Answer :
Certainly! Let's solve the system of equations:
[tex]\[ \begin{cases} X - Y = 2 \\ X^2 + Y^2 = 1 \end{cases} \][/tex]
### Step 1: Solve the first equation for [tex]\(X\)[/tex]
From the first equation:
[tex]\[ X - Y = 2 \][/tex]
We can solve for [tex]\(X\)[/tex]:
[tex]\[ X = Y + 2 \][/tex]
### Step 2: Substitute [tex]\(X\)[/tex] in the second equation
Now we substitute this expression for [tex]\(X\)[/tex] into the second equation:
[tex]\[ (Y + 2)^2 + Y^2 = 1 \][/tex]
Expand the squared term:
[tex]\[ (Y + 2)(Y + 2) + Y^2 = 1 \][/tex]
[tex]\[ (Y^2 + 4Y + 4) + Y^2 = 1 \][/tex]
[tex]\[ 2Y^2 + 4Y + 4 = 1 \][/tex]
### Step 3: Simplify the quadratic equation
Bring all terms to one side to set the equation to 0:
[tex]\[ 2Y^2 + 4Y + 4 - 1 = 0 \][/tex]
[tex]\[ 2Y^2 + 4Y + 3 = 0 \][/tex]
### Step 4: Solve the quadratic equation
Now solve the quadratic equation using the quadratic formula [tex]\(Y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 3\)[/tex]:
[tex]\[ Y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{16 - 24}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{-8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{-1 \cdot 8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{-1} \cdot \sqrt{8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm i\sqrt{8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm 2i\sqrt{2}}{4} \][/tex]
[tex]\[ Y = -1 \pm \frac{i\sqrt{2}}{2} \][/tex]
So the solutions for [tex]\(Y\)[/tex] are:
[tex]\[ Y = -1 + \frac{\sqrt{2}i}{2} \][/tex]
[tex]\[ Y = -1 - \frac{\sqrt{2}i}{2} \][/tex]
### Step 5: Find the corresponding [tex]\(X\)[/tex] values
Using the expression [tex]\(X = Y + 2\)[/tex] for each [tex]\(Y\)[/tex]:
1. For [tex]\( Y = -1 + \frac{\sqrt{2}i}{2} \)[/tex]:
[tex]\[ X = \left(-1 + \frac{\sqrt{2}i}{2}\right) + 2 \][/tex]
[tex]\[ X = 1 + \frac{\sqrt{2}i}{2} \][/tex]
2. For [tex]\( Y = -1 - \frac{\sqrt{2}i}{2} \)[/tex]:
[tex]\[ X = \left(-1 - \frac{\sqrt{2}i}{2}\right) + 2 \][/tex]
[tex]\[ X = 1 - \frac{\sqrt{2}i}{2} \][/tex]
### Step 6: Compile the solutions
So, the solutions to the system of equations are:
[tex]\[ (X, Y) = \left(1 - \frac{\sqrt{2}i}{2}, -1 - \frac{\sqrt{2}i}{2}\right) \][/tex]
[tex]\[ (X, Y) = \left(1 + \frac{\sqrt{2}i}{2}, -1 + \frac{\sqrt{2}i}{2}\right) \][/tex]
Hence, the solutions to the system of equations are:
[tex]\[ \left(1 - \frac{\sqrt{2}i}{2}, -1 - \frac{\sqrt{2}i}{2}\right) \][/tex]
[tex]\[ \left(1 + \frac{\sqrt{2}i}{2}, -1 + \frac{\sqrt{2}i}{2}\right) \][/tex]
[tex]\[ \begin{cases} X - Y = 2 \\ X^2 + Y^2 = 1 \end{cases} \][/tex]
### Step 1: Solve the first equation for [tex]\(X\)[/tex]
From the first equation:
[tex]\[ X - Y = 2 \][/tex]
We can solve for [tex]\(X\)[/tex]:
[tex]\[ X = Y + 2 \][/tex]
### Step 2: Substitute [tex]\(X\)[/tex] in the second equation
Now we substitute this expression for [tex]\(X\)[/tex] into the second equation:
[tex]\[ (Y + 2)^2 + Y^2 = 1 \][/tex]
Expand the squared term:
[tex]\[ (Y + 2)(Y + 2) + Y^2 = 1 \][/tex]
[tex]\[ (Y^2 + 4Y + 4) + Y^2 = 1 \][/tex]
[tex]\[ 2Y^2 + 4Y + 4 = 1 \][/tex]
### Step 3: Simplify the quadratic equation
Bring all terms to one side to set the equation to 0:
[tex]\[ 2Y^2 + 4Y + 4 - 1 = 0 \][/tex]
[tex]\[ 2Y^2 + 4Y + 3 = 0 \][/tex]
### Step 4: Solve the quadratic equation
Now solve the quadratic equation using the quadratic formula [tex]\(Y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 3\)[/tex]:
[tex]\[ Y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{16 - 24}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{-8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{-1 \cdot 8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm \sqrt{-1} \cdot \sqrt{8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm i\sqrt{8}}{4} \][/tex]
[tex]\[ Y = \frac{-4 \pm 2i\sqrt{2}}{4} \][/tex]
[tex]\[ Y = -1 \pm \frac{i\sqrt{2}}{2} \][/tex]
So the solutions for [tex]\(Y\)[/tex] are:
[tex]\[ Y = -1 + \frac{\sqrt{2}i}{2} \][/tex]
[tex]\[ Y = -1 - \frac{\sqrt{2}i}{2} \][/tex]
### Step 5: Find the corresponding [tex]\(X\)[/tex] values
Using the expression [tex]\(X = Y + 2\)[/tex] for each [tex]\(Y\)[/tex]:
1. For [tex]\( Y = -1 + \frac{\sqrt{2}i}{2} \)[/tex]:
[tex]\[ X = \left(-1 + \frac{\sqrt{2}i}{2}\right) + 2 \][/tex]
[tex]\[ X = 1 + \frac{\sqrt{2}i}{2} \][/tex]
2. For [tex]\( Y = -1 - \frac{\sqrt{2}i}{2} \)[/tex]:
[tex]\[ X = \left(-1 - \frac{\sqrt{2}i}{2}\right) + 2 \][/tex]
[tex]\[ X = 1 - \frac{\sqrt{2}i}{2} \][/tex]
### Step 6: Compile the solutions
So, the solutions to the system of equations are:
[tex]\[ (X, Y) = \left(1 - \frac{\sqrt{2}i}{2}, -1 - \frac{\sqrt{2}i}{2}\right) \][/tex]
[tex]\[ (X, Y) = \left(1 + \frac{\sqrt{2}i}{2}, -1 + \frac{\sqrt{2}i}{2}\right) \][/tex]
Hence, the solutions to the system of equations are:
[tex]\[ \left(1 - \frac{\sqrt{2}i}{2}, -1 - \frac{\sqrt{2}i}{2}\right) \][/tex]
[tex]\[ \left(1 + \frac{\sqrt{2}i}{2}, -1 + \frac{\sqrt{2}i}{2}\right) \][/tex]