Question

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Which of the following is the simplified form of the quantity 4 y squared minus 1 over the quantity 2 y squared plus 3 y minus 2 ?

2y minus 1 over y minus 2 , with the restrictions y ≠ –1 over 2 and y ≠ –2

2y plus 1 over y plus 2 , with the restrictions y ≠ 1 over 2 and y ≠ 2

2y minus 1 over y minus 2, with the restriction y ≠ –1 over 2 and y ≠ 2

2y plus 1 over y plus 2, with the restriction y ≠ 1 over 2 and y ≠ –2

Answer 1

Lets write out the original equation as \( \large \frac{ 4y^2-1}{2y^2+3y-2}\)

Answer 2

Okay,

Answer 3

So now we have to factor the numerator and the denomanator
So lets start with the numerator
\(4y^2-1\) which we know has the form \((a^2-b^2)=(a+b)(a-b)\)
So in our case a^2=4y^2 so a=2y and b^2=1 so b=1
Therefore \(4y^2-1=(2y+1)(2y-1)\)

Answer 4

Okai

Answer 5

Next we factor the denomanator \(2y^2+3y-2\)
A=2 B=3 and C=-2
A*C=-4
So what 2 numbers when you multiply them it equals -4 and when you add them it equals 3
so the numbers are 4 and -1
4*-1=-4
4-1=3
So \( 2y^2+4y-y-2\)
\(2y(y+2)-1(y+2)\)
\( (2y-1)(y+2)\)
So now we can rewrite our equation

Answer 6

OK.

Answer 7

\(\huge \frac{ 4y^2-1}{2y^2+3y-2}=\frac{(2y+1)(2y-1)}{(2y-1)(y+2)}=\frac{(2y+1)}{(y+2)}\)

Answer 8

Now you gotta choose the right sentence

Answer 9

Okay thank you.