Question

How to solve question 2(a,b,c)?​

Answer 1

Answer:

a) For h(2)=5 then k=4

b) For h(3)=k then k=3

c) For h(k)=k then k=3

Step-by-step explanation:

Given that the function h is defined by

$h(x)=\frac{kx-3}{x-1}$ where $x\neq1$

a) To find for h(2)=5:

$h(x)=\frac{kx-3}{x-1}$ where $x\neq1$

that is put x=2 in the above function

$h(2)=\frac{k(2)-3}{2-1}=5$

Now $\frac{2k-3}{1}=5$

$2k-3=5$

$2k=5+3$

$k=\frac{8}{2}$

k=4

Therefore k=4

b) To find for h(3)=k:

$h(x)=\frac{kx-3}{x-1}$  where $x\neq1$

that is put x=3 in the above function

$h(3)=\frac{k(3)-3}{3-1}=k$

Now $\frac{3k-3}{2}=k$

$3k-3=2k$

$3k-2k-3=0$

$k-3=0$

k=3

Therefore k=3

c) h(k)=k

To find for h(3)=K:

$h(x)=\frac{kx-3}{x-1}$ where $x\neq1$

that is put x=k in the above function

$h(k)=\frac{k(k)-3}{k-1}=k$  here $k\neq1$

Now $\frac{k^2-3}{k-1}=k$

$k^2-3=k(k-1)$

$k^2-3=k^2-k$

$k^2-3-k^2+k=0$

k-3=0

k=3

Therefore k=3