Solve by the addition method. (If there is no solution, enter NO SOLUTION. Use the parameters *x* and *y* as necessary.)

5*x*

6

 + 

*y*

3

 = 

23

6

2*x*

3

 − 

*y*

2

 = 

23

6

We are given the following system of equations:

\[
\frac{5x}{6} + \frac{y}{3} = \frac{23}{6}
\]
\[
\frac{2x}{3} - \frac{y}{2} = \frac{23}{6}
\]

### Step 1: Eliminate fractions by multiplying both equations by the least common denominator (LCD).

For the first equation, the LCD of 6 is 6:
\[
6 \times \left(\frac{5x}{6} + \frac{y}{3}\right) = 6 \times \frac{23}{6}
\]
\[
5x + 2y = 23
\]

For the second equation, the LCD of 6 is 6:
\[
6 \times \left(\frac{2x}{3} - \frac{y}{2}\right) = 6 \times \frac{23}{6}
\]
\[
4x - 3y = 23
\]

Now the system is:
\[
5x + 2y = 23
\]
\[
4x - 3y = 23
\]

### Step 2: Solve by the addition method.

We will multiply the first equation by 3 and the second equation by 2, so that the coefficients of \(y\) become opposites:

First equation:
\[
3(5x + 2y) = 3(23)
\]
\[
15x + 6y = 69
\]

Second equation:
\[
2(4x - 3y) = 2(23)
\]
\[
8x - 6y = 46
\]

Now we have:
\[
15x + 6y = 69
\]
\[
8x - 6y = 46
\]

### Step 3: Add the equations to eliminate \(y\).

\[
(15x + 6y) + (8x - 6y) = 69 + 46
\]
\[
23x = 115
\]

### Step 4: Solve for \(x\).

\[
x = \frac{115}{23} = 5
\]

### Step 5: Substitute \(x = 5\) into one of the original equations to solve for \(y\).

Using the first equation:
\[
5x + 2y = 23
\]
Substitute \(x = 5\):
\[
5(5) + 2y = 23
\]
\[
25 + 2y = 23
\]
\[
2y = 23 - 25
\]
\[
2y = -2
\]
\[
y = \frac{-2}{2} = -1
\]

### Final Solution:

\[
x = 5, \quad y = -1
\]

Would you like further details on any step or have questions?

### Related Questions:
1. What happens if we use substitution instead of addition to solve this system?
2. Can you solve a system where both variables cancel out and lead to no solution?
3. How does the LCD help in clearing fractions in the system of equations?
4. What happens if both variables cancel out and lead to an identity?
5. How can you verify the solution of a system of equations?

#### Tip:
When solving systems of equations, always check if it's easier to eliminate fractions first by multiplying through by the LCD.

Solve by the addition method. (If there is no solution, enter NO SOLUTION. Use the parameters *x* and *y* as necessary.) 5x/6 + y/3 = 23/6 and 2x/3 - y/2 = 23/6.

Learn how to solve the system of linear equations 5x/6 + y/3 = 23/6 and 2x/3 - y/2 = 23/6 using the addition method. This step-by-step solution covers multiplying to eliminate fractions, adding equations, and solving for variables. Suitable for Grades 8-10 students.

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