[(8x)/(xto the 2nd power-4x-5)]+[6/(xto the 2nd power-7x+10)]?

Answer 1

[ (8x) / (x^2 - 4x - 5) ] + [ 6 / (x^2 - 7x +10) ]
[ (8x) / (x-5)(x+1) ] + [ 6 / (x-5)(x-2) ]
you have a common factor of (x-5) in the denominator of both terms.
[ (8x)(x-2) / (x-5)(x+1)(x-2) ] + [ 6(x+1) / (x-5)(x-2)(x+1) ]
now that you have a common denominator, you can add them.
[ (8x)(x-2) + 6(x+1) ] / (x-5)(x-1)(x-2)
(8x^2 - 16x + 6x +6) / (x-5)(x-1)(x-2)
(8x^2 - 10x + 6) / (x-5)(x-1)(x-2)
2(4x^2 -5x + 3) / (x-5)(x-1)(x-2)

Answer 2

nicely, you would possibly want to attempt this 2 approaches. both easily take the powers and then plug them to their bases (5), and multiply both products, or a really effortless shortcut, is once you're multiplying 2 numbers with exponents, AND in the experience that they have got an same base (5) then you actually can take the bottom, and upload both exponents at the same time: 5^10+2, and calculate that. with any success you've a calculator to attempt this, that could want to be extremely stressful on paper.. you need to pop out to 244,one hundred and forty,625...yet try it your self too.

Answer 3

This is a standard addition of fractions, although it is a little more tedious.

Hint: The denominators have a common factor.

Answer 4

How can this be answered? Usually you solve for X. But now there are two variables, X and the answer of the equation.

Answer 5

(8x)/(x^2-4x-5)+6/(x^2-7x+10)
=8x/(x-5)(x+1)+6/(x-5)(x-2)
LCD is (x+1)(x+2)(x-5)
8x(x-2)+6(x+1)/(x+1)(x-2)(x-5)
=8x^2-16x+6x+6/(x+1)(x-2)(x-5)
8x^2-10x+6/(x+1)(x-2)(x-5)

Answer 6

=You tell us and your teacher