What is the standard form of $y = 3 x (x - 3) + (2 - x) (x - \frac{4}{5})$ $y= 3x(x-3)+(2-x)(x-4/5)$ ?

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Answer 1 · 2016-01-23T15:23:04

$10 x^{2} - 31 x - 5 y - 8 = 0$ $10x^2-31x-5y-8=0$

$y = 3 x (x - 3) + (2 - x) (x - \frac{4}{5})$ $y=3x(x-3)+(2-x)(x-4/5)$

Use the distributive property

$y = 3 x^{2} - 9 x + 2 x - \frac{8}{5} - x^{2} + \frac{4 x}{5}$ $y=3x^2-9x+2x-8/5-x^2+(4x)/5$

$y = 2 x^{2} - \frac{31}{5} x - \frac{8}{5}$ $y=2x^2-31/5x-8/5$

Then multiply the whole equation with 5

$(y = 2 x^{2} - \frac{31}{5} x - \frac{8}{5}) \cdot 5$ $(y=2x^2-31/5x-8/5)*5$

$5 y = 10 x^{2} - 31 x - 8$ $5y=10x^2-31x-8$

$5 y - 10 x^{2} + 31 x + 8 = 0$ $5y-10x^2+31x+8=0$

$10 x^{2} - 31 x - 5 y - 8 = 0$ $10x^2-31x-5y-8=0$ And this is the final standard answer

What is the standard form of y=3x(x−3)+(2−x)(x−45) y= 3x(x-3)+(2-x)(x-4/5) ?