13. \( \frac{(2x+7)}{5}+\frac{(x-3)}{2}=\frac{3(x+1)}{10} \)

To solve the equation \( \frac{(2x+7)}{5}+\frac{(x-3)}{2}=\frac{3(x+1)}{10} \), we start by finding a common denominator. The least common multiple of the denominators \(5\), \(2\), and \(10\) is \(10\). Rewriting each term with a denominator of \(10\), we get: \[ \frac{2(2x+7)}{10} + \frac{5(x-3)}{10} = \frac{3(x+1)}{10} \] Now, combining the fractions on the left side: \[ \frac{4x + 14 + 5x - 15}{10} = \frac{3(x+1)}{10} \] Simplifying the left side: \[ \frac{(4x + 5x + 14 - 15)}{10} = \frac{3(x+1)}{10} \] This simplifies to: \[ \frac{(9x - 1)}{10} = \frac{3(x+1)}{10} \] Now, remove the common denominator (multiply both sides by \(10\)): \[ 9x - 1 = 3(x + 1) \] Expanding the right side: \[ 9x - 1 = 3x + 3 \] Now, collect terms involving \(x\) on one side and constants on the other side: \[ 9x - 3x = 3 + 1 \] This gives: \[ 6x = 4 \] Dividing both sides by \(6\): \[ x = \frac{4}{6} = \frac{2}{3} \] Thus, the solution to the equation is: \[ \boxed{\frac{2}{3}} \]