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Question-204994




Question Number 204994 by LuisTony last updated on 05/Mar/24
Answered by TonyCWX08 last updated on 05/Mar/24
3x−5=x+3  3x−x=3+5  2x=8  x=4    y=x+3  y=4+3  y=7
$$\mathrm{3}{x}−\mathrm{5}={x}+\mathrm{3} \\ $$$$\mathrm{3}{x}−{x}=\mathrm{3}+\mathrm{5} \\ $$$$\mathrm{2}{x}=\mathrm{8} \\ $$$${x}=\mathrm{4} \\ $$$$ \\ $$$${y}={x}+\mathrm{3} \\ $$$${y}=\mathrm{4}+\mathrm{3} \\ $$$${y}=\mathrm{7} \\ $$
Answered by Rasheed.Sindhi last updated on 05/Mar/24
y=3x−5 ∧ y=x+3⇒  y=3x−5 ∧ 3y=3x+9⇒  ((3x−5)/(3x+9))=(y/(3y))=(1/3)  (((3x−5)+(3x+9))/((3x−5)−(3x+9)))=((1+3)/(1−3))  ((6x+4)/(−14))=(4/(−2))=−2  ((3x+2)/7)=2  3x=14−2=12⇒x=4  y=x+3=4+3=7
$${y}=\mathrm{3}{x}−\mathrm{5}\:\wedge\:{y}={x}+\mathrm{3}\Rightarrow \\ $$$${y}=\mathrm{3}{x}−\mathrm{5}\:\wedge\:\mathrm{3}{y}=\mathrm{3}{x}+\mathrm{9}\Rightarrow \\ $$$$\frac{\mathrm{3}{x}−\mathrm{5}}{\mathrm{3}{x}+\mathrm{9}}=\frac{{y}}{\mathrm{3}{y}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\left(\mathrm{3}{x}−\mathrm{5}\right)+\left(\mathrm{3}{x}+\mathrm{9}\right)}{\left(\mathrm{3}{x}−\mathrm{5}\right)−\left(\mathrm{3}{x}+\mathrm{9}\right)}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{1}−\mathrm{3}} \\ $$$$\frac{\mathrm{6}{x}+\mathrm{4}}{−\mathrm{14}}=\frac{\mathrm{4}}{−\mathrm{2}}=−\mathrm{2} \\ $$$$\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{7}}=\mathrm{2} \\ $$$$\mathrm{3}{x}=\mathrm{14}−\mathrm{2}=\mathrm{12}\Rightarrow{x}=\mathrm{4} \\ $$$${y}={x}+\mathrm{3}=\mathrm{4}+\mathrm{3}=\mathrm{7} \\ $$

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