Question

1/3x-y + 1/3x-y = 12/35 ; 1/2(3x+y) - 1/2(3x-y) = 2/35

Answer 1

$\textbf{Given:}$

$\displaystyle\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{12}{35}$

$\displaystyle\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{2}{35}$

$\textbf{To find: x and y}$

$\text{The given equations can be written as}$

$\displaystyle\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{12}{35}$ ........(1)

$\displaystyle\frac{1}{3x+y}-\frac{1}{3x-y}=\frac{4}{35}$...(2)

$\text{Adding (1) and (2), we get}$

$\displaystyle\frac{2}{3x+y}=\frac{16}{35}$

$\displaystyle\frac{1}{3x+y}=\frac{8}{35}$

$\text{(1) gives}$

$\displaystyle\frac{8}{35}+\frac{1}{3x-y}=\frac{12}{35}$

$\displaystyle\frac{1}{3x-y}=\frac{12}{35}-\frac{8}{35}$

$\displaystyle\frac{1}{3x-y}=\frac{4}{35}$

$\text{Now, we have}$

$3x+y=\frac{35}{8}$

$3x-y=\frac{35}{4}$

$\text{Adding these equations, we get}$

$6x=\frac{70+35}{8}$

$6x=\frac{105}{8}$

$2x=\frac{35}{8}$

$\implies\boxed{\bf,x=\frac{35}{16}}$

$3x+y=\frac{35}{8}\implies\,3(\frac{35}{16})+y=\frac{35}{8}$

$\frac{105}{16}+y=\frac{35}{8}$

$\implies\,y=\frac{35}{8}-\frac{105}{16}$

$\implies\,y=\frac{70-105}{16}$

$\implies\boxed{\bf\,y=\frac{-35}{16}}$

$\therefore\textbf{The solution is \bf\,x=\frac{35}{16} and \bf\,y=\frac{-35}{16} }$

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Answer 2

The final answer is x= 1.24 and y = -2.11

Step-by-step explanation:

$\ Given \ value:\\\\\frac{1}{3x-y}+\frac{1}{3x-y} = \frac{12}{35}\ and \ \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)}= \frac{2}{35} \\\\\ Find:\\\\x \ and \ y = ?\\\\ \ Solution: \\\\\frac{1}{3x-y}+\frac{1}{3x-y} = \frac{12}{35}........(i)\\\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)}= \frac{2}{35}.......(ii) \\\\\ Solve \ equation (i)\\\\\ Equation: \\\\ \frac{1}{3x-y}+\frac{1}{3x-y} = \frac{12}{35}\\\\\rightarrow \frac{1+1}{(3x-y)} = \frac{12}{35}\\\\\rightarrow \frac{2}{(3x-y)} = \frac{12}{35}$

$\ cross \ multiply:\\\\\rightarrow 2\times 35 = (3x-y)12\\\\\rightarrow70 = 36x-12y\\\\\rightarrow70 +12y= 36x\\\\\rightarrow 36x = 70 +12y\\\\\rightarrow x = \frac{70 +12y}{36}\\\\\rightarrow x = \frac{(35 +6y)}{18}$

$\ Solve \ equation \ (ii)$

$\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)}= \frac{2}{35} \\\\\rightarrow \frac{(3x-y)-(3x+y)}{2(3x+y)(3x-y)} = \frac{2}{35}\\\\\rightarrow \frac{(3x-y-3x-y)}{2((3x)^2-(y)^2)} = \frac{2}{35}\\\\\rightarrow \frac{(-2y)}{(18x^2-2y^2)} = \frac{2}{35}\\\\\rightarrow -70y = 36x^2-4y^2\\\\\rightarrow 4y^2-70y = 36x^2\\\\\rightarrow x^2= \frac{4y^2-70y}{36}\\\\\rightarrow x^2= \frac{(2y^2-35y)}{18}$

$\ put \ the \ calculated \ value \ of \ in \ above \ equation:\\\\x = \frac{35+6y}{18}\\\\x^2 =(\frac{35+6y}{18})^2\\\\x^2= \frac{1225+36y^2+420}{324}\\\\\ compare \ both \ value:\\\\\frac{1225+36y^2+420}{324} = \frac{2y^2-35y}{18}$

$(1225+36y^2+420) = \frac{(2y^2-35y)}{18} \times 324\\\\(1225+36y^2+420) = (2y^2-35y) \times 18\\\\1225+36y^2+420 = 36y^2-630y\\\\1645=-630y\\\\y= - \frac{1645}{630}\\\\y= -2.11\\\\\ put \ the \ value \ y \ in \ equation \ (i) \\\\\ equation\\\\ x= \frac{35+6y}{18}\\\\x= \frac{35+6(-2.11)}{18}\\\\x= \frac{22.34}{18}\\\\x= 1.24$

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