ແກ້ x/y+z/y+2(x+z)=10x+y

$$\frac{ x }{ y } + \frac{ z }{ y } +2(x+z)=10x+y$$

ແກ້ສຳລັບ x (complex solution)

$$\left\{\begin{matrix}x=-\frac{z+2yz-y^{2}}{1-8y}\text{, }&y\neq \frac{1}{8}\text{ and }y\neq 0\\x\in \mathrm{C}\text{, }&y=\frac{1}{8}\text{ and }z=\frac{1}{80}\end{matrix}\right,$$

ຂັ້ນຕອນສຳລັບການແກ້ສົມຜົນເສັ້ນຊື່

ຄູນທັງສອງຂ້າງຂອງສົມຜົນດ້ວຍ y.

$$x+z+2\left(x+z\right)y=10xy+yy$$

ຄູນ y ກັບ y ເພື່ອໃຫ້ໄດ້ y^{2}.

$$x+z+2\left(x+z\right)y=10xy+y^{2}$$

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ 2 ດ້ວຍ x+z.

$$x+z+\left(2x+2z\right)y=10xy+y^{2}$$

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ 2x+2z ດ້ວຍ y.

$$x+z+2xy+2zy=10xy+y^{2}$$

ລົບ 10xy ອອກຈາກທັງສອງຂ້າງ.

$$x+z+2xy+2zy-10xy=y^{2}$$

ຮວມ 2xy ແລະ -10xy ເພື່ອຮັບ -8xy.

$$x+z-8xy+2zy=y^{2}$$

ລົບ z ອອກຈາກທັງສອງຂ້າງ.

$$x-8xy+2zy=y^{2}-z$$

ລົບ 2zy ອອກຈາກທັງສອງຂ້າງ.

$$x-8xy=y^{2}-z-2zy$$

ຮວມທຸກຄຳສັບທີ່ມີ x.

$$\left(1-8y\right)x=y^{2}-z-2zy$$

ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.

$$\left(1-8y\right)x=y^{2}-2yz-z$$

ຫານທັງສອງຂ້າງດ້ວຍ -8y+1.

$$\frac{\left(1-8y\right)x}{1-8y}=\frac{y^{2}-2yz-z}{1-8y}$$

ການຫານດ້ວຍ -8y+1 ຈະຍົກເລີກການຄູນດ້ວຍ -8y+1.

$$x=\frac{y^{2}-2yz-z}{1-8y}$$

ແກ້ສຳລັບ x

$$\left\{\begin{matrix}x=-\frac{z+2yz-y^{2}}{1-8y}\text{, }&y\neq \frac{1}{8}\text{ and }y\neq 0\\x\in \mathrm{R}\text{, }&y=\frac{1}{8}\text{ and }z=\frac{1}{80}\end{matrix}\right,$$

ຂັ້ນຕອນສຳລັບການແກ້ສົມຜົນເສັ້ນຊື່

ຄູນທັງສອງຂ້າງຂອງສົມຜົນດ້ວຍ y.

$$x+z+2\left(x+z\right)y=10xy+yy$$

ຄູນ y ກັບ y ເພື່ອໃຫ້ໄດ້ y^{2}.

$$x+z+2\left(x+z\right)y=10xy+y^{2}$$

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ 2 ດ້ວຍ x+z.

$$x+z+\left(2x+2z\right)y=10xy+y^{2}$$

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ 2x+2z ດ້ວຍ y.

$$x+z+2xy+2zy=10xy+y^{2}$$

ລົບ 10xy ອອກຈາກທັງສອງຂ້າງ.

$$x+z+2xy+2zy-10xy=y^{2}$$

ຮວມ 2xy ແລະ -10xy ເພື່ອຮັບ -8xy.

$$x+z-8xy+2zy=y^{2}$$

ລົບ z ອອກຈາກທັງສອງຂ້າງ.

$$x-8xy+2zy=y^{2}-z$$

ລົບ 2zy ອອກຈາກທັງສອງຂ້າງ.

$$x-8xy=y^{2}-z-2zy$$

ຮວມທຸກຄຳສັບທີ່ມີ x.

$$\left(1-8y\right)x=y^{2}-z-2zy$$

ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.

$$\left(1-8y\right)x=y^{2}-2yz-z$$

ຫານທັງສອງຂ້າງດ້ວຍ -8y+1.

$$\frac{\left(1-8y\right)x}{1-8y}=\frac{y^{2}-2yz-z}{1-8y}$$

ການຫານດ້ວຍ -8y+1 ຈະຍົກເລີກການຄູນດ້ວຍ -8y+1.

$$x=\frac{y^{2}-2yz-z}{1-8y}$$

ແກ້ສຳລັບ y (complex solution)

$$\left\{\begin{matrix}y=\sqrt{16x^{2}-8xz+x+z^{2}+z}+z-4x\text{, }&\left(z\neq 0\text{ and }arg(z)<\pi \right)\text{ or }x\neq -z\\y=-\sqrt{16x^{2}-8xz+x+z^{2}+z}+z-4x\text{, }&\left(z\neq 0\text{ and }arg(z)\geq \pi \right)\text{ or }x\neq -z\end{matrix}\right,$$

ແກ້ສຳລັບ y

$$\left\{\begin{matrix}y=\sqrt{16x^{2}-8xz+x+z^{2}+z}+z-4x\text{, }&z\geq \frac{1}{80}\text{ or }\left(z>0\text{ and }x\leq \frac{z}{4}-\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z\leq \frac{1}{80}\right)\text{ or }\left(z>0\text{ and }x\geq \frac{z}{4}+\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z\leq \frac{1}{80}\right)\text{ or }\left(x\neq -z\text{ and }x\leq \frac{z}{4}-\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z\leq \frac{1}{80}\right)\text{ or }\left(x\neq -z\text{ and }x\geq \frac{z}{4}+\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z\leq \frac{1}{80}\right)\\y=-\sqrt{16x^{2}-8xz+x+z^{2}+z}+z-4x\text{, }&\left(x\leq \frac{z}{4}-\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z<0\right)\text{ or }\left(x\geq \frac{z}{4}+\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z<0\right)\text{ or }\left(x\neq -z\text{ and }z\geq \frac{1}{80}\right)\text{ or }\left(x\neq -z\text{ and }x\leq \frac{z}{4}-\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z\leq \frac{1}{80}\right)\text{ or }\left(x\neq -z\text{ and }x\geq \frac{z}{4}+\frac{\sqrt{1-80z}}{32}-\frac{1}{32}\text{ and }z\leq \frac{1}{80}\right)\end{matrix}\right,$$