Question Number 39072 by ajfour last updated on 02/Jul/18
Commented by ajfour last updated on 02/Jul/18
$${Calculate}\:{x}\:. \\ $$
Answered by MrW3 last updated on 02/Jul/18
![(c/x)=((c+4.8)/(10)) ⇒10c=xc+4.8x ⇒c=((4.8x)/(10−x)) ((4.8)/x)=((4.8+c+x)/(10)) 48=x(4.8+c+x) 48=x(4.8+x+((4.8x)/(10−x))) 48(10−x)=x[(4.8+x)(10−x)+4.8x] 480−48x=x(48+10x−4.8x−x^2 +4.8x) 480−48x=48x+10x^2 −x^3 ⇒x^3 −10x^2 −96x+480=0 ⇒x=4.0](Q39084.png)
$$\frac{{c}}{{x}}=\frac{{c}+\mathrm{4}.\mathrm{8}}{\mathrm{10}} \\ $$$$\Rightarrow\mathrm{10}{c}={xc}+\mathrm{4}.\mathrm{8}{x} \\ $$$$\Rightarrow{c}=\frac{\mathrm{4}.\mathrm{8}{x}}{\mathrm{10}−{x}} \\ $$$$\frac{\mathrm{4}.\mathrm{8}}{{x}}=\frac{\mathrm{4}.\mathrm{8}+{c}+{x}}{\mathrm{10}} \\ $$$$\mathrm{48}={x}\left(\mathrm{4}.\mathrm{8}+{c}+{x}\right) \\ $$$$\mathrm{48}={x}\left(\mathrm{4}.\mathrm{8}+{x}+\frac{\mathrm{4}.\mathrm{8}{x}}{\mathrm{10}−{x}}\right) \\ $$$$\mathrm{48}\left(\mathrm{10}−{x}\right)={x}\left[\left(\mathrm{4}.\mathrm{8}+{x}\right)\left(\mathrm{10}−{x}\right)+\mathrm{4}.\mathrm{8}{x}\right] \\ $$$$\mathrm{480}−\mathrm{48}{x}={x}\left(\mathrm{48}+\mathrm{10}{x}−\mathrm{4}.\mathrm{8}{x}−{x}^{\mathrm{2}} +\mathrm{4}.\mathrm{8}{x}\right) \\ $$$$\mathrm{480}−\mathrm{48}{x}=\mathrm{48}{x}+\mathrm{10}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{96}{x}+\mathrm{480}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{4}.\mathrm{0} \\ $$
Commented by MJS last updated on 02/Jul/18
$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{somewhere},\:\mathrm{MrW} \\ $$$$\mathrm{from}\:\frac{{c}}{{x}}=\frac{{c}+\mathrm{4}.\mathrm{8}}{\mathrm{10}}\:\mathrm{and}\:\frac{\mathrm{4}.\mathrm{8}}{{x}}=\frac{\mathrm{4}.\mathrm{8}+{c}+{x}}{\mathrm{10}}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{96}{x}+\mathrm{480}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{3}−\sqrt{\mathrm{129}};\:{c}_{\mathrm{1}} =−\mathrm{9}+\frac{\mathrm{3}}{\mathrm{5}}\sqrt{\mathrm{129}} \\ $$$${x}_{\mathrm{2}} =\mathrm{3}+\sqrt{\mathrm{129}};\:{c}_{\mathrm{2}} =−\mathrm{9}−\frac{\mathrm{3}}{\mathrm{5}}\sqrt{\mathrm{129}} \\ $$$${x}_{\mathrm{3}} =\mathrm{4};\:{c}_{\mathrm{3}} =\frac{\mathrm{16}}{\mathrm{5}}\approx\mathrm{3}.\mathrm{2} \\ $$
Commented by MrW3 last updated on 02/Jul/18
$${Thank}\:{you}\:{both}\:{sirs}.\:{I}\:{have}\:{fixed}\:{the} \\ $$$${mistake}. \\ $$
Commented by MrW3 last updated on 02/Jul/18
Commented by ajfour last updated on 02/Jul/18
$${Thank}\:{you}\:{Sir}\:{MjS}. \\ $$
Answered by ajfour last updated on 02/Jul/18
$$\frac{\mathrm{4}.\mathrm{8}}{{x}}=\frac{\mathrm{4}.\mathrm{8}+{c}+{x}}{\mathrm{10}}\:\:\:\:\:....\left({i}\right) \\ $$$$\frac{\mathrm{10}−{x}}{\mathrm{4}.\mathrm{8}}=\frac{\mathrm{10}}{\mathrm{4}.\mathrm{8}+{c}}\:\:\:\:\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{4}.\mathrm{8}+{c}\:=\:\frac{\mathrm{48}}{\mathrm{10}−{x}} \\ $$$${So}\:{using}\:{this}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\frac{\mathrm{48}}{{x}}\:=\:\frac{\mathrm{48}}{\mathrm{10}−{x}}+{x} \\ $$$$\:\:\mathrm{480}−\mathrm{48}{x}=\mathrm{48}{x}+{x}^{\mathrm{2}} \left(\mathrm{10}−{x}\right) \\ $$$$\:\:{x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{96}{x}+\mathrm{480}=\mathrm{0} \\ $$$$\:\:{x}=\:\mathrm{4}\:. \\ $$