To solve the given system of equations:
[tex]\[
\left\{
\begin{array}{l}
\frac{4}{x+y-1}-\frac{3}{2 x-y+3}=-\frac{5}{2} \\
\frac{3}{x+y-1}+\frac{1}{2 x-y+3}=-\frac{7}{5}
\end{array}
\right.
\][/tex]
we need to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex] such that both equations are satisfied. Finally, we will add the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find [tex]\( x + y \)[/tex].
### Steps to Approach:
1. Label the Equations:
Let's label the equations for convenience:
[tex]\[
\text{(1)} \quad \frac{4}{x+y-1} - \frac{3}{2x-y+3} = -\frac{5}{2}
\][/tex]
[tex]\[
\text{(2)} \quad \frac{3}{x+y-1} + \frac{1}{2x-y+3} = -\frac{7}{5}
\][/tex]
2. Introduce Substitution Variables:
Let:
[tex]\[
A = \frac{1}{x+y-1}
\][/tex]
[tex]\[
B = \frac{1}{2x-y+3}
\][/tex]
Rewrite the equations using [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[
4A - 3B = -\frac{5}{2} \quad \text{(3)}
\][/tex]
[tex]\[
3A + B = -\frac{7}{5} \quad \text{(4)}
\][/tex]
3. Solve the Linear System:
From equation (4):
[tex]\[
3A + B = -\frac{7}{5}
\][/tex]
Solving for [tex]\( B \)[/tex]:
[tex]\[
B = -\frac{7}{5} - 3A
\][/tex]
Substitute this expression for [tex]\( B \)[/tex] into equation (3):
[tex]\[
4A - 3\left(-\frac{7}{5} - 3A\right) = -\frac{5}{2}
\][/tex]
Simplify and solve for [tex]\( A \)[/tex]:
[tex]\[
4A + \frac{21}{5} + 9A = -\frac{5}{2}
\][/tex]
[tex]\[
13A + \frac{21}{5} = -\frac{5}{2}
\][/tex]
Multiply through by 10 to clear the fractions:
[tex]\[
130A + 42 = -25
\][/tex]
[tex]\[
130A = -67
\][/tex]
[tex]\[
A = -\frac{67}{130}
\][/tex]
Substitute [tex]\( A \)[/tex] back into the expression for [tex]\( B \)[/tex]:
[tex]\[
B = -\frac{7}{5} - 3\left(-\frac{67}{130}\right)
\][/tex]
Simplify [tex]\( B \)[/tex]:
[tex]\[
B = -\frac{7}{5} + \frac{201}{130}
\][/tex]
[tex]\[
B = -\frac{7 \cdot 26}{130} + \frac{201}{130}
\][/tex]
[tex]\[
B = -\frac{182}{130} + \frac{201}{130}
\][/tex]
[tex]\[
B = \frac{201-182}{130} = \frac{19}{130}
\][/tex]
4. Back-substitute for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
From [tex]\( A = \frac{1}{x+y-1} \)[/tex]:
[tex]\[
x+y-1 = \frac{1}{A} = \frac{130}{-67}
\][/tex]
[tex]\[
x + y - 1 = -\frac{130}{67}
\][/tex]
Therefore:
[tex]\[
x + y = 1 - \frac{130}{67}
\][/tex]
Finally:
[tex]\[
x + y = 1 - 1.940298507462687
\][/tex]
[tex]\[
x + y = -0.940298507462687
\][/tex]
Thus, the value of [tex]\( x + y \)[/tex] is closest to [tex]\(-1\)[/tex].
So the correct answer is:
[tex]\[
\boxed{-1}
\][/tex]
