Whakaoti i te {l}{x-y=-1}{z=x+2y}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { x - y = -1 }\\ { z = x + 2 y }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Whakaoti mō x, y, z

$$x=\frac{a-2}{3}$$

$$y=\frac{a+1}{3}$$

$$z=a$$

Ngā Hipanga Poto E Whakamahi Whakakapinga Ana

Me raupapa anō ngā whārite.

$$z=x+2y x-y=-1 a=z$$

Whakakapia te x+2y mō te z i te whārite a=z.

$$a=x+2y$$

Me whakaoti te whārite tuarua mō y me te whārite tuatoru mō x.

$$y=x+1 x=a-2y$$

Whakakapia te x+1 mō te y i te whārite x=a-2y.

$$x=a-2\left(x+1\right)$$

Me whakaoti te x=a-2\left(x+1\right) mō x.

$$x=-\frac{2}{3}+\frac{1}{3}a$$

Whakakapia te -\frac{2}{3}+\frac{1}{3}a mō te x i te whārite y=x+1.

$$y=-\frac{2}{3}+\frac{1}{3}a+1$$

Tātaitia te y i te y=-\frac{2}{3}+\frac{1}{3}a+1.

$$y=\frac{1}{3}+\frac{1}{3}a$$

Whakakapia te \frac{1}{3}+\frac{1}{3}a mō te y me te -\frac{2}{3}+\frac{1}{3}a mō x i te whārite z=x+2y.

$$z=-\frac{2}{3}+\frac{1}{3}a+2\left(\frac{1}{3}+\frac{1}{3}a\right)$$

Tātaitia te z i te z=-\frac{2}{3}+\frac{1}{3}a+2\left(\frac{1}{3}+\frac{1}{3}a\right).

$$z=a$$

Kua oti te pūnaha te whakatau.

$$x=-\frac{2}{3}+\frac{1}{3}a y=\frac{1}{3}+\frac{1}{3}a z=a$$