# 2 Methods

• Sep 15th 2011, 12:03 PM
vaironxxrd
2 Methods
Hello forum Vaironl here, I have to solve some equations using two methods.
1 Substitute x and then solve for y.
2 Solve for y and then substitute x.

Here is my first problem I can do the first method but I don't understand exactly how to solve the second method.
$\4x+& 9y& =30& ; x=3$
For the first method I got

12+9y=30 =
9y=18 =
y= 2
• Sep 15th 2011, 12:05 PM
e^(i*pi)
Re: 2 Methods
Quote:

Originally Posted by vaironxxrd
Hello forum Vaironl here, I have to solve some equations using two methods.
1 Substitute x and then solve for y.
2 Solve for y and then substitute x.

Here is my first problem I can do the first method but I don't understand exactly how to solve the second method.
$\4x& 9y& =30& ; x=3$
For the first method I got

12+9y=30 =
9y=18 =
y= 2

$4x9y = 36xy$

Is the equation meant to read $4x+9y =30\ ,\ x=3$ per chance?
• Sep 15th 2011, 12:12 PM
vaironxxrd
Re: 2 Methods
Quote:

Originally Posted by e^(i*pi)
$4x9y = 36xy$

Is the equation meant to read $4x+9y =30\ ,\ x=3$ per chance?

Yes sorry for the incorrect symbols
• Sep 15th 2011, 12:13 PM
e^(i*pi)
Re: 2 Methods
Quote:

Originally Posted by vaironxxrd
Yes sorry for the incorrect symbols

In that case solve the equation again using the value of y=2 that you just obtained to get x (x should be 3)
• Sep 15th 2011, 12:18 PM
vaironxxrd
Re: 2 Methods
Quote:

Originally Posted by e^(i*pi)
In that case solve the equation again using the value of y=2 that you just obtained to get x (x should be 3)

Wow. I can't believe it was that simple.

$\4*x&+ 9y& = 30$
$\4x& + 9*2&=30$
$\4x& + 18& =30$
$\4x& =12$
$x& = 3$
• Sep 16th 2011, 11:47 AM
vaironxxrd
Re: 2 Methods
Quote:

Originally Posted by e^(i*pi)
In that case solve the equation again using the value of y=2 that you just obtained to get x (x should be 3)

Sorry to bother you e^(i*pi) ,
but after spending sometime yesterday doing my homework I notice that you would solve for x first without entering y and then finding the value of y.
Do you know how could I do that by any chance?
• Sep 16th 2011, 11:58 AM
e^(i*pi)
Re: 2 Methods
Quote:

Originally Posted by vaironxxrd
Sorry to bother you e^(i*pi) ,
but after spending sometime yesterday doing my homework I notice that you would solve for x first without entering y and then finding the value of y.
Do you know how could I do that by any chance?

Normally you would but you cannot in this case because you're told the value of x and since you only have one equation with two unknowns you can't solve simultaneously either.