{4x - y=2 {x^2 + y^2 =3
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14 января 2025 08:44
To solve the system of equations given by:
- ( 4x - y = 2 )
- ( x^2 + y^2 = 3 )
we can follow these steps:
Step 1: Solve the first equation for ( y ).
From the first equation, we can express ( y ) in terms of ( x ):
[ y = 4x - 2 ]
Step 2: Substitute ( y ) into the second equation.
Now, we substitute ( y ) from the first equation into the second equation:
[ x^2 + (4x - 2)^2 = 3 ]
Step 3: Expand and simplify.
Expanding ( (4x - 2)^2 ):
[ (4x - 2)^2 = 16x^2 - 16x + 4 ]
Now, substitute this into the second equation:
[ x^2 + 16x^2 - 16x + 4 = 3 ]
Combine like terms:
[ 17x^2 - 16x + 4 = 3 ]
Subtract 3 from both sides:
[ 17x^2 - 16x + 1 = 0 ]
Step 4: Use the quadratic formula.
To solve the quadratic equation ( 17x^2 - 16x + 1 = 0 ), we apply the quadratic formula:
[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Here, ( a = 17 ), ( b = -16 ), and ( c = 1 ):
[ b^2 - 4ac = (-16)^2 - 4 \cdot 17 \cdot 1 = 256 - 68 = 188 ]
Now, substitute values into the quadratic formula:
[ x = \frac{16 \pm \sqrt{188}}{34} ]
Step 5: Simplify ( \sqrt{188} ).
Finding ( \sqrt{188} ):
[ \sqrt{188} = \sqrt{4 \cdot 47} = 2\sqrt{47} ]
So our expression for ( x ) becomes:
[ x = \frac{16 \pm 2\sqrt{47}}{34} = \frac{8 \pm \sqrt{47}}{17} ]
Step 6: Find corresponding ( y ) values.
Now substitute back to find ( y ) values using ( y = 4x - 2 ):
- For ( x = \frac{8 + \sqrt{47}}{17} ):
[ y = 4\left(\frac{8 + \sqrt{47}}{17}\right) - 2 = \frac{32 + 4\sqrt{47}}{17} - 2 = \frac{32 + 4\sqrt{47} - 34}{17} = \frac{-2 + 4\sqrt{47}}{17} ]
- For ( x = \frac{8 - \sqrt{47}}{17} ):
[ y = 4\left(\frac{8 - \sqrt{47}}{17}\right) - 2 = \frac{32 - 4\sqrt{47}}{17} - 2 = \frac{32 - 4\sqrt{47} - 34}{17} = \frac{-2 - 4\sqrt{47}}{17} ]
Step 7: Final solutions.
Thus, the solutions to the system of equations are:
- ( \left( \frac{8 + \sqrt{47}}{17}, \frac{-2 + 4\sqrt{47}}{17} \right) )
- ( \left( \frac{8 - \sqrt{47}}{17}, \frac{-2 - 4\sqrt{47}}{17} \right) )
These are the points of intersection of the line and the circle defined by the given equations.
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