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Sagot :
Certainly! Let's solve the system of equations step-by-step:
[tex]\[ \begin{array}{l} \frac{10}{x+y}+\frac{2}{x-y}=4 \quad \text{(Equation 1)} \\ \frac{15}{x+y}-\frac{9}{x-y}=-2 \quad \text{(Equation 2)} \end{array} \][/tex]
To solve this system, we will use substitution or elimination techniques. Let's first introduce new variables to simplify the equations:
Let:
[tex]\[ u = \frac{1}{x+y} \][/tex]
[tex]\[ v = \frac{1}{x-y} \][/tex]
Now, rewriting the original equations using [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
### Step 1: Substitute the new variables
[tex]\[ 10u + 2v = 4 \quad \text{(Transformed from Equation 1)} \][/tex]
[tex]\[ 15u - 9v = -2 \quad \text{(Transformed from Equation 2)} \][/tex]
### Step 2: Solve the transformed system of linear equations
We now have a simpler system of linear equations:
[tex]\[ \begin{array}{l} 10u + 2v = 4 \quad \text{(Equation 3)} \\ 15u - 9v = -2 \quad \text{(Equation 4)} \end{array} \][/tex]
First, let's solve Equation 3 for [tex]\(u\)[/tex]:
[tex]\[ 10u + 2v = 4 \][/tex]
[tex]\[ 10u = 4 - 2v \][/tex]
[tex]\[ u = \frac{4 - 2v}{10} \][/tex]
[tex]\[ u = \frac{2 - v}{5} \quad \text{(Equation 5)} \][/tex]
Next, we'll substitute Equation 5 into Equation 4:
[tex]\[ 15\left(\frac{2 - v}{5}\right) - 9v = -2 \][/tex]
[tex]\[ 15 \cdot \frac{2 - v}{5} - 9v = -2 \][/tex]
[tex]\[ 3(2 - v) - 9v = -2 \][/tex]
[tex]\[ 6 - 3v - 9v = -2 \][/tex]
[tex]\[ 6 - 12v = -2 \][/tex]
[tex]\[ -12v = -2 - 6 \][/tex]
[tex]\[ -12v = -8 \][/tex]
[tex]\[ v = \frac{8}{12} \][/tex]
[tex]\[ v = \frac{2}{3} \][/tex]
Now, substitute [tex]\( v = \frac{2}{3} \)[/tex] back into Equation 5 to find [tex]\(u\)[/tex]:
[tex]\[ u = \frac{2 - v}{5} \][/tex]
[tex]\[ u = \frac{2 - \frac{2}{3}}{5} \][/tex]
[tex]\[ u = \frac{\frac{6}{3} - \frac{2}{3}}{5} \][/tex]
[tex]\[ u = \frac{\frac{4}{3}}{5} \][/tex]
[tex]\[ u = \frac{4}{15} \][/tex]
### Step 3: Substitute back to find [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
We originally set:
[tex]\[ u = \frac{1}{x+y} \Rightarrow \frac{1}{x+y} = \frac{4}{15} \Rightarrow x+y = \frac{15}{4} \][/tex]
[tex]\[ v = \frac{1}{x-y} \Rightarrow \frac{1}{x-y} = \frac{2}{3} \Rightarrow x-y = \frac{3}{2} \][/tex]
Now we solve these two equations:
[tex]\[ \begin{cases} x + y = \frac{15}{4} \\ x - y = \frac{3}{2} \end{cases} \][/tex]
Add the two equations together to eliminate [tex]\( y \)[/tex]:
[tex]\[ (x + y) + (x - y) = \frac{15}{4} + \frac{3}{2} \][/tex]
[tex]\[ 2x = \frac{15}{4} + \frac{6}{4} \][/tex]
[tex]\[ 2x = \frac{21}{4} \][/tex]
[tex]\[ x = \frac{21}{8} \][/tex]
Next, substitute [tex]\( x \)[/tex] back to find [tex]\( y \)[/tex]:
[tex]\[ x + y = \frac{15}{4} \Rightarrow \frac{21}{8} + y = \frac{15}{4} \][/tex]
[tex]\[ y = \frac{15}{4} - \frac{21}{8} \][/tex]
[tex]\[ y = \frac{30}{8} - \frac{21}{8} \][/tex]
[tex]\[ y = \frac{9}{8} \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = \left( \frac{21}{8}, \frac{9}{8} \right) \][/tex]
