Solve the equation

Question

3(2ā +5)+4(3ā+1)= 109

Answer 1

To solve the equation \(3(2ā + 5) + 4(3ā + 1) = 109\), we will distribute the terms and combine like terms.

1. Start by distributing the \(3\) and \(4\):

\[ 3(2ā) + 3(5) + 4(3ā) + 4(1) = 109 \]

This simplifies to:

\[ 6ā + 15 + 12ā + 4 = 109 \]

2. Combine like terms (\(6ā\) and \(12ā\), and \(15\) and \(4\)):

\[ (6ā + 12ā) + (15 + 4) = 109 \]

This gives:

\[ 18ā + 19 = 109 \]

3. Next, isolate \(18ā\) by subtracting \(19\) from both sides:

\[ 18ā = 109 - 19 \] \[ 18ā = 90 \]

4. Now, divide both sides by \(18\):

\[ ā = \frac{90}{18} \] \[ ā = 5 \]

Thus, the solution to the equation is

\[ ā = 5 \]