# determine the value of k can i cross multiply ?2x^2 -9x+k= 2x+5 _________ ______ 3x^2-22x+7 3x-1

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(2x^2 -9x +k)/(3x^2-22x+7) = (2x+5)/(3x-1)

But (3x^2 -22x +7)= (3x-1)(x-7)

==> (2x^2 -9x +k)/ (3x-1)(x-7)= (2x+5)/(3x-1)

Now multiply by (3x-1):

==> (2x^2 -9x + k)/(x-7)= (2x+5)

Now let us try and complete the square:

==> (2x^2 -9x -35 + 35 + k)/(x-7) = (2x+5)

==> [(2x^2 -9x -35) + (k+35)]/(x-7)= (2x+5)

Factorize:

==>[ (2x+5)(x-7)+ (k+35)]/(x-7)= 2x+5

==> (2x+5) + (k+35)/(x-7) = (2x+5)

Now comparing both sides, we conclude that in order for the equality to be true, then (k+35)/(x-7) must equal zero.

Then k+ 35 =0

==> **k= -35**

To check:

(2x^2 -9x -35)/ (3x^2 -22x + 7) = (2x+5)/(3x-1)

(2x+5)(x-7)/(3x-1)(x-7) = (2x+5)/(3x-1)

Reduce similar:

(2x+5)/(3x-1) = (2x+5)/(3x-1)

(2x^2-9x+k)/(3x^2-22x+7) =(2x+5)/(3x-1). To solve for k you can cross multiply and solve for k if possible only.

(2x^2-9x+k)(3x-1) = (3x^2-22x+7)(2x+5)

6x^3-27x^2+3kx-2x^2+9x-k = 6x^3-44x^2+14x+15x^2-110x+35. Simplify by collecting the like terms.

6x^3+(-27-2)x^2 +(3k+9)x-k = 6x^3+(-44+15)x^2+(14-110)x+35.

6x^3-29x^2+(3k+9)x-k = 6x^3 -29x^2-96x+35. If this is an identy, then we each terms on both sides must be equal. Therefore,

(3k+9)x = -96x ....(1) and -k = 35....(-2) .

So k =-35 satisfies both (1) and (2).

Therefore k =-35.