# Thread: Linear Law - Unclear of solving part (b)

1. ## Linear Law - Unclear of solving part (b)

Variables x and y are related by the equation $y = \frac{ax}{x-b}$. The graph of y against $\frac{y}{x}$ has gradient 3 and y-intercept 2.

(a) Find the value of a and of b.
(b) Find the values of x for which x = y.

Attempted Solution

(a) $y = \frac{ax}{x-b}$
$yx-yb = ax$
$\frac{yx}{x} - \frac{yb}{x} = \frac{ax}{x}$
$y - \frac{yb}{x} = a$
$y = b\frac{y}{x} + a$
b=3 , a=2

Unclear of doing part B

2. Whoops.

3. $y = \frac{ax}{x-b}$ is not equals to
$y = \frac{ax}{xb}$ therefore, the x cannot be cancelled

in any case i don't really know what you are talking about, perhaps you meant this

4. Originally Posted by Punch
$y = \frac{ax}{x-b}$ is not equals to
$y = \frac{ax}{xb}$ therefore, the x cannot be cancelled
YES; BUT you originally showed ax / x + b

5. False accuse! Your previous post showed ax / x - b, lol main point is solving part (b), thanks for your effort anyway.

6. Originally Posted by Punch
Variables x and y are related by the equation $y = \frac{ax}{x-b}$. The graph of y against $\frac{y}{x}$ has gradient 3 and y-intercept 2.

(a) Find the value of a and of b.
(b) Find the values of x for which x = y.

Attempted Solution

(a) $y = \frac{ax}{x-b}$
$yx-yb = ax$
$\frac{yx}{x} - \frac{yb}{x} = \frac{ax}{x}$
$y - \frac{yb}{x} = a$
$y = b\frac{y}{x} + a$
b=3 , a=2

Unclear of doing part B
Assuming that what you have so far is correct, that is, that $y= \frac{2x}{y- 3}$, then "y= x" gives $x= \frac{2x}{x- 3}$. Multiplying on both sides of the equation gives $x(x-3)= x^2- 3x= 2x$ which is the same as $x^2- 5x= x(x- 5)= 0$.; There are two values of x that make that true.

7. Lol it was like i suddenly got it, thanks!