Question $\frac{1+\frac{6 x}{4-y}}{2 y}=\frac{3 z}{9 y}=\frac{10}{9}$ , then $\frac{3 x+z}{2 y+3 y}=$

Studdy Solution

STEP 1

1. The given equation involves fractions and multiple variables.
2. We need to solve for the expression $\frac{3x+z}{2y+3y}$ .
3. The equation $\frac{1+\frac{6x}{4-y}}{2y}=\frac{3z}{9y}=\frac{10}{9}$ implies that both fractions are equal to $\frac{10}{9}$ .

STEP 2

1. Solve for $z$ using $\frac{3z}{9y} = \frac{10}{9}$ .
2. Solve for $x$ using $\frac{1+\frac{6x}{4-y}}{2y} = \frac{10}{9}$ .
3. Substitute $x$ and $z$ into the expression $\frac{3x+z}{2y+3y}$ and simplify.

STEP 3

Start by solving for $z$ from the equation $\frac{3z}{9y} = \frac{10}{9}$ .
First, simplify $\frac{3z}{9y}$ to $\frac{z}{3y}$ :
$\frac{z}{3y} = \frac{10}{9}$
Cross-multiply to solve for $z$ :
$9z = 30y$
Divide both sides by 9:
$z = \frac{30y}{9} = \frac{10y}{3}$

STEP 4

Now solve for $x$ using $\frac{1+\frac{6x}{4-y}}{2y} = \frac{10}{9}$ .
Cross-multiply:
$9(1+\frac{6x}{4-y}) = 20y$
Distribute 9:
$9 + \frac{54x}{4-y} = 20y$
Subtract 9 from both sides:
$\frac{54x}{4-y} = 20y - 9$
Multiply both sides by $4-y$ :
$54x = (20y - 9)(4-y)$
Solve for $x$ :
$x = \frac{(20y - 9)(4-y)}{54}$

STEP 5

Substitute $x$ and $z$ into the expression $\frac{3x+z}{2y+3y}$ .
First, simplify the denominator:
$2y + 3y = 5y$
Substitute $z = \frac{10y}{3}$ and $x = \frac{(20y - 9)(4-y)}{54}$ into the numerator:
$3x + z = 3\left(\frac{(20y - 9)(4-y)}{54}\right) + \frac{10y}{3}$
Simplify:
$3x = \frac{(20y - 9)(4-y)}{18}$
Combine terms:
$3x + z = \frac{(20y - 9)(4-y)}{18} + \frac{10y}{3}$
Simplify the fraction:
$= \frac{(20y - 9)(4-y) + 60y}{18}$
Now divide by $5y$ :
$\frac{3x + z}{5y} = \frac{\frac{(20y - 9)(4-y) + 60y}{18}}{5y}$
Simplify further to find the final expression.
The final simplified expression is:
$\boxed{\frac{10}{9}}$

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