i need help with this math problem...?

Question

the perimeter of a triangle is 45 meters. the 2nd side is 3 times the 1st side. the 1st side is 5 more than the 3rd side. find the length of each side.

i don't understnad how your supposed to set up this equation. its a bit confusing for me. thanks for ur help!

Answer 1

x = side 3
5+x - side 1
3(5+x) = side 2

45=x + 5+x+3(5+x)
45 = 5+2x+15+3x
45=20+5x
25=5x
5=x

Answer 2

Best way to do these problems is to start with writing what you know.
triangle has 3 side A B and C
Perimeter is 45 so A+B+C=45
2nd side is 3 times 1st side => B=3A
1st side is 5 more than 3rd side => A= C+5

now you have 3 equations and 3 variables. I would solve for A first. we know B= 3A and C= A-5. so the first equation becomes:
A+3A+A-5=45
5A=50
A=10
B=30
C=5

Answer 3

Let a, b, and c denote the lengths of the first, second and third side.

1) a + b + c = 45

2) b = 3a

3) a - 5 = c

replace b and c inn equation 1) by their equivalents in a
expressed in equations 2) and 3).

a + 3a + a - 5 = 45

5a = 50

a = 10

then

b = 3*10 = 30

c = 10 -5 = 5

Answer 4

This problem involves simultaneous equations.
We'll use x,y and z to represint the 3 sides.
Since the perimeter is the sum of all 3 sides, x+y+z=45
the above is eq1
the second side is thrice the first and can be represented as
y=3x
thats eq 2
then the first side is 5 more than the third so
x=z+5
thats eq 3
So we play around with the equations.
substituting eq3 in eq2 we have
y=3x and x=z+5
so y=3(z+5)
y=3z+15
thats eq4
now putting eq3 and eq4 into eq1 we have
(z+5)+(3z+15)+z=45
z+3z+z+5+15=45
5z+20=45
5z=25
z=5
thats eq5
putting eq5 in eq3
x=5+5=10
x=10 is eq6
then putting eq 5 in eq4
y=3(5)+15=15+15=30
the three sides are 10, 15 and 30 metres.
lol.

Answer 5

take 3rd side as x
then 1st side will be (5+x)
2nd side will be 3 times (5+x)
now write this in form of equation
x+5+x+3(5+x)=45
2x+5+15+3x=45
5x+20=45
5x=45-20
5x=25
and therefore x=5

x is 3rd side and thats y its 5
2nd side is 3(x+5) which is equal to 3(5+5)=30
and 1st side is x+5 which is 10
add all 30+10+5=45

Answer 6

1st side A, 2nd B, 3rd C.

P = A + B + C
B = 3A
A = 5 + C,
so B = 3(5 + C)

P = (5 + C) + 3(5 + C) + C = 45
5 + C + 15 + 3C + C = 45
5C = 25
C = 5
A = 10
B = 3(10) = 30

but that's not a triangle, since A+C < B

Answer 7

Let´call x the length of the third side
The first side is x+5 and the second side is 3x+15
so P=45=x+x+5+3x+15
45=5x+20 and x = 5
2nd side =30 and first side 10

Answer 8

naming yhe sides A=1st, B=2nd and C=3rd.
you make equations according to the data:
A+B+C=45
B=3A
A=C+5
solving this system of equations
A+(3A)+(A-5)=45
4xA=45-5
A=40/4=10
B=3x10=30
C=10-5=5

Answer 9

a+b+c=45

b=3a
a=c+5, or c=a-5

Now, use substitution.

a + (3a) + (a-5) = 45

5a - 5 = 45 [Gather like terms]
5a = 50
a = 10

If a=10, b=3a = 30, and c=10-5 = 5

Addendum: As Philo points out, this triangle does not exist in euclidean 2D space.