We are given a diagram where $AOB$ is a straight line. We are also given the following angle measures: $\angle AOC = 3(x+y)^{\circ}$, $\angle COB = 45^{\circ}$, $\angle AOD = (5x+y)^{\circ}$, and $\angle DOB = y^{\circ}$. The problem asks us to find the values of $x$ and $y$.

1. Problem Description

We are given a diagram where $AOB$ is a straight line. We are also given the following angle measures: $\angle AOC = 3(x+y)^{\circ}$, $\angle COB = 45^{\circ}$, $\angle AOD = (5x+y)^{\circ}$, and $\angle DOB = y^{\circ}$. The problem asks us to find the values of $x$ and $y$.

2. Solution Steps

Since $AOB$ is a straight line, the sum of the angles on one side of the line is $180^{\circ}$. We can write the equation

$\angle AOC + \angle COB + \angle BOD = 180^{\circ}$.

Substituting the given angle measures, we have

$3(x+y) + 45 + y = 180$.

Simplifying the equation, we get

$3x + 3y + 45 + y = 180$

$3x + 4y = 180 - 45$

$3x + 4y = 135$ (Equation 1)

Also, since $AOB$ is a straight line, $\angle AOD + \angle DOB = 180 - \angle AOD - \angle DOB = 180^{\circ}$.

So $\angle AOD + \angle DOB = 180^{\circ}$

$(5x+y) + y = 180$

$5x + 2y = 180$ (Equation 2)

Now we have a system of two equations with two variables:

$3x + 4y = 135$ (Equation 1)

$5x + 2y = 180$ (Equation 2)

We can solve for $x$ and $y$ using substitution or elimination. Let's use elimination.

Multiply Equation 2 by 2:

$10x + 4y = 360$ (Equation 3)

Subtract Equation 1 from Equation 3:

$(10x + 4y) - (3x + 4y) = 360 - 135$

$7x = 225$

$x = \frac{225}{7}$

Substitute $x = \frac{225}{7}$ into Equation 2:

$5(\frac{225}{7}) + 2y = 180$

$\frac{1125}{7} + 2y = 180$

$2y = 180 - \frac{1125}{7}$

$2y = \frac{1260 - 1125}{7}$

$2y = \frac{135}{7}$

$y = \frac{135}{14}$

3. Final Answer

$x = \frac{225}{7}$

$y = \frac{135}{14}$