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Question Number 88160 by john santu last updated on 08/Apr/20

$$solve:x^2=3x+6y;xy=5x+4y$$

Commented by john santu last updated on 08/Apr/20

$$y=\frac{x^2-3x}{6};y(x-4)=5x$$$$\Rightarrow \frac{(x^2-3x)(x-4)}{6}=5x$$$$x^3-7x^2+12x=30x$$$$x(x^2-7x-18)=0$$$$x(x-9)(x+2)=0$$$$x=0\wedge y=0$$$$x=9\wedge y=9$$$$x=-2\wedge y=\frac{5}{3}$$$$$$

Answered by ajfour last updated on 08/Apr/20

$$x^2-3x=\frac{6\times 5x}{x-4}$$$$x\{(x-3)(x-4)-30\}=0$$$$x(x^2-7x-18)=0$$$$x=-2,0,9,$$$$y=\frac{5x}{x-4}.$$

Answered by $@ty@m123 last updated on 08/Apr/20

$$3x+6y=x^2...\left(1\right)$$$$5x+4y=xy...\left(2\right)$$$$\left(1\right)\times 2-\left(2\right)\times 3$$$$6x+12y=2x^2$$$${}_{(-)}15x+{}_{(-)}12y={}_{(-)}3xy$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$-9x=2x^2-3xy$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$y=\frac{2x^2+9x}{3x}$$$$y=\frac{2}{3}x+3...\left(3\right)$$$$\therefore from\left(1\right)$$$$x^2-3x-6(\frac{2}{3}x+3)=0$$$$x^2-7x-18=0$$$$(x-9)(x+2)=0$$$$x=-2,9$$$$\therefore from\left(3\right),$$$$y=\frac{5}{3},9$$

Commented by jagoll last updated on 08/Apr/20

$$\mathrm{x}=0?\mathrm{not}\mathrm{solution}?$$

Commented by jagoll last updated on 08/Apr/20

$$\mathrm{i}\mathrm{think}\mathrm{x}=0\mathrm{is}\mathrm{solution}.$$

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