[6t-5 / 3t+1] + [4t+3 / 3t+1] help :[ please!?

Question

alqebra homework.
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Answer 1

Since the denominator is the same, just add up the numerators on top:

[(6t-5) + (4t+3)] / (3t + 1)

= (10t - 2) / (3t +1)

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

Answer 2

You already have a common denominator, so just add the two fractions together, and you get

(6t-5+4t+3)/(3t+1)

Now just combine common terms.

(10t-2)/(3t+1)

You could even go further and factor the top portion of the fraction.

2(5t-1) all over 3t+1

Is that what you were supposed to do?

Answer 3

Answer :
(10t - 2)/(3t + 1) or 2(5t - 1)/(3t + 1)

Process :

Since 3t+1 is your common denominator, simply add the numerator and copy the common denominator.

6t - 5 + 4t + 3
= --------------------
3t +1

(6t + 4t) + (-5 + 3)
= -------------------------
3t + 1

10t - 2
= -----------
3t + 1

Answer 4

(6t - 5)/(3t + 1) + (4t + 3)/(3t + 1)
= (6t - 5 + 4t + 3) / (3t + 1)
= (10t - 2) / (3t + 1)
= 2.(5t - 1) / (3t + 1)

Answer 5

[6t-5/3t+1]+[4t+3/3t+1]=
(10t-2)/(3t+1)=
2(5t-1)/(3t+1)

Answer 6

a million. t / 8tz * 4z / 3t = 4tz / 24t^2z = a million / 6t 2. 4x-8 / x-a million * x^2-a million / x^2-2 = 4(x-2)/(x-a million) * (x-a million)(x+a million) / (x-2)(x+2) = 4(x-2)(x-a million)(x+a million) / (x-a million)(x-2)(x+2)= 4(x+a million)/(x+2) 3. (z/12tx) / (3z/4t) = z/12tz * 4t/3z = 4tz/36tz^2 = a million/9z 4. (15n^5 / 7x^3y^4) / (5mn^2/14x^2y) = 15n^5 / 7x^3y^4 * 14x^2y/5mn^2 = 15n^5 * 14 x^2y / 7x^3y^4 * 5mn^2 = 6n^3/xy^3m i wish the numbers are precise. reliable success !