Question

2/3x+2y + 3/3x-2y = 17/5 and 5/3x+2y + 1/3x-2y = 2
solve by elimination method

Answer 1

The final value of x and y is 1,1

Step-by-step explanation:

$\ Given \ value:\\\\\frac{2}{3x + 2y}+ \frac{3}{3x -2y} = \frac{17}{5} \\\\\frac{5}{3x + 2y}+ \frac{1}{3x -2y} = 2\\\\\ find:\\\\\ x \ and \ y= ?\\\\\ Solution:\\\\ \frac{2}{3x + 2y}+ \frac{3}{3x -2y} = \frac{17}{5} ......(i) \\\\\frac{5}{3x + 2y}+ \frac{1}{3x-2y} = 2......(ii)\\\\\ In \ equation \ (i) \ we \ multiply \ by \ 5 \ and \ in \\ \ equation \ (ii)\ 2\ then \ subtract \ from \ each \ other\\\\( \frac{2}{3x + 2y}+ \frac{3}{3x -2y} = \frac{17}{5}) \times 5$

$(\frac{10}{3x + 2y}+ \frac{15}{3x-2y} = \frac{17 \times 5}{5}) \\\\(\frac{10}{3x + 2y}+ \frac{15}{3x-2y} = 17) ....(i) \\\\(\frac{5}{3x + 2y}+ \frac{1}{3x-2y} = 2) \times 2\\\\(\frac{10}{3x + 2y}+ \frac{2}{3x-2y} = 4) .....(ii)\\\\\ equation:\\\\\frac{10}{3x + 2y}+ \frac{15}{3x-2y} = 17 \\ \frac{10}{3x + 2y}+ \frac{2}{3x-2y} = 4\\\ subtract \\\\\frac{15}{3x-2y} - \frac{2}{3x-2y} = 13 \\\\\frac{15-2}{3x-2y} = 13 \\\\\frac{13}{3x-2y} = 13 \\\\13 = 13 \times 3x-2y \\\\3x-2y =1....(b)$

$\frac{2}{3x + 2y}+ \frac{3}{3x -2y} = \frac{17}{5} ......(i) \\\\\frac{5}{3x + 2y}+ \frac{1}{3x-2y} = 2......(ii)\\\\\ In \ equation \ ii \ we \ multiply \ by 3 \ then \ subtract \ from \ equation (i) \\\\(\frac{5}{3x + 2y}+ \frac{1}{3x-2y} = 2) \times 3 \\\\\frac{15}{3x + 2y}+ \frac{3}{3x-2y} = 6 \\\\ \frac{2}{3x + 2y}+ \frac{3}{3x -2y} = \frac{17}{5}\\\\\ subtract\\ \frac{15}{3x + 2y}- \frac{2}{3x+2y} = 6-\frac{17}{5}\\\\ \frac{15-2}{3x + 2y}= \frac{30-17}{5}$

$\frac{13}{3x + 2y}= \frac{13}{5}\\\\3x+2y=5.....(a)$

$\ when \ we \ subtract \ and \ add \ both \ equation \\ \ so, \ we \ get \ x \ and \ y \ value \ i.e. 1\\\\a+b)\\\\(3x+2y)+(3x-2y)=5+1\\\\3x+2y+3x-2y= 6\\\\6x=6\\\\x= 1\\\\a-b)\\\\(3x+2y)-(3x-2y)=5-1\\\\3x+2y-3x+2y= 4\\\\4y=4\\\\y= 1$

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