28.7k views
3 votes
G'day!
Can anyone please explain taking LCM of 2/t + 1/1+t = -3/2+t

User Ixrock
by
8.1k points

1 Answer

6 votes

Answer:

To solve the equation 2/t + 1/(1+t) = -3/(2+t), we first need to find the least common multiple (LCM) of the denominators, which are t and 1+t, and then rewrite each fraction with the LCM as its denominator.

The LCM of t and 1+t is (t)(1+t) or t(t+1). To rewrite the fractions with this common denominator, we need to multiply the first fraction by (t+1)/(t+1) and the second fraction by t/t:

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

Simplifying each fraction, we get:

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

Combining the fractions on the left side, we get:

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

Simplifying further:

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

Now, we can cross-multiply and simplify:

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

6t^2 + 11t + 4 = -3t^2 - 3t

9t^2 + 14t + 4 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 9, b = 14, and c = 4.

Plugging in these values, we get:

t = (-14 ± sqrt(14^2 - 4(9)(4))) / 2(9)

t = (-14 ± sqrt(136)) / 18

t = (-14 ± 2sqrt(34)) / 18

Simplifying the expression, we get:

t = (-7 ± sqrt(34)) / 9

These are the two possible solutions for t that satisfy the original equation.

User Sam Meldrum
by
8.0k points

Related questions

2 answers
4 votes
111k views