How do you solve $Y = 2 x^{2} - 3$ $Y=2x^2-3$ and $Y = 3 x - 1$ $Y=3x-1$ ?

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How do you solve $Y = 2 x^{2} - 3$ $Y=2x^2-3$ and $Y = 3 x - 1$ $Y=3x-1$ ?

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Answer 1 · 2015-09-10T10:30:59

$(x, y) = (- \frac{1}{2}, - 2 \frac{1}{2}) or (2, 5)$ $(x,y) = (-1/2, -2 1/2) or (2,5)$

Explanation:

If
[1] $XXX y = 2 x^{2} - 3$ $color(white)("XXX")y = 2x^2-3$
and
[2] $XXX y = 3 x - 1$ $color(white)("XXX")y=3x-1$
then
[3] $XXX 2 x^{2} - 3 = 3 x - 1$ $color(white)("XXX")2x^2-3=3x-1$

Simplifying
[4] $XXX 2 x^{2} - 3 x - 2 = 0$ $color(white)("XXX")2x^2-3x-2=0$
Factoring
[5] $XXX (2 x + 1) (x - 2) = 0$ $color(white)("XXX")(2x+1)(x-2)=0$

[6] $XXX x = - \frac{1}{2} XXX or [7] XXX x = 2$ $color(white)("XXX")x=-1/2color(white)("XXX")or [7]color(white)("XXX")x=2$

${: (color(white)("XXX"),x=-1/2,color(white)("XXXXXXX"),x=2), (color(white)("XXX"),y=2x^2-3,color(white)("XXXXXXX"),y=2x^2-3), (color(white)("XXX"),color(white)("X")=1/2-3,color(white)("XXXXXXX"),color(white)("X")=8-3), (color(white)("XXX"),color(white)("X")=-2 1/2,color(white)("XXXXXXX"),color(white)("X")=5) :}$

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How do you solve $Y = 2 x^{2} - 3$ $Y=2x^2-3$ and $Y = 3 x - 1$ $Y=3x-1$ ?

1 Answer

Explanation:

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How do you solve Y=2x^2-3Y=2x2−3Y=2x^2-3 and Y=3x-1Y=3x−1Y=3x-1?

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Explanation:

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How do you solve $Y = 2 x^{2} - 3$ $Y=2x^2-3$ and $Y = 3 x - 1$ $Y=3x-1$ ?