Whakaoti i te (-3-y)^2-(2-y)^2=(y+4)*(4-y)+y(y+1)-2

$$( - 3 - y ) ^ { 2 } - ( 2 - y ) ^ { 2 } = ( y + 4 ) \cdot ( 4 - y ) + y ( y + 1 ) - 2$$

Whakaoti mō y

$$y=1$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(-3-y\right)^{2}.

$$9+6y+y^{2}-\left(2-y\right)^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2-y\right)^{2}.

$$9+6y+y^{2}-\left(4-4y+y^{2}\right)=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Hei kimi i te tauaro o 4-4y+y^{2}, kimihia te tauaro o ia taurangi.

$$9+6y+y^{2}-4+4y-y^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Tangohia te 4 i te 9, ka 5.

$$5+6y+y^{2}+4y-y^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Pahekotia te 6y me 4y, ka 10y.

$$5+10y+y^{2}-y^{2}=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Pahekotia te y^{2} me -y^{2}, ka 0.

$$5+10y=\left(y+4\right)\left(4-y\right)+y\left(y+1\right)-2$$

Whakaarohia te \left(y+4\right)\left(4-y\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Pūrua 4.

$$5+10y=16-y^{2}+y\left(y+1\right)-2$$

Whakamahia te āhuatanga tohatoha hei whakarea te y ki te y+1.

$$5+10y=16-y^{2}+y^{2}+y-2$$

Pahekotia te -y^{2} me y^{2}, ka 0.

$$5+10y=16+y-2$$

Tangohia te 2 i te 16, ka 14.

$$5+10y=14+y$$

Tangohia te y mai i ngā taha e rua.

$$5+10y-y=14$$

Pahekotia te 10y me -y, ka 9y.

$$5+9y=14$$

Tangohia te 5 mai i ngā taha e rua.

$$9y=14-5$$

Tangohia te 5 i te 14, ka 9.

$$9y=9$$

Whakawehea ngā taha e rua ki te 9.

$$y=\frac{9}{9}$$

Whakawehea te 9 ki te 9, kia riro ko 1.

$$y=1$$