How do you solve and check the solutions of:
x^2 + x - 12 / x - 3 = 15
and
2x^2 + 8x - 10 / 2x^2 + 14x + 20 = 4
3 answers
If those are fractions in your equations... you might want to clear them out first by using a common denominator : )
I will do the hard one, you do the easier one
2x^2 + 8x - 10 / 2x^2 + 14x + 20 = 4
You probably meant:
(2x^2 + 8x - 10) / (2x^2 + 14x + 20) = 4 , those brackets are essential
2(x^2 + 4x - 5)/(2(x^2 + 7x + 10) = 4
(x+5)(x-1)/((x+5)(x+2)) = 4
(x-1)/(x+2) = 4 , as long as x ≠ -5
4x + 8 = x - 1
3x = -9
x = -3
2x^2 + 8x - 10 / 2x^2 + 14x + 20 = 4
You probably meant:
(2x^2 + 8x - 10) / (2x^2 + 14x + 20) = 4 , those brackets are essential
2(x^2 + 4x - 5)/(2(x^2 + 7x + 10) = 4
(x+5)(x-1)/((x+5)(x+2)) = 4
(x-1)/(x+2) = 4 , as long as x ≠ -5
4x + 8 = x - 1
3x = -9
x = -3
LOL! Thanks Reiny! I totally missed where those brackets would be : )
1 answer