# Two reacting substances which obey the law of mass action starting with 80 kg and 60 kg after chemical reaction leads to the equation.y=1/(80-x)(60-x)=k+ct

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We'll integrate to get a logarithmic formula for x and to determine the amount of reaction product.

Int dx/(80-x)(60-x) = k Int dt

We'll solve the integral using the partial fractions method.

We'll write the integrand as an algebraic sum of irreducible ratios.

1/(80-x)(60-x) = A/(80-x) + B/(60-x)

We'll calculate LCD of the ratios from the right side:

A/(80-x) + B/(60-x) = [A(60-x) + B(80-x)]/(80-x)(60-x)

We'll remove the brackets:

1 = 60A - Ax + 80B - Bx

We'll combine like terms:

1 = x(-A-B) + 60A + 80B

The coefficients of x from both sides have to be equal:

-A - B = 0

A = -B

and free terms:

60A + 80B = 1

-60B + 80B = 1

20B = 1

B = 1/20

A = -1/20

1/(80-x)(60-x) = -1/20(80-x) + 1/20(60-x)

Int dx/(80-x)(60-x) = Int-dx/20(80-x) + Int dx/20(60-x)

We'll substitute 80 - x = t

We'll differentiate:

-dx = dt

Int-dx/20(80-x) = Int dt/t

(Int dt/t)/20 = (ln |t|)/20 + C

Int-dx/20(80-x) = (ln |80 - x|)/20 + C

Int dx/20(60-x) = (-ln |60 - x|)/20 + C

Int dx/(80-x)(60-x) = (ln |80 - x| - ln |60 - x|)/20 + C

We'll use the quotient property of logarithms:

**Int dx/(80-x)(60-x) = (1/20)*ln |(80-x)/(60-x)| + C**