To solve the system of linear equations:

\[ 
\begin{cases} 
A + B = 11 \\
3A + 2B = 30 
\end{cases}
\]

we can use the substitution or elimination method. Here, we'll use the substitution method.

1. From the first equation, solve for \( B \):

\[ 
B = 11 - A 
\]

2. Substitute \( B \) into the second equation:

\[ 
3A + 2(11 - A) = 30 
\]

3. Simplify and solve for \( A \):

\[ 
3A + 22 - 2A = 30 
\]
\[ 
A + 22 = 30 
\]
\[ 
A = 30 - 22 
\]
\[ 
A = 8 
\]

4. Substitute \( A = 8 \) back into the equation \( B = 11 - A \):

\[ 
B = 11 - 8 
\]
\[ 
B = 3 
\]

So, the solution to the system is:

\[ 
A = 8 \quad \text{and} \quad B = 3 
\]

Would you like more details or have any questions about this solution?

Here are 5 more questions you might want to ask:
1. How do you solve a system of equations using the elimination method?
2. Can you explain the steps to graphically solve a system of linear equations?
3. What are the methods to solve a quadratic equation?
4. How can matrix algebra be used to solve systems of linear equations?
5. What is the significance of the determinant in solving linear equations?

**Tip:** When solving systems of linear equations, it's often helpful to check your solution by substituting the values back into the original equations to ensure they satisfy both equations.

Discover how to solve the system of equations A + B = 11 and 3A + 2B = 30 step-by-step. This solution covers key algebraic concepts and methods, including substitution and elimination techniques, suitable for students in Grades 7-9.

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