What is the answer to solvefor z,p(1+q^2)=q(z-1) ?

The answer to solvefor z,p(1+q^2)=q(z-1) is z=(p(1+q^2))/q+1;\quad q\ne 0

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solvefor z,p(1+q^2)=q(z-1)

solve for

z, p (1 + q^{2}) = q (z - 1)

z = \frac{p ( 1 + q ^{2} )}{q} + 1; q \neq = 0

Solution steps

p (1 + q^{2}) = q (z - 1)

Switch sides

q (z - 1) = p (1 + q^{2})

Divide both sides by

q; q \neq = 0

z - 1 = \frac{p ( 1 + q ^{2} )}{q}; q \neq = 0

Move

1

to the right side

z = \frac{p ( 1 + q ^{2} )}{q} + 1; q \neq = 0

p, 3 q + p = 10

0.12 (2 x - 5) + 3 x = 81

y, 2 y + 10 y = x

x, 2 x + y = 24.5

x, 3 y - 2 x + 12 = 0

What is the answer to solvefor z,p(1+q^2)=q(z-1) ?
The answer to solvefor z,p(1+q^2)=q(z-1) is z=(p(1+q^2))/q+1;\quad q\ne 0