Whakaoti i te (a+b)x+a-b+1=x+6

$$( a + b ) x + a - b + 1 = x + 6$$

Whakaoti mō a (complex solution)

$$\left\{\begin{matrix}a=-\frac{bx-x-b-5}{x+1}\text{, }&x\neq -1\\a\in \mathrm{C}\text{, }&x=-1\text{ and }b=-2\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te a+b ki te x.

$$ax+bx+a-b+1=x+6$$

Tangohia te bx mai i ngā taha e rua.

$$ax+a-b+1=x+6-bx$$

Me tāpiri te b ki ngā taha e rua.

$$ax+a+1=x+6-bx+b$$

Tangohia te 1 mai i ngā taha e rua.

$$ax+a=x+6-bx+b-1$$

Tangohia te 1 i te 6, ka 5.

$$ax+a=x+5-bx+b$$

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

$$\left(x+1\right)a=x+5-bx+b$$

He hanga arowhānui tō te whārite.

$$\left(x+1\right)a=5+b+x-bx$$

Whakawehea ngā taha e rua ki te x+1.

$$\frac{\left(x+1\right)a}{x+1}=\frac{5+b+x-bx}{x+1}$$

Mā te whakawehe ki te x+1 ka wetekia te whakareanga ki te x+1.

$$a=\frac{5+b+x-bx}{x+1}$$

Whakaoti mō b (complex solution)

$$\left\{\begin{matrix}b=-\frac{ax-x+a-5}{x-1}\text{, }&x\neq 1\\b\in \mathrm{C}\text{, }&x=1\text{ and }a=3\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te a+b ki te x.

$$ax+bx+a-b+1=x+6$$

Tangohia te ax mai i ngā taha e rua.

$$bx+a-b+1=x+6-ax$$

Tangohia te a mai i ngā taha e rua.

$$bx-b+1=x+6-ax-a$$

Tangohia te 1 mai i ngā taha e rua.

$$bx-b=x+6-ax-a-1$$

Tangohia te 1 i te 6, ka 5.

$$bx-b=x+5-ax-a$$

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

$$\left(x-1\right)b=x+5-ax-a$$

He hanga arowhānui tō te whārite.

$$\left(x-1\right)b=5-a+x-ax$$

Whakawehea ngā taha e rua ki te x-1.

$$\frac{\left(x-1\right)b}{x-1}=\frac{5-a+x-ax}{x-1}$$

Mā te whakawehe ki te x-1 ka wetekia te whakareanga ki te x-1.

$$b=\frac{5-a+x-ax}{x-1}$$

Whakaoti mō a

$$\left\{\begin{matrix}a=-\frac{bx-x-b-5}{x+1}\text{, }&x\neq -1\\a\in \mathrm{R}\text{, }&x=-1\text{ and }b=-2\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te a+b ki te x.

$$ax+bx+a-b+1=x+6$$

Tangohia te bx mai i ngā taha e rua.

$$ax+a-b+1=x+6-bx$$

Me tāpiri te b ki ngā taha e rua.

$$ax+a+1=x+6-bx+b$$

Tangohia te 1 mai i ngā taha e rua.

$$ax+a=x+6-bx+b-1$$

Tangohia te 1 i te 6, ka 5.

$$ax+a=x+5-bx+b$$

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

$$\left(x+1\right)a=x+5-bx+b$$

He hanga arowhānui tō te whārite.

$$\left(x+1\right)a=5+b+x-bx$$

Whakawehea ngā taha e rua ki te x+1.

$$\frac{\left(x+1\right)a}{x+1}=\frac{5+b+x-bx}{x+1}$$

Mā te whakawehe ki te x+1 ka wetekia te whakareanga ki te x+1.

$$a=\frac{5+b+x-bx}{x+1}$$

Whakaoti mō b

$$\left\{\begin{matrix}b=-\frac{ax-x+a-5}{x-1}\text{, }&x\neq 1\\b\in \mathrm{R}\text{, }&x=1\text{ and }a=3\end{matrix}\right.$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tohatoha hei whakarea te a+b ki te x.

$$ax+bx+a-b+1=x+6$$

Tangohia te ax mai i ngā taha e rua.

$$bx+a-b+1=x+6-ax$$

Tangohia te a mai i ngā taha e rua.

$$bx-b+1=x+6-ax-a$$

Tangohia te 1 mai i ngā taha e rua.

$$bx-b=x+6-ax-a-1$$

Tangohia te 1 i te 6, ka 5.

$$bx-b=x+5-ax-a$$

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

$$\left(x-1\right)b=x+5-ax-a$$

He hanga arowhānui tō te whārite.

$$\left(x-1\right)b=5-a+x-ax$$

Whakawehea ngā taha e rua ki te x-1.

$$\frac{\left(x-1\right)b}{x-1}=\frac{5-a+x-ax}{x-1}$$

Mā te whakawehe ki te x-1 ka wetekia te whakareanga ki te x-1.

$$b=\frac{5-a+x-ax}{x-1}$$