What is the answer to solvefor a,19ab+1+9+a+b=1987 ?

The answer to solvefor a,19ab+1+9+a+b=1987 is a=(-b+1977)/(19b+1);\quad b\ne-1/19

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solvefor a,19ab+1+9+a+b=1987

solve for

a, 19 ab + 1 + 9 + a + b = 1987

a = \frac{- b + 1977}{19 b + 1}; b \neq = - \frac{1}{19}

Solution steps

19 ab + 1 + 9 + a + b = 1987

Add the numbers:

1 + 9 = 10

19 ab + b + a + 10 = 1987

Move

b

to the right side

19 ab + a + 10 = 1987 - b

Move

10

to the right side

19 ab + a = - b + 1977

Factor

19 ab + a : a (19 b + 1)

a (19 b + 1) = - b + 1977

Divide both sides by

19 b + 1; b \neq = - \frac{1}{19}

a = \frac{- b + 1977}{19 b + 1}; b \neq = - \frac{1}{19}

q, p = 2 q (r^{2} - s^{2})

\frac{\frac{( \frac{2}{7} x + \frac{3}{7} )}{11}}{5} - \frac{3}{7} = 1

b, 1.5 b + 0.4 d = 1.8

a, 4 a - 4 + c = 0

3 x + 7 = 7 x + 15

What is the answer to solvefor a,19ab+1+9+a+b=1987 ?
The answer to solvefor a,19ab+1+9+a+b=1987 is a=(-b+1977)/(19b+1);\quad b\ne-1/19