To solve the system of equations
- \( 2x + y = 1 \) (Equation 1)
- \( 2x^2 - xy + y^2 = 4 \) (Equation 2)
we will first express \( y \) in terms of \( x \) using Equation 1, and then substitute it into Equation 2.
From Equation 1, we can solve for \( y \): \[ y = 1 - 2x \]
Next, we substitute this expression for \( y \) into Equation 2:
Substituting \( y = 1 - 2x \) into Equation 2:
\[ 2x^2 - x(1 - 2x) + (1 - 2x)^2 = 4 \]
Now, we expand the equation:
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First, calculate \( -x(1 - 2x) \): \[ -x(1 - 2x) = -x + 2x^2 \]
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Next, calculate \( (1 - 2x)^2 \): \[ (1 - 2x)^2 = 1 - 4x + 4x^2 \]
Now substituting back into the equation:
\[ 2x^2 - x + 2x^2 + 1 - 4x + 4x^2 = 4 \]
Combine like terms:
\[ (2x^2 + 2x^2 + 4x^2) - x - 4x + 1 = 4 \]
This simplifies to:
\[ 8x^2 - 5x + 1 = 4 \]
Next, we move 4 to the left side:
\[ 8x^2 - 5x + 1 - 4 = 0 \]
This simplifies to:
\[ 8x^2 - 5x - 3 = 0 \]
Now, we use the quadratic formula to solve for \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 8 \), \( b = -5 \), and \( c = -3 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-5)^2 - 4(8)(-3) = 25 + 96 = 121 \]
Now substituting into the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{121}}{2 \cdot 8} \] \[ x = \frac{5 \pm 11}{16} \]
Calculating the two possible values for \( x \):
- \( x = \frac{5 + 11}{16} = \frac{16}{16} = 1 \)
- \( x = \frac{5 - 11}{16} = \frac{-6}{16} = \frac{-3}{8} \)
Now we'll substitute these values of \( x \) back into Equation 1 to find the corresponding \( y \) values.
For \( x = 1 \): \[ y = 1 - 2(1) = 1 - 2 = -1 \]
So, one solution is \( (1, -1) \).
For \( x = -\frac{3}{8} \): \[ y = 1 - 2\left(-\frac{3}{8}\right) = 1 + \frac{6}{8} = 1 + \frac{3}{4} = \frac{7}{4} \]
So, the second solution is \( \left(-\frac{3}{8}, \frac{7}{4}\right) \).
In conclusion, the solutions for the system of equations are:
- \( (1, -1) \)
- \( \left(-\frac{3}{8}, \frac{7}{4}\right) \)
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