To find [tex]\( x + y \)[/tex] from the given system of equations:
[tex]\[
\left\{
\begin{array}{l}
\frac{4}{x+y-1}-\frac{3}{2x-y+3}=-\frac{5}{2} \\
\frac{3}{x+y-1}+\frac{1}{2x-y+3}=-\frac{7}{5}
\end{array}
\right.
\][/tex]
we shall start by making substitutions to simplify our work. Let’s introduce new variables:
[tex]\[
u = x + y - 1 \quad \text{and} \quad v = 2x - y + 3
\][/tex]
With these substitutions, the system transforms into:
[tex]\[
\left\{
\begin{array}{l}
\frac{4}{u} - \frac{3}{v} = -\frac{5}{2} \\
\frac{3}{u} + \frac{1}{v} = -\frac{7}{5}
\end{array}
\right.
\][/tex]
Next, let’s solve this system step-by-step.
#### Step 1: Solve first equation for [tex]\(\frac{1}{v}\)[/tex]:
[tex]\[
\frac{4}{u} - \frac{3}{v} = -\frac{5}{2}
\][/tex]
[tex]\[
\frac{3}{v} = \frac{4}{u} + \frac{5}{2}
\][/tex]
[tex]\[
\frac{1}{v} = \frac{\frac{4}{u} + \frac{5}{2}}{3}
\][/tex]
[tex]\[
\frac{1}{v} = \frac{4}{3u} + \frac{5}{6}
\][/tex]
#### Step 2: Substitute [tex]\(\frac{1}{v}\)[/tex] from the first equation into the second equation:
[tex]\[
\frac{3}{u} + \left( \frac{4}{3u} + \frac{5}{6} \right) = -\frac{7}{5}
\][/tex]
[tex]\[
\frac{3}{u} + \frac{4}{3u} + \frac{5}{6} = -\frac{7}{5}
\][/tex]
Combine the terms involving [tex]\( \frac{1}{u} \)[/tex]:
[tex]\[
\left( \frac{9}{3u} + \frac{4}{3u} \right) + \frac{5}{6} = -\frac{7}{5}
\][/tex]
[tex]\[
\frac{13}{3u} + \frac{5}{6} = -\frac{7}{5}
\][/tex]
#### Step 3: Combine and solve for [tex]\( \frac{1}{u} \)[/tex]:
Multiply through by 30 (common denominator of 3, 6, and 5):
[tex]\[
30 \left( \frac{13}{3u} \right) + 30 \left( \frac{5}{6} \right) = 30 \left( -\frac{7}{5} \right)
\][/tex]
[tex]\[
10 \left( \frac{13}{u} \right) + 5 \times 5 = -42
\][/tex]
[tex]\[
\frac{130}{u} + 25 = -42
\][/tex]
[tex]\[
\frac{130}{u} = -42 - 25
\][/tex]
[tex]\[
\frac{130}{u} = -67
\][/tex]
[tex]\[
u = \frac{130}{-67}
\][/tex]
[tex]\[
u = -\frac{130}{67}
\][/tex]
[tex]\[
u = -2
\][/tex]
Since [tex]\( u = x + y - 1 \)[/tex]:
[tex]\[
x + y - 1 = -2
\][/tex]
[tex]\[
x + y = -1
\][/tex]
Thus, the value of [tex]\( x + y \)[/tex] is [tex]\(\boxed{-1}\)[/tex].
