x²-9x-367 and Y²-6y+9 and x
Answers
Step-by-step explanation:
Method 1: Dividing (1) by (2), we get
[math]\frac{y^2+3y}{y^2-2y}=\frac{9x/2}{3x}[/math]
[math]\frac{y^2+3y}{y^2-2y}=\frac{3}{2}[/math]
[math]2y^2+6y=3y^2-6y[/math]
[math]y^2-12y=0[/math]
[math]y(y-12)=0[/math]
[math]y=0, 12\implies x=0, 40[/math]
Method 2: subtracting (2) from (1), we get
[math]y^2+3y-y^2+2y=\frac92x-3x[/math]
[math]5y=\frac32x[/math]
[math]y=\frac{3x}{10}[/math]
Substituting value of [math]y=\frac{3x}{10}[/math] in (2), we get
[math]\left(\frac{3x}{10}\right)^2-2\left(\frac{3x}{10}\right)=3x[/math]
[math]3x^3-20x=100x[/math]
[math]3x^2-120x=0[/math]
[math]3x(x-40)=0[/math]
[math]x=0, 40[/math]
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· Answer requested by Diable
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Step-by-step explanation:
This quadratic equation cannot be factorised.
Hence, factorised.
I hope it helps you. If you have any doubts, then don't hesitate to ask.