all factors to get

9side2 = 90

side2 = 10

use this to get the other sides by substituition into the equations above for each side

so side1 = 17
side 3 = 51

Answer 3

let x be the second side
the first side is 3 less than twice the 2nd. 2x - 3
third side: 3(2x-3)

x + 2x - 3 + 3(2x-3) = 78
3x - 3 + 6x - 9 = 78
9x - 12 = 78
9x = 90
x= 10

first side: 17
second side: 10
third side: 51

Answer 4

1st leg =13
2nd leg = 26
3rd leg = 39

am i close..? i'll show my work. i really don't know math too well. 78 is what it all equals to.

so 78/2= 39 (3rd leg)
39/3= 13 (1st leg)
13x2= 26 (2nd leg)