Question

x^2+4y^2-15x=10(3y-8);xy=6

Answer 1

The value of x= 3 and y= 2

Step-by-step explanation:

$\ Given \ value:\\\\x^2+4y^2-15x=10(3y-8) \\\\\ xy=6\\\\\ Find: \\\\ x=? \ and\ y =?\\\\\ Solution:\\\\\ x^2+4y^2-15x=10(3y-8) \\\\\rightarrow x^2-15x+4y^2= 30y-80\\\\\rightarrow x^2-15x+4y^2 -30y+80 =0......(i)\\\\\therefore xy= 6........(i)\\\\ \because x=\frac{6}{y}\\\\\ put \ the\ value \ of \ x \ in \ (i) \ equation:\\\\\ Equation: \\\\ \rightarrow x^2-15x+4y^2 -30y+80 =0\\\\(\frac{6}{y})^2-15\times \frac{6}{y}+ 4y^2 - 30y+80 =0\\\\\frac{36}{y^2}- \frac{90}{y}+4y^2-30y+80=0$

$\frac{36-90y+ 4y^4-30y^3+80y^2}{y^2}=0\\\\36-90y+4y^4-30y^3+80y^2=0$

$4y^4-30y^3+80y^2-90y+36=0\\\\2(2y^4-15y^3+40y^2-45y+18)=0\\\\2y^4-15y^3+40y^2-45y+18=0\\\\\ by \ hit \ and \ trail \ method \ we \ put \ value \ y=2 \ in \ above \ equation:\\\\ 2(2)^4-15(2)^3+40(2)^2-45(2)+18=0\\\\2(16)-15(8)+40(4)-45(2)+18=0\\\\32-120+160-90+18=0$

$210 -210 =0\\\\\ so, \ the \ value \of\ y\ is\ 2\\\\\ put \ the \ value \ of \ in \ equation \ (i) \\\\\ equation: \\ xy= 6\\\\x\times 2 =6\\\\x= \frac{6}{2}\\\\x= 3\\\\\ x= 3 \ and \ y = 2$

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