Réitigh (7-x)^2+(1-y)^2=(3-x)^2+(5-y^2)

$$( 7 - x ) ^ { 2 } + ( 1 - y ) ^ { 2 } = ( 3 - x ) ^ { 2 } + ( 5 - y ^ { 2 } )$$

Réitigh do x.

$$x=\frac{y^{2}-y+18}{4}$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(7-x\right)^{2} a leathnú.

$$49-14x+x^{2}+\left(1-y\right)^{2}=\left(3-x\right)^{2}+5-y^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(1-y\right)^{2} a leathnú.

$$49-14x+x^{2}+1-2y+y^{2}=\left(3-x\right)^{2}+5-y^{2}$$

Suimigh 49 agus 1 chun 50 a fháil.

$$50-14x+x^{2}-2y+y^{2}=\left(3-x\right)^{2}+5-y^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(3-x\right)^{2} a leathnú.

$$50-14x+x^{2}-2y+y^{2}=9-6x+x^{2}+5-y^{2}$$

Suimigh 9 agus 5 chun 14 a fháil.

$$50-14x+x^{2}-2y+y^{2}=14-6x+x^{2}-y^{2}$$

Cuir 6x leis an dá thaobh.

$$50-14x+x^{2}-2y+y^{2}+6x=14+x^{2}-y^{2}$$

Comhcheangail -14x agus 6x chun -8x a fháil.

$$50-8x+x^{2}-2y+y^{2}=14+x^{2}-y^{2}$$

Bain x^{2} ón dá thaobh.

$$50-8x+x^{2}-2y+y^{2}-x^{2}=14-y^{2}$$

Comhcheangail x^{2} agus -x^{2} chun 0 a fháil.

$$50-8x-2y+y^{2}=14-y^{2}$$

Bain 50 ón dá thaobh.

$$-8x-2y+y^{2}=14-y^{2}-50$$

Dealaigh 50 ó 14 chun -36 a fháil.

$$-8x-2y+y^{2}=-36-y^{2}$$

Cuir 2y leis an dá thaobh.

$$-8x+y^{2}=-36-y^{2}+2y$$

Bain y^{2} ón dá thaobh.

$$-8x=-36-y^{2}+2y-y^{2}$$

Comhcheangail -y^{2} agus -y^{2} chun -2y^{2} a fháil.

$$-8x=-36-2y^{2}+2y$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$-8x=-2y^{2}+2y-36$$

Roinn an dá thaobh faoi -8.

$$\frac{-8x}{-8}=\frac{-2y^{2}+2y-36}{-8}$$

Má roinntear é faoi -8 cuirtear an iolrúchán faoi -8 ar ceal.

$$x=\frac{-2y^{2}+2y-36}{-8}$$

Roinn -36-2y^{2}+2y faoi -8.

$$x=\frac{y^{2}}{4}-\frac{y}{4}+\frac{9}{2}$$

Réitigh do y. (complex solution)

$$y=\frac{\sqrt{16x-71}+1}{2}$$

$$y=\frac{-\sqrt{16x-71}+1}{2}$$

Réitigh do y.

$$y=\frac{\sqrt{16x-71}+1}{2}$$

$$y=\frac{-\sqrt{16x-71}+1}{2}\text{, }x\geq \frac{71}{16}$$