What is the answer to solvefor x,(y-x)p+(2-x)q=x-y ?

The answer to solvefor x,(y-x)p+(2-x)q=x-y is x=-(-y-py-2q)/(1+p+q);\quad p\ne-q-1

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solvefor x,(y-x)p+(2-x)q=x-y

solve for

x, (y - x) p + (2 - x) q = x - y

x = - \frac{- y - p y - 2 q}{1 + p + q}; p \neq = - q - 1

Solution steps

(y - x) p + (2 - x) q = x - y

Expand

(y - x) p + (2 - x) q : p y - p x + 2 q - x q

p y - p x + 2 q - x q = x - y

Move

p y

to the right side

- p x + 2 q - x q = x - y - p y

Move

2 q

to the right side

- p x - x q = x - y - p y - 2 q

Move

x

to the left side

- p x - x q - x = - y - p y - 2 q

Factor

- p x - x q - x : - x (p + q + 1)

- x (p + q + 1) = - y - p y - 2 q

Divide both sides by

- (p + q + 1); p \neq = - q - 1

x = - \frac{- y - p y - 2 q}{1 + p + q}; p \neq = - q - 1

4 x - 15 = 33

a, 5.2 \cdot 6.1 = \frac{a}{b}

x, 3 x + 4 y - \frac{20}{3} = 0

y, 2 y + 3 (2 y - 5) + 4 = y + 3 (2 y - z) - 5

3.3 x + 3^{2} x = 90

What is the answer to solvefor x,(y-x)p+(2-x)q=x-y ?
The answer to solvefor x,(y-x)p+(2-x)q=x-y is x=-(-y-py-2q)/(1+p+q);\quad p\ne-q-1