To solve the system of equations:
- \( 9a + 3b = 30 \) (Equation 1)
- \( 8a + 4b = 28 \) (Equation 2)
We can simplify both equations.
Starting with Equation 1, we can divide every term by 3:
\[ 3a + b = 10 \quad \text{(Equation 3)} \]
Now, let's simplify Equation 2 by dividing every term by 4:
\[ 2a + b = 7 \quad \text{(Equation 4)} \]
Now we have the simplified system:
- \( 3a + b = 10 \) (Equation 3)
- \( 2a + b = 7 \) (Equation 4)
Next, we can eliminate \( b \) by subtracting Equation 4 from Equation 3:
\[ (3a + b) - (2a + b) = 10 - 7 \]
This simplifies to:
\[ 3a - 2a + b - b = 3 \]
So, we have:
\[ a = 3 \]
Now, we can substitute this value of \( a \) back into either Equation 3 or Equation 4 to find \( b \). Using Equation 4:
\[ 2(3) + b = 7 \]
This becomes:
\[ 6 + b = 7 \]
Subtracting 6 from both sides gives:
\[ b = 1 \]
Thus, the solution to the system of equations is:
\[ a = 3, \quad b = 1 \]