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

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

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

solve for

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

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

Solution steps

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

Expand

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

Expand

(x + y) : x + y

p y + p z + 2 q + x q = x + y

Move

p y

to the right side

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

Move

p z

to the right side

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

Move

2 q

to the right side

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

Move

x

to the left side

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

Factor

x q - x : x (q - 1)

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

Divide both sides by

q - 1; q \neq = 1

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

x, 0 = 8 + x^{9} - z^{-}

x, 5 x + 3 y = 137

x, (x - 3) (y + 4) = 6

7 = 3 e^{- 2} \cdot x

k = \frac{2.303}{100} \frac{lo g _{10} ( 0.5 )}{0.488}

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