Question

x²-9x-367 and Y²-6y+9 and x​

Answer 1

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 2

Step-by-step explanation:

${x}^{2} - 9x - 367$

This quadratic equation cannot be factorised.

${y}^{2} - 6y + 9$

$= {y}^{2} - 3y - 3y + 9$

$= y(y - 3) - 3(y - 3)$

$= (y - 3)(y - 3)$

Hence, factorised.

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