Whakaoti i te (x+4)*(x+2y)=8x

$$( x + 4 ) \times ( x + 2 y ) = 8 x$$

Whakaoti mō y

$$y=-\frac{x\left(x-4\right)}{2\left(x+4\right)}$$

$$x\neq -4$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x+4 ki te x+2y.

$$x^{2}+2xy+4x+8y=8x$$

Tangohia te x^{2} mai i ngā taha e rua.

$$2xy+4x+8y=8x-x^{2}$$

Tangohia te 4x mai i ngā taha e rua.

$$2xy+8y=8x-x^{2}-4x$$

Pahekotia te 8x me -4x, ka 4x.

$$2xy+8y=4x-x^{2}$$

Pahekotia ngā kīanga tau katoa e whai ana i te y.

$$\left(2x+8\right)y=4x-x^{2}$$

Whakawehea ngā taha e rua ki te 2x+8.

$$\frac{\left(2x+8\right)y}{2x+8}=\frac{x\left(4-x\right)}{2x+8}$$

Mā te whakawehe ki te 2x+8 ka wetekia te whakareanga ki te 2x+8.

$$y=\frac{x\left(4-x\right)}{2x+8}$$

Whakawehe x\left(4-x\right) ki te 2x+8.

$$y=\frac{x\left(4-x\right)}{2\left(x+4\right)}$$

Whakaoti mō x (complex solution)

$$x=\sqrt{y^{2}-12y+4}-y+2$$

$$x=-\sqrt{y^{2}-12y+4}-y+2$$

Whakaoti mō x

$$x=\sqrt{y^{2}-12y+4}-y+2$$

$$x=-\sqrt{y^{2}-12y+4}-y+2\text{, }y\geq 4\sqrt{2}+6\text{ or }y\leq 6-4\sqrt{2}$$