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Sagot :
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]
[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]
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