# Angle relationships?

1. Mar 14, 2006

### Richay

Can somebody take me through the steps of solving this problem?

http://img56.imageshack.us/img56/7987/anglerelationships13ot.jpg [Broken]

I just want to know how to do it. This is the appropriate forum right? Kind of lost here

Last edited by a moderator: May 2, 2017
2. Mar 14, 2006

### z-component

What I did is I set up a system of equations using geometry and solved simultaneously. For example, I can set some of those expressions equal to known angle measurements using theorems of geometry, i.e. vertical angles and straight lines.

3. Mar 14, 2006

### Richay

Hmmmm I see. I kinda know how to solve the problem but.. I don't know what the answer is. What I mean is how do i answer it? Like how is the solution suppose to be layed out like? I solve for x and that all right?

4. Mar 14, 2006

### z-component

Since you're given angle measures both in terms of x and y, you should solve for both. If you have a system of equations set up, it's easy to substitute one variable back in to find the other. Would you like to show us what you've done so far and go on our way to finishing this problem off?

5. Mar 18, 2006

### Richay

Yes.

20x + 5y = 110

2x + 5y = 40

x= ? y= ?

I don't understand the steps in solving this problem anymore. Knowing 5y is in both equations makes it difficult because I'm not sure how to solve for x then?

6. Mar 18, 2006

### Mulder

Knowing 5y is in both equations makes it easy. Just take both sides of one equation away from both sides of the other and get a single equation with just "x" in - (have you come across solving simultaneous equations before?)

7. Mar 20, 2006

### Richay

No, i've never been taught simultaneous equations before.
To solve for 'x' i have to go through elimination right?

8. Mar 20, 2006

### z-component

You have to take one of your equations and solve for one of the variables, then substitute that value into the second equation. You'll end up with the value for one of the two unknown variables, and with that you can substitute the known variable back into either equation to find the other. Make sense?

9. Mar 20, 2006

### Richay

It does make sense, but I understand better with an visual please?

10. Mar 20, 2006

### z-component

Say we have two equations with two common unknowns:

2y=10x+4 (1)
x+5y=15 (2)

*Note: The (1) and (2) after the equations are labels, not part of the equations themselves.

We solve for one unknown in either equation. For example, we solve for y in equation (1):

y=5x+2

Notice the equation still has two unknowns, so we can't do anything unless we substitute it into equation (2). Substituting equation (1) into equation (2) gives:

x+5(5x+2)=15

Now we have only one unknown left because we put y into terms of x from equation (1). Solving for x in the above equation, we get x=5/26.

So we have one variable left to find, so we take the x value we just calculated and substitute that back into one of the ORIGINAL equations, either (1) or (2), doesn't matter. Let's put it back into equation (1). Remember we already put equation (1) in terms of y, so it's easy to use that:

y=5x+2
y=5(5/26)+2
y=77/26

To verify we did the substitutions correctly, we make sure the system of equations still holds true.

2y=10x+4 (1)
x+5y=15 (2)

becomes

2(77/26)=10(5/26)+4 (1)
(5/26)+5(77/26)=15 (2)

Are BOTH of these equations true? Yes, they have to be. If they're not, you made an arithmetic error.

11. Mar 21, 2006

### Richay

So that's how it's done. It was confusing at first, but the visual really helped me understand it better. Thanks (I don't understand it completely yet though..)

Now, let's see if i'm doing this problem correctly?

Equations
20x+5y=110
2x+5y = 40

x=5y+20
Or
y=20x+2

(I freeze here, but i'm guessing when you got your answer that you rounded to the nearest number when you got here?)

2x+5(5x+20?)=40
Or
2x+5(20x+2)=40
x=1.5
??

The reason i stop here, is because your equations didn't have two of the same euqations like I do with "5y" So I become confused with how to subsiture x and y.

I'm still not positive how to solve for y when they're the same.
I feel like a pain in the ass, but i'm just trying to learn......!