Math Question and Answers


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
Question Number 193235 by mathlove last updated on 08/Jun/23

	Answered by a.lgnaoui last updated on 08/Jun/23
	$Totale Area=22+A (A=xy) 22+xy=6x+(8−x)y 2xy=8y+6x−22 xy=4y+3x−11 (1) { (((8−x)y=10)),(((6−y)x=12 ⇒10−8y=12−6x)) :} x=((4y+1)/3) (2) (1)⇒x(y−3)=4y−11 x=((4y−11)/(y−3)) (1)⇔ ((y(4y+1))/3)=8y−10 (1)=(2)⇒ ((4y+1)/3)=((4y−11)/(y−3)) 4y^2 −12y+y−3=12y−33 4y^2 −23y+30=0 y^2 −((23)/4)y+((15)/2)=0 △=((23^2 )/(16))−30=((23^2 −480)/(16))=((7/4))^2 y=((((23)/4)±(7/4))/2)= y=(((15)/4),2) a− y=2 x=3 (x;y)= (3,2) b− y=((15)/4) x=((16)/3) (x;y)= (((16)/3) ; ((15)/4)) ⇒xy= { ((6 pour (x,y)=(3,2))),((20 pour (x,y)=(((16)/3),((15)/4)))) :}$
	$T otale Area = 22 + A (A = xy)$ $22 + xy = 6 x + (8 - x) y$ $2 xy = 8 y + 6 x - 22$ $xy = 4 y + 3 x - 11 (1)$ ${\begin{cases} (8 - x) y = 10 \\ (6 - y) x = 12 \Rightarrow 10 - 8 y = 12 - 6 x \end{cases}$ $x = \frac{4 y + 1}{3} (2)$ $(1) \Rightarrow x (y - 3) = 4 y - 11 x = \frac{4 y - 11}{y - 3}$ $(1) \Leftrightarrow \frac{y (4 y + 1)}{3} = 8 y - 10$ $(1) = (2) \Rightarrow \frac{4 y + 1}{3} = \frac{4 y - 11}{y - 3}$ ${4 y}^{2} - 12 y + y - 3 = 12 y - 33$ ${4 y}^{2} - 23 y + 30 = 0$ $y^{2} - \frac{23}{4} y + \frac{15}{2} = 0$ $△ = \frac{23^{2}}{16} - 30 = \frac{23^{2} - 480}{16} = {(\frac{7}{4})}^{2}$ $y = \frac{\frac{23}{4} \pm \frac{7}{4}}{2} = y = (\frac{15}{4}, 2)$ $a - y = 2 x = 3 (x; y) = (3, 2)$ $b - y = \frac{15}{4} x = \frac{16}{3} (x; y) = (\frac{16}{3}; \frac{15}{4})$ $\Rightarrow xy = {\begin{cases} 6 pour (x, y) = (3, 2) \\ 20 pour (x, y) = (\frac{16}{3}, \frac{15}{4}) \end{cases}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum