SOLUTION: (m + n) / (m - n) + (m - n) / (m + n) Find the sum. Not sure how to continue.

Algebra.Com
Question 1133895: (m + n) / (m - n) + (m - n) / (m + n)
Find the sum.

Not sure how to continue.

Found 2 solutions by algebrahouse.com, MathTherapy:
Answer by algebrahouse.com(1659)   (Show Source): You can put this solution on YOUR website!
(m + n)/(m - n) + (m - n)(m + n)
(m + n)(m + n)/(m + n)(m - n) + (m - n)(m - n)/(m + n)(m - n) {got common denominator of (m + n)(m - n)}
= (m² + 2mn + n²)/(m + n)(m - n) + (m² - 2mn + n²)/(m + n)(m - n) {multiplied in numerators}
= (2m² + 2n²)/(m + n)(m - n) {combined numerators over common denominator}

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

(m + n) / (m - n) + (m - n) / (m + n)
Find the sum.

Not sure how to continue.

------- Multiplying each expression by LCD, (m - n)(m + n)
<======= CORRECT ANSWER
RELATED QUESTIONS

(m + n) /(m - n) + ( m - n) / (m + n) Determine sum. m^2 + n^2 + m^2 - n^2 (cross... (answered by greenestamps)
m-n/n-m (answered by jim_thompson5910,mananth)
(m-n)-(m+n) (answered by Alan3354)
m/(m-n)(m+n)/n+1/(m+n)^2 (answered by mangopeeler07)
(a^m+n)^m-n (answered by Fombitz)
The sum m & n ,divided by the difference m &... (answered by fractalier)
(-5)(m)(m)(m)(n) (answered by edjones)
Find the sum. 5/m^2n^2 + m/n (answered by Alan3354)
the expression (m-n)- (n-m) is equivalent to (answered by vleith,MathLover1)