Question

Решите систему уравнений
2x+y=3
3x^2+4xy+7y^2+X+8y=5

Answer 1

$(2; -1) \quad ; \quad (1\dfrac{18}{23} ;-\dfrac{13}{23}) \quad ;$

Объяснение:

$$ \displaystyle \left \{ {{2x+y=3} \atop {3x^{2}+4xy+7y^{2}+x+8y=5}} \right. \Leftrightarrow \left \{ {{2x+y=3} \atop {4x^{2}+4xy+y^{2}-x^{2}+6y^{2}+x+8y=5}} \right. \Leftrightarrow $$

$$ \displaystyle \left \{ {{2x+y=3} \atop {(2x+y)^{2}-x^{2}+6y^{2}+x+8y=5}} \right. \Leftrightarrow \left \{ {{2x+y=3} \atop {9-x^{2}+6y^{2}+x+8y=5}} \right. \Leftrightarrow $$

$$ \displaystyle \Leftrightarrow \left \{ {{y=3-2x} \atop {9-(x^{2}-6y^{2}-x-8y)=5}} \right. \Leftrightarrow $$

$$ \displaystyle \Leftrightarrow \left \{ {{y=3-2x} \atop {x^{2}-6 \cdot (3-2x)^{2}-x-8 \cdot (3-2x)=4}} \right. \Leftrightarrow $$

$$ \displaystyle \Leftrightarrow \left \{ {{y=3-2x} \atop {x^{2}-6 \cdot (9-12x+4x^{2})-x-24+16x-4=0}} \right. \Leftrightarrow $$

$$ \displaystyle \Leftrightarrow \left \{ {{y=3-2x} \atop {x^{2}-54+72x-24x^{2}+15x-28=0}} \right. \Leftrightarrow \left \{ {{y=3-2x} \atop {-23x^{2}+87x-82=0}} \right. ; $$

$-23x^{2}+87x-82=0 \quad | \quad \cdot (-1)$

$23x^{2}-87x+82=0;$

$D=b^{2}-4ac;$

$D=(-87)^{2}-4 \cdot 23 \cdot 82=87^{2}-4 \cdot (20+3) \cdot (80+2)=(90-3)^{2}-4 \cdot$

$\cdot (20 \cdot 80+20 \cdot 2+3 \cdot 80+3 \cdot 2)=8100-540+9-4 \cdot (1600+40+240+6)=$

$=8100-540+9-6400-160-960-24=1700-700-960-15=1000-960-15=$

$=40-15=25;$

$x_{1}=\dfrac{-b+\sqrt{D}}{2a} \Rightarrow x_{1}=\dfrac{-(-87)+\sqrt{25}}{2 \cdot 23}=\dfrac{87+5}{46}=\dfrac{92}{46}=2;$

$x_{2}=\dfrac{-b-\sqrt{D}}{2a} \Rightarrow x_{2}=\dfrac{-(-87)-\sqrt{25}}{2 \cdot 23}=\dfrac{87-5}{46}=\dfrac{82}{46}=\dfrac{41}{23}=1\dfrac{18}{23};$

$x_{1}=2 \Rightarrow y_{1}=3-2 \cdot 2=3-4=-1;$

$x_{2}=1\dfrac{18}{23} \Rightarrow y_{2}=3-2 \cdot 1\dfrac{18}{23}=3-2\dfrac{36}{23}=3-3\dfrac{13}{23}=-\dfrac{13}{23};$

$(2; -1) \quad ; \quad (1\dfrac{18}{23} ;-\dfrac{13}{23}) \quad ;$