ok i know how to this but i donno if i did right

3y-x=3 and 4x+3y-18=0

-solve by graphing

-solve using algebra

-and formal check

Results 1 to 4 of 4

- Feb 20th 2006, 04:15 PM #1

- Joined
- Dec 2005
- Posts
- 21

- Feb 20th 2006, 08:56 PM #2Originally Posted by
**usm_67**

to graphing: I've attached a drawing. The coordinates of the intersection is the solution you're looking for.

to using algebra: There are at least 4 to 5 different method to get a solution:

$\displaystyle 3y-x=3\ \Longleftrightarrow \ y= \frac{1}{3} x+1$

$\displaystyle 4x+3y-18=0\ \Longleftrightarrow \ y= - \frac{4}{3} x+6$

Both values of y must be equal:

$\displaystyle \frac{1}{3} x+1 = - \frac{4}{3} x+6$

$\displaystyle \frac{5}{3} x =5$ and that means: x = 3

Put x = 3 in one the equations: $\displaystyle y= \frac{1}{3} 3+1=2$

So you get the intersection at (3,2)

to formal check: Put x=3 and y=2 in your original equations:

$\displaystyle 3y-x=3\ \Longrightarrow \ 3 \cdot 2-3=3$ you'll get 3=3 and that's true.

$\displaystyle 4x+3y-18=0\ \Longrightarrow \ 4 \cdot 2 + 3 \cdot 3 -18=0$ You'll get: 12 + 6 -18=0 and that's true.

Greetings

EB

- Feb 21st 2006, 04:41 PM #3

- Joined
- Dec 2005
- Posts
- 21

- Feb 22nd 2006, 10:31 AM #4Originally Posted by
**usm_67**

here I'm again. Of course you're right. But as I mentioned before there are many different methods. So if you have to use elimination and substitution, you can do it like this:

$\displaystyle 3y-x=3\ \Longleftrightarrow \ y= \frac{1}{3} x+1$

Put this value of y into the 2nd equation to eliminate the variable y:

$\displaystyle 4x+3\cdot \left( \frac{1}{3} x+1 \right)-18=0\ \Longrightarrow \ 4x+x+3-18=0$

Simplify this equation and you'll get: 5x - 15 = 0. That means: 5x = 15 and so x = 3

Put this value into the 1rst equation and you'll get y = 2

And that's the result you've got before.

It doesn't matter which method you use, all of them come to the same solution.

Greetings

EB