Word problem

[GypsySkies]

I have this word problem:

The perimiter of a right triangle is 30 ft. One leg is 2 ft longer than the twice the shortest leg. The hypotenuse is 2 ft less than 3 times the shortest leg. Find the lengths of the sides of this triangle.

So I set up the problem:
a+b+c=30
a=2z+2
b=3z-2

But when I tried to slove it, it didn't turn out. I went over and over how I had worked it and couldn't find an error so I asumed that I had set it up wrong.

 

[mmm4444bot]

Hi Gypsy Skies:

Your equation a + b + c = 30 is a good start.

Your other two equations show that you've got the right idea, yet we do not need to introduce a fourth variable (z) because a triangle only has three sides (heh, heh). In other words, the exercise gives relational information between three sides, so once we've assigned variables a, b, and c to represent these three lengths, all of the relationships can be expressed in terms of equations containing only the symbols a, b, and c.

First, we can decide to let the symbol c represent the length of the hypotenuse (that's standard), and then we decide which symbol represents the length of the short leg (a or b).

One way:

c = hypotenuse
b = long leg
a = short leg

Now we can write equations from the two given relationships involving the "shortest" leg using the variable a.

I'll do one: "The hypotenuse is 2 ft less than 3 times the shortest leg".

c = 3a - 2

You do the other, and there will be a system of three equations in three variables for you to solve. If you'd like more help with this exercise, please show whatever work you're able to complete or explain why you're stuck.

Cheers,

~ Mark

 

[GypsySkies]

The z was a mistake. Originally I had x, y, and z. I actually meant c. So really, everything is the same except no z. I think all you did was switch around the variables. I'll still end up with:

a+b+c=30
b=2a+2
c=3a-2

Same thing really. Is that right?

 

[mmm4444bot]

a + b + c = 30
b = 2a + 2
c = 3a - 2

This system is correct for the way that I listed above.

Its solution is {a = 5, b = 12, c = 13}.

We need to see what you've done, if you would like more help. I understand that you cannot locate your mistake(s), but, until they see your work, neither can anybody else.

 

[Denis]

The z was a mistake. Originally I had x, y, and z. I actually meant c. So really, everything is the same except no z. I think all you did was switch around the variables. I'll still end up with:
a+b+c=30 [1]
b=2a+2 [2]
c=3a-2 [3]
Same thing really. Is that right?

So what's your problem?
3 unknowns, 3 equations: just solve; substitute [2] and [3] in [1].