Question

May I have some help here please, I got a bit confused.
There are two triangles one has a perimeter of 14 (sides are y, y, and base x) and the other has a perimeter of 21(sides are 5/4y , 5/4y and base 3x). What are x and y?

Answer 1

Write two perimeter equations and then you will have two equations and two unknowns that you can solve for.
The first perimeter equation will be x+2y=14.

Answer 2

I got that Stacy, thanks for your help! I know the second triangle's perimeter is : 5/4y +5/4y +3x =21
Could you help me some more here?

Answer 3

2y + x = 14
5/2y + 3x = 21

2y + x = 14
-2(5/2y + 3x = 21)

6(2y + x = 14)
-5y - 6x = -42

12y + 6x = 84
-5y - 6x = -42

7y = 42
y = 6
Plug it into an equation.
2(6) + x = 14

12 + x = 14
x = 2

Therefore:
x = 2
y = 6
(2, 6)

Answer 4

@Calcmathlete, thanks a lot for your explanation, but how did you get 5/2, shouldn't it be 5/4?

Answer 5

I simplified. 5/4 + 5/4 = 10/4 = 5/2

Answer 6

Get it?

Answer 7

ok, I tried that and erased it thinking I was wrong. Thanks man! Yes but not the negative.

Answer 8

Oh ok.

Answer 9

You're welcome. :)

Answer 10

Why do you have -5/2 though?

Answer 11

Where? It says -2(5/2y + 3x = 21). I'm multiplying the equation by -2.

Answer 12

oh ok! You are a live saver man! I appreciate you taking the time to explain it to me. :)

Answer 13

You're welcome and no problem. I'm here to assist. :)