1. Simple solving for X..

Its been a while since i took a math course, and i completely forgot how to solve for X in this equation.... y=4x-1/2x+3. I looked it up and found this from yahoo answers...

1. (x) = (4x-1) / (2x+3)
2 y = (4x-1) / (2x+3)
3. y(2x+3) = 4x-1
4. 2xy+3y = 4x-1
5. 2xy - 4x = -3y-1
6. 2x(y-2) = -3y-1
7. x =(-3y-1)/[2(x-2)]
8. f^-1(y) = (-3y-1)/[2(x-2)]

...my question is in step 5 above what did he do to get 2xy-4x from 2xy-3y, and also the procedure from steps 5 to 6 and 6 to 7.

Thanks.

2. In 4 -> 5, he subtracted 3y from both sides. Then he subtracted 4x from both sides. Notice that the other side changed as well. In 5 -> 6, he factored by grouping. If you still don't get it after looking at the link I can explain it to you but I think it goes over the topic pretty well. In 6 -> 7, he divided both sides by (y-2) to get 2x = (-3y-1)/(y-2) and then divided both sides by 2 to get just x on the left side. The author made a slight mistake when he wrote [2(x-2)] when is should be [2(y-2)]. Were you trying to solve for x or find the inverse function?

3. Originally Posted by evankimia
Its been a while since i took a math course, and i completely forgot how to solve for X in this equation.... y=4x-1/2x+3. I looked it up and found this from yahoo answers...

1. (x) = (4x-1) / (2x+3)
2 y = (4x-1) / (2x+3)
3. y(2x+3) = 4x-1
4. 2xy+3y = 4x-1
5. 2xy - 4x = -3y-1
6. 2x(y-2) = -3y-1
7. x =(-3y-1)/[2(x-2)]
8. f^-1(y) = (-3y-1)/[2(x-2)]

...my question is in step 5 above what did he do to get 2xy-4x from 2xy-3y, and also the procedure from steps 5 to 6 and 6 to 7.

Thanks.
4. $2xy+3y = 4x-1$

5. $2xy - 4x = -3y-1$

6. $2x(y-2) = -3y-1$

7. $x = \frac{-3y-1}{2(x-2)}$

4-5: He collected the x terms on one side and the non x terms on the other side in order to factorise in step 6. This entailed taking $4x$ from both sides and taking $3y$ from both sides.

5-6: He factored out a common term on the left which is $2x$ since both $2$ and $x$ are factors.

6-7: He should have divided through by $2(y-2)$ which would give $x = -\frac{3y+1}{2(y-2)}$. That is what makes x the subject.

Step 8 and parts of step 7 introduce the idea of finding the inverse function, is this what you wanted to do?

4. Yeah This Is For Solving For An Inverse Equation

5. Originally Posted by evankimia
Yeah This Is For Solving For An Inverse Equation
For inverse functions we set the equation solved for x and replace that x with $f^{-1}(x)$ and y with x.

$
f^{-1}(x) = -\frac{3x+1}{2(x-2)}
$

The domain is $x \in \mathbb{R} \: \: , \: \: x \neq 2$

The range is anything that satisfies the domain in $f(x)$. $x \in \mathbb{R} \: \: , \: \: x \neq -\frac{3}{2}$

edit: see the attached graph of $f(x)$ (blue), $g(x) = f^{-1}(x)$ (green) and $h(x) = x$ (red). Since g(x) and f(x) are reflections in the line y=x they are inverse

6. Thank you so much...learn something new every day. Just curious, what program did you use to make that function? -Evan

7. I use KMPlot which is part of the KDE desktop usually associated with GNU/Linux

It's licened under the GPL but the homepage is down at the moment :|

8. I will check their homepage in a bit to see if its back up. In the meantime... heres another example.....

Find the inverse of f(x)= (x+4)/(x-3) Use this to determine the domain and range of f and f-1.

1. f(x)= (x+4)/(x-3) = y=(x+4)/(x-3)
2. y(x-3)=x+4
3. 1yx-3y=x+4
how would i get the x out of there? just divide it out normally on both sides so it would be -2y=x+4/x?

9. Originally Posted by evankimia
I will check their homepage in a bit to see if its back up. In the meantime... heres another example.....

Find the inverse of f(x)= (x+4)/(x-3) Use this to determine the domain and range of f and f-1.

1. f(x)= (x+4)/(x-3) = y=(x+4)/(x-3)
2. y(x-3)=x+4
3. 1yx-3y=x+4
how would i get the x out of there? just divide it out normally on both sides so it would be -2y=x+4/x?
yx - 3y = x + 4

=> yx - x = 4 + 3y => x(y - 1) = 4 + 3y => x = (4 + 3y)/(y - 1).

So the inverse function is $f^{-1} = \frac{4 + 3x}{x - 1}$.