[tex]\[ \begin{array}{l} \frac{10}{x+y}+\frac{2}{x-y}=4 \quad \text{(Equation 1)} \\ \frac{15}{x+y}-\frac{9}{x-y}=-2 \quad \text{(Equation 2)} \end{array} \][/tex]
To solve this system, we will use substitution or elimination techniques. Let's first introduce new variables to simplify the equations:
Let:
[tex]\[ u = \frac{1}{x+y} \][/tex]
[tex]\[ v = \frac{1}{x-y} \][/tex]
Now, rewriting the original equations using [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
### Step 1: Substitute the new variables
[tex]\[ 10u + 2v = 4 \quad \text{(Transformed from Equation 1)} \][/tex]
[tex]\[ 15u - 9v = -2 \quad \text{(Transformed from Equation 2)} \][/tex]
### Step 2: Solve the transformed system of linear equations
We now have a simpler system of linear equations:
[tex]\[ \begin{array}{l} 10u + 2v = 4 \quad \text{(Equation 3)} \\ 15u - 9v = -2 \quad \text{(Equation 4)} \end{array} \][/tex]
First, let's solve Equation 3 for [tex]\(u\)[/tex]:
[tex]\[ 10u + 2v = 4 \][/tex]
[tex]\[ 10u = 4 - 2v \][/tex]
[tex]\[ u = \frac{4 - 2v}{10} \][/tex]
[tex]\[ u = \frac{2 - v}{5} \quad \text{(Equation 5)} \][/tex]
Next, we'll substitute Equation 5 into Equation 4:
[tex]\[ 15\left(\frac{2 - v}{5}\right) - 9v = -2 \][/tex]
[tex]\[ 15 \cdot \frac{2 - v}{5} - 9v = -2 \][/tex]
[tex]\[ 3(2 - v) - 9v = -2 \][/tex]
[tex]\[ 6 - 3v - 9v = -2 \][/tex]
[tex]\[ 6 - 12v = -2 \][/tex]
[tex]\[ -12v = -2 - 6 \][/tex]
[tex]\[ -12v = -8 \][/tex]
[tex]\[ v = \frac{8}{12} \][/tex]
[tex]\[ v = \frac{2}{3} \][/tex]
Now, substitute [tex]\( v = \frac{2}{3} \)[/tex] back into Equation 5 to find [tex]\(u\)[/tex]:
[tex]\[ u = \frac{2 - v}{5} \][/tex]
[tex]\[ u = \frac{2 - \frac{2}{3}}{5} \][/tex]
[tex]\[ u = \frac{\frac{6}{3} - \frac{2}{3}}{5} \][/tex]
[tex]\[ u = \frac{\frac{4}{3}}{5} \][/tex]
[tex]\[ u = \frac{4}{15} \][/tex]
### Step 3: Substitute back to find [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
We originally set:
[tex]\[ u = \frac{1}{x+y} \Rightarrow \frac{1}{x+y} = \frac{4}{15} \Rightarrow x+y = \frac{15}{4} \][/tex]
[tex]\[ v = \frac{1}{x-y} \Rightarrow \frac{1}{x-y} = \frac{2}{3} \Rightarrow x-y = \frac{3}{2} \][/tex]
Now we solve these two equations:
[tex]\[ \begin{cases} x + y = \frac{15}{4} \\ x - y = \frac{3}{2} \end{cases} \][/tex]
Add the two equations together to eliminate [tex]\( y \)[/tex]:
[tex]\[ (x + y) + (x - y) = \frac{15}{4} + \frac{3}{2} \][/tex]
[tex]\[ 2x = \frac{15}{4} + \frac{6}{4} \][/tex]
[tex]\[ 2x = \frac{21}{4} \][/tex]
[tex]\[ x = \frac{21}{8} \][/tex]
Next, substitute [tex]\( x \)[/tex] back to find [tex]\( y \)[/tex]:
[tex]\[ x + y = \frac{15}{4} \Rightarrow \frac{21}{8} + y = \frac{15}{4} \][/tex]
[tex]\[ y = \frac{15}{4} - \frac{21}{8} \][/tex]
[tex]\[ y = \frac{30}{8} - \frac{21}{8} \][/tex]
[tex]\[ y = \frac{9}{8} \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = \left( \frac{21}{8}, \frac{9}{8} \right) \][/tex]
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