How do you combine #t/(6+5t+t^2)-2/(2+3t+t^2)#?

Question

Nathan Axel · Accepted Answer

#(t-3)/(t^2+4t+3)# We know that, #t^3+5t+6=t^2+3t+2t+6=t(t+3)+2(t+3)=(t+3)(t+2)#And, #t^2+3t+2=t^2+2t+t+2=t(t+2)+1(t+2)=(t+2)(t+1)#We have, #t/(t^2+5t+6)-2/(t^2+3t+2)##=t/((t+3)(t+2))-2/((t+2)(t+1))# #=(t(t+1)-2(t+3))/((t+1)(t+2)(t+3))=(t^2+t-2t-6)/((t+1)(t+2)(t+3))##=(t^2-t-6)/((t+1)(t+2)(t+3))# #=(t^2-3t+2t-6)/((t+1)(t+2)(t+3))=((t-3)cancel((t+2)))/((t+1)cancel((t+2))(t+3))=(t-3)/((t+1)(t+3))=(t-3)/(t^2+4t+3)#

Eli Assouline · Answer

undefined To combine the given expressions, we need to find a common denominator and then combine the fractions. The common denominator for both fractions is (t^2 + 3t + 2).

The first fraction, t/(6 + 5t + t^2), can be multiplied by (2 + 3t + t^2)/(2 + 3t + t^2) to obtain (t * (2 + 3t + t^2))/(6 + 5t + t^2).

The second fraction, -2/(2 + 3t + t^2), can be multiplied by (6 + 5t + t^2)/(6 + 5t + t^2) to obtain (-2 * (6 + 5t + t^2))/(2 + 3t + t^2).

Combining the fractions, we have (t * (2 + 3t + t^2))/(6 + 5t + t^2) - (-2 * (6 + 5t + t^2))/(2 + 3t + t^2).

Simplifying further, we get (2t + 3t^2 + t^3 - (-12 - 10t - 2t^2))/(6 + 5t + t^2).

Combining like terms, we have (2t + 3t^2 + t^3 + 12 + 10t + 2t^2)/(6 + 5t + t^2).

Finally, simplifying the numerator and denominator, we get (t^3 + 5t^2 + 12t + 12)/(t^2 + 5t + 6).