We are given two conditions relating two numbers $x$ and $y$. First, their sum is 35, so $x + y = 35$. Second, when $x$ is divided by 4 and $y$ is divided by 3, the sum of the quotients is 8, so $\frac{x}{4} + \frac{y}{3} = 8$. We need to solve this system of two equations for $x$ and $y$.

1. Problem Description

We are given two conditions relating two numbers $x$ and $y$.

First, their sum is 35, so $x + y = 35$.

Second, when $x$ is divided by 4 and $y$ is divided by 3, the sum of the quotients is 8, so $\frac{x}{4} + \frac{y}{3} = 8$.

We need to solve this system of two equations for $x$ and $y$.

2. Solution Steps

We have the system of equations:

$x + y = 35$

$\frac{x}{4} + \frac{y}{3} = 8$

From the first equation, we can express $y$ in terms of $x$:

$y = 35 - x$

Substitute this expression for $y$ into the second equation:

$\frac{x}{4} + \frac{35 - x}{3} = 8$

Multiply both sides of the equation by 12 (the least common multiple of 4 and 3) to eliminate the fractions:

$12 (\frac{x}{4} + \frac{35 - x}{3}) = 12(8)$

$3x + 4(35 - x) = 96$

$3x + 140 - 4x = 96$

$-x + 140 = 96$

$-x = 96 - 140$

$-x = -44$

$x = 44$

Now, substitute the value of $x$ back into the equation $y = 35 - x$:

$y = 35 - 44$

$y = -9$

3. Final Answer

$x = 44$ and $y = -9$