Whakaoti i te (x-y)(400+5y)=4250

$$( x - y ) ( 400 + 5 y ) = 4250$$

Whakaoti mō x

$$x=\frac{y^{2}+80y+850}{y+80}$$

$$y\neq -80$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te x-y ki te 400+5y.

$$400x+5xy-400y-5y^{2}=4250$$

Me tāpiri te 400y ki ngā taha e rua.

$$400x+5xy-5y^{2}=4250+400y$$

Me tāpiri te 5y^{2} ki ngā taha e rua.

$$400x+5xy=4250+400y+5y^{2}$$

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

$$\left(400+5y\right)x=4250+400y+5y^{2}$$

He hanga arowhānui tō te whārite.

$$\left(5y+400\right)x=5y^{2}+400y+4250$$

Whakawehea ngā taha e rua ki te 400+5y.

$$\frac{\left(5y+400\right)x}{5y+400}=\frac{5y^{2}+400y+4250}{5y+400}$$

Mā te whakawehe ki te 400+5y ka wetekia te whakareanga ki te 400+5y.

$$x=\frac{5y^{2}+400y+4250}{5y+400}$$

Whakawehe 4250+400y+5y^{2} ki te 400+5y.

$$x=\frac{y^{2}+80y+850}{y+80}$$

Whakaoti mō y (complex solution)

$$y=\frac{-\sqrt{x^{2}+160x+3000}+x-80}{2}$$

$$y=\frac{\sqrt{x^{2}+160x+3000}+x-80}{2}$$

Whakaoti mō y

$$y=\frac{-\sqrt{x^{2}+160x+3000}+x-80}{2}$$

$$y=\frac{\sqrt{x^{2}+160x+3000}+x-80}{2}\text{, }x\geq 10\sqrt{34}-80\text{ or }x\leq -10\sqrt{34}-80$$