Math  /  Algebra

Question1+6x4y2y=3z9y=109\frac{1+\frac{6 x}{4-y}}{2 y}=\frac{3 z}{9 y}=\frac{10}{9}, then 3x+z2y+3y=\frac{3 x+z}{2 y+3 y}=

Studdy Solution
Substitute xx and zz into the expression 3x+z2y+3y\frac{3x+z}{2y+3y}.
First, simplify the denominator:
2y+3y=5y2y + 3y = 5y
Substitute z=10y3z = \frac{10y}{3} and x=(20y9)(4y)54x = \frac{(20y - 9)(4-y)}{54} into the numerator:
3x+z=3((20y9)(4y)54)+10y33x + z = 3\left(\frac{(20y - 9)(4-y)}{54}\right) + \frac{10y}{3}
Simplify:
3x=(20y9)(4y)183x = \frac{(20y - 9)(4-y)}{18}
Combine terms:
3x+z=(20y9)(4y)18+10y33x + z = \frac{(20y - 9)(4-y)}{18} + \frac{10y}{3}
Simplify the fraction:
=(20y9)(4y)+60y18= \frac{(20y - 9)(4-y) + 60y}{18}
Now divide by 5y5y:
3x+z5y=(20y9)(4y)+60y185y\frac{3x + z}{5y} = \frac{\frac{(20y - 9)(4-y) + 60y}{18}}{5y}
Simplify further to find the final expression.
The final simplified expression is:
109\boxed{\frac{10}{9}}

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