3f-x-2f-x-59-x-1-10x-30-for-all-rael-x-1-faind-volue-of-f-7-

Question Number 168698 by mathlove last updated on 16/Apr/22

$$\mathrm{3}{f}\left({x}\right)+\mathrm{2}{f}\left(\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}\right)=\mathrm{10}{x}+\mathrm{30}\:{for}\:{all} \\ $$$${rael}\:\:{x}\cancel{=}\mathrm{1}\:{faind}\:{volue}\:{of} \\ $$$${f}\left(\mathrm{7}\right)=? \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 16/Apr/22

3f(x)+2f(((x+59)/(x−1)))=10x+30 { ((x=7:3f(7)+2f(((7+59)/(7−1)))=10(7)+30)),((x=11:3f(11)+2f(((11+59)/(11−1)))=10(11)+30)) :} { ((3f(7)+2f(11)=100)),((3f(11)+2f(7)=140)) :} { ((3×( 3f(7)+2f(11)=100 ))),((2×( 3f(11)+2f(7)=140 ))) :} { ((9f(7)+6f(11)=300...(i))),((6f(11)+4f(7)=280..(ii))) :} ⇒(i)−(ii):5f(7)=20 f(7)=((20)/5)=4

$$\mathrm{3}{f}\left({x}\right)+\mathrm{2}{f}\left(\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}\right)=\mathrm{10}{x}+\mathrm{30} \\ $$$$\begin{cases}{{x}=\mathrm{7}:\mathrm{3}{f}\left(\mathrm{7}\right)+\mathrm{2}{f}\left(\frac{\mathrm{7}+\mathrm{59}}{\mathrm{7}−\mathrm{1}}\right)=\mathrm{10}\left(\mathrm{7}\right)+\mathrm{30}}\\{{x}=\mathrm{11}:\mathrm{3}{f}\left(\mathrm{11}\right)+\mathrm{2}{f}\left(\frac{\mathrm{11}+\mathrm{59}}{\mathrm{11}−\mathrm{1}}\right)=\mathrm{10}\left(\mathrm{11}\right)+\mathrm{30}}\end{cases} \\ $$$$\begin{cases}{\mathrm{3}{f}\left(\mathrm{7}\right)+\mathrm{2}{f}\left(\mathrm{11}\right)=\mathrm{100}}\\{\mathrm{3}{f}\left(\mathrm{11}\right)+\mathrm{2}{f}\left(\mathrm{7}\right)=\mathrm{140}}\end{cases}\:\: \\ $$$$\begin{cases}{\mathrm{3}×\left(\:\mathrm{3}{f}\left(\mathrm{7}\right)+\mathrm{2}{f}\left(\mathrm{11}\right)=\mathrm{100}\:\right)}\\{\mathrm{2}×\left(\:\mathrm{3}{f}\left(\mathrm{11}\right)+\mathrm{2}{f}\left(\mathrm{7}\right)=\mathrm{140}\:\right)}\end{cases}\:\: \\ $$$$\begin{cases}{\mathrm{9}{f}\left(\mathrm{7}\right)+\mathrm{6}{f}\left(\mathrm{11}\right)=\mathrm{300}…\left(\mathrm{i}\right)}\\{\mathrm{6}{f}\left(\mathrm{11}\right)+\mathrm{4}{f}\left(\mathrm{7}\right)=\mathrm{280}..\left(\mathrm{ii}\right)}\end{cases}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{i}\right)−\left(\mathrm{ii}\right):\mathrm{5}{f}\left(\mathrm{7}\right)=\mathrm{20} \\ $$$${f}\left(\mathrm{7}\right)=\frac{\mathrm{20}}{\mathrm{5}}=\mathrm{4} \\ $$

Commented by mathlove last updated on 16/Apr/22

$${thank}\:{bro} \\ $$

Commented by Rasheed.Sindhi last updated on 16/Apr/22

$$\mathcal{T}{hanks}\:{for}\:{feedback}! \\ $$

Answered by mr W last updated on 16/Apr/22

t=((x+59)/(x−1)) ⇒x=((t+59)/(t−1)) 3f(((t+59)/(t−1)))+2f(t)=10×((t+59)/(t−1))+30 2f(x)+3f(((x+59)/(x−1)))=10×((x+59)/(x−1))+30 ...(ii) 3×(i)−2×(ii): 5f(x)=30x+90−20×((x+59)/(x−1))−60 ⇒f(x)=6(x+1)−((4(x+59))/(x−1))=((2(3x^2 −2x−121))/(x−1)) ⇒f(7)=6(7+1)−((4(7+59))/(7−1))=4 ✓

$${t}=\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}\:\Rightarrow{x}=\frac{{t}+\mathrm{59}}{{t}−\mathrm{1}} \\ $$$$\mathrm{3}{f}\left(\frac{{t}+\mathrm{59}}{{t}−\mathrm{1}}\right)+\mathrm{2}{f}\left({t}\right)=\mathrm{10}×\frac{{t}+\mathrm{59}}{{t}−\mathrm{1}}+\mathrm{30} \\ $$$$\mathrm{2}{f}\left({x}\right)+\mathrm{3}{f}\left(\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}\right)=\mathrm{10}×\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}+\mathrm{30}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{3}×\left({i}\right)−\mathrm{2}×\left({ii}\right): \\ $$$$\mathrm{5}{f}\left({x}\right)=\mathrm{30}{x}+\mathrm{90}−\mathrm{20}×\frac{{x}+\mathrm{59}}{{x}−\mathrm{1}}−\mathrm{60} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{6}\left({x}+\mathrm{1}\right)−\frac{\mathrm{4}\left({x}+\mathrm{59}\right)}{{x}−\mathrm{1}}=\frac{\mathrm{2}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{121}\right)}{{x}−\mathrm{1}} \\ $$$$\Rightarrow{f}\left(\mathrm{7}\right)=\mathrm{6}\left(\mathrm{7}+\mathrm{1}\right)−\frac{\mathrm{4}\left(\mathrm{7}+\mathrm{59}\right)}{\mathrm{7}−\mathrm{1}}=\mathrm{4}\:\checkmark \\ $$

Commented by mathlove last updated on 16/Apr/22