Oops. What is ##\frac {120} H * H##Physiona said:And then it'll equal:
4H + 120/H * H = 100.
The H will cancel, leaving:
4H + 120 = 100
Oh, is it 40H/H?tnich said:Oops. What is ##\frac {120} H * H##
Sorry, I meant what is ##\frac {40} H * H##
Sorry, I messed that up. Let's go back to post #27 where you say correctly thatPhysiona said:Is it 120H/H?
What is ##\frac {40} H * H##?Physiona said:4H + 80/H + 40/H * H = 100
OK, what is ##\frac H H##? It's not a trick question.Physiona said:Yes would it equal 40H/H?
Right. So now what does your equation 4H + 80/H + 40/H * H = 100 become?Physiona said:That's 1 isn't it?
Sorry, having a rough day.
So it's 40?
OK. Now solve it for H.Physiona said:4H + 80/H + 40 = 100
Try multiplying the whole equation by H.Physiona said:The fraction bit of 4H and 80/H is confusing. Do I multiply 4H by H and divide 80?
Nope. You need to multiply each term in the equation (on both sides) by H. Also, what is ##\frac H H##?Physiona said:Will I get 4H2 + 120 = 100
Right. Can you explain to me why if ##x + y + z = c##, then ##ax + ay + az = ac##?Physiona said:4H * H = 4H2
80/H = 80H/H or 80 I presume
40 * H = 40H
100 * H = 100H
Is this right?
Right. And why does that work?Physiona said:You're multiplying by a coefficient of a for all the terms aren't you?
If you start with ##x=3## and then say if you multiply both sides by a, then why is ##ax = 3a##?Physiona said:Because they all have a common like term in them? I'm not sure..
Say you have a number. Whatever you decide to call it, it's the same number, right? So whether you call it 3 or you call it x, it's the same number. That's what it means when we say ##x=3##. If you take that number and multiply it by a, then you get the same result whether you call it ##x## or 3. So you can say ##ax = 3a##. The point is, when you have an equation, if you want both sides to stay equal, you can't multiply one side by a number (or a variable) without multiplying the other side by the same number. In the same way, if you want to maintain equality, you can't multiple one term in an equation by a number without multiplying all of the other terms in the equation on both sides by the same number.Physiona said:To make it balanced with the same coefficient. Right? As you multiply both sides by a..
Physiona said:To make it balanced with the same coefficient. Right? As you multiply both sides by a..
I'm sorry but I did do that, and you haven't gave feedback on it. (Refer to post 44)tnich said:Say you have a number. Whatever you decide to call it, it's the same number, right? So whether you call it 3 or you call it x, it's the same number. That's what it means when we say ##x=3##. If you take that number and multiply it by a, then you get the same result whether you call it ##x## or 3. So you can say ##ax = 3a##. The point is, when you have an equation, if you want both sides to stay equal, you can't multiply one side by a number (or a variable) without multiplying the other side by the same number. In the same way, if you want to maintain equality, you can't multiple one term in an equation by a number without multiplying all of the other terms in the equation on both sides by the same number.
So now, take your equation 4H + 80/H + 40 = 100 and multiply by H.
You have not written the equation, yet. Multiply both sides of 4H + 80/H + 40 = 100 by H and show me what you get.Physiona said:I'm sorry but I did do that, and you haven't gave feedback on it. (Refer to post 44)
You should get a single equation as a result.tnich said:You have not written the equation, yet. Multiply both sides of 4H + 80/H + 40 = 100 by H and show me what you get.
Great! Now solve that for H.Physiona said:I get 4H2 + 80 + 40H = 100H
Go ahead.Physiona said:Okay thank you!
So do I collect like terms from both side:
4H2 + 80 = 100H - 40H
4H2+80= 60H
Hang on, can I solve it as a quadratic?
In the form of:
4H2-60H+80=0
I think so. Now that you have H, how are you going to find L?Physiona said:I end up with two solutions of:
13.520797289396
1.4792027106039
Is this correct?