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Question Number 157972 by MathsFan last updated on 30/Oct/21

given 2x=3y and 4y=3z find ((3x−4y)/(2x−y+5z))

$$\:{given}\:\:\mathrm{2}{x}=\mathrm{3}{y}\:\:{and}\:\:\mathrm{4}{y}=\mathrm{3}{z} \\ $$$$\:{find}\:\:\:\frac{\mathrm{3}{x}−\mathrm{4}{y}}{\mathrm{2}{x}−{y}+\mathrm{5}{z}} \\ $$

Commented by tounghoungko last updated on 30/Oct/21

let 3y=k⇒y=(k/3) ∧ 2x=k⇒x=(k/2) ⇒z=((4k)/9) ((((3k)/2)−((4k)/3))/(k−(k/3)+((20k)/9))) = (((3/2)−(4/3))/((2/3)+((20)/9))) = ((1/6)/((26)/9)) = (1/6)×(9/(26))=(3/(52))

$${let}\:\mathrm{3}{y}={k}\Rightarrow{y}=\frac{{k}}{\mathrm{3}}\:\wedge\:\mathrm{2}{x}={k}\Rightarrow{x}=\frac{{k}}{\mathrm{2}} \\ $$$$\Rightarrow{z}=\frac{\mathrm{4}{k}}{\mathrm{9}} \\ $$$$\frac{\frac{\mathrm{3}{k}}{\mathrm{2}}−\frac{\mathrm{4}{k}}{\mathrm{3}}}{{k}−\frac{{k}}{\mathrm{3}}+\frac{\mathrm{20}{k}}{\mathrm{9}}}\:=\:\frac{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{4}}{\mathrm{3}}}{\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{20}}{\mathrm{9}}}\:=\:\frac{\frac{\mathrm{1}}{\mathrm{6}}}{\frac{\mathrm{26}}{\mathrm{9}}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{9}}{\mathrm{26}}=\frac{\mathrm{3}}{\mathrm{52}} \\ $$

Commented by MathsFan last updated on 30/Oct/21

thanks sir

$${thanks}\:{sir} \\ $$

Commented by MathsFan last updated on 30/Oct/21

sir, how did you get z=((4k)/3)

$${sir},\:{how}\:{did}\:{you}\:{get}\:\:{z}=\frac{\mathrm{4}{k}}{\mathrm{3}} \\ $$

Commented by Rasheed.Sindhi last updated on 30/Oct/21

z=(4/3)y⇒z=(4/3)×(k/3)=((4k)/9)

$${z}=\frac{\mathrm{4}}{\mathrm{3}}{y}\Rightarrow{z}=\frac{\mathrm{4}}{\mathrm{3}}×\frac{{k}}{\mathrm{3}}=\frac{\mathrm{4}{k}}{\mathrm{9}} \\ $$

Commented by MathsFan last updated on 30/Oct/21

yeah thanks

$${yeah} \\ $$$${thanks} \\ $$

Commented by tounghoungko last updated on 30/Oct/21

given codition 3z=4y and 3y=k or y=(k/3) so z=((4y)/3)=(4/3)×(k/3)=((4k)/9)

$${given}\:{codition}\:\mathrm{3}{z}=\mathrm{4}{y}\:{and}\:\mathrm{3}{y}={k}\:{or}\:{y}=\frac{{k}}{\mathrm{3}} \\ $$$${so}\:{z}=\frac{\mathrm{4}{y}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}}×\frac{{k}}{\mathrm{3}}=\frac{\mathrm{4}{k}}{\mathrm{9}} \\ $$

Answered by Rasheed.Sindhi last updated on 30/Oct/21

2x=3y and 4y=3z; ((3x−4y)/(2x−y+5z))=? 2x=3y⇒(x/y)=(3/2) 4y=3z⇒(z/y)=(4/3) ((3x−4y)/(2x−y+5z))=((3((x/y))−4)/(2((x/y))−1+5((z/y)))) =((3((3/2))−4)/(2((3/2))−1+5((4/3))))=(( ((9−8)/2) )/((18−6+40)/6)) =(1/2_1 )×(6^3 /(52))=(3/(52))

$$\mathrm{2}{x}=\mathrm{3}{y}\:\:{and}\:\:\mathrm{4}{y}=\mathrm{3}{z};\:\:\frac{\mathrm{3}{x}−\mathrm{4}{y}}{\mathrm{2}{x}−{y}+\mathrm{5}{z}}=? \\ $$$$\mathrm{2}{x}=\mathrm{3}{y}\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{4}{y}=\mathrm{3}{z}\Rightarrow\frac{{z}}{{y}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\:\frac{\mathrm{3}{x}−\mathrm{4}{y}}{\mathrm{2}{x}−{y}+\mathrm{5}{z}}=\frac{\mathrm{3}\left(\frac{{x}}{{y}}\right)−\mathrm{4}}{\mathrm{2}\left(\frac{{x}}{{y}}\right)−\mathrm{1}+\mathrm{5}\left(\frac{{z}}{{y}}\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{4}}{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{1}+\mathrm{5}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}=\frac{\:\:\frac{\mathrm{9}−\mathrm{8}}{\mathrm{2}}\:\:}{\frac{\mathrm{18}−\mathrm{6}+\mathrm{40}}{\mathrm{6}}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\cancel{\underset{\mathrm{1}} {\mathrm{2}}}}×\frac{\cancel{\overset{\mathrm{3}} {\mathrm{6}}}}{\mathrm{52}}=\frac{\mathrm{3}}{\mathrm{52}} \\ $$

Commented by MathsFan last updated on 30/Oct/21

thanks sir

$${thanks}\:{sir} \\ $$

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