Resolver 2ap+p^2d-pd=2a^2+a^2d-ad

$$2 a p + p ^ { 2 } d - p d = 2 a ^ { 2 } + a ^ { 2 } d - a d$$

Ebatzi: d

$$\left\{\begin{matrix}d=\frac{2a}{1-a-p}\text{, }&a\neq 1-p\\d\in \mathrm{R}\text{, }&a=p\text{ or }\left(a=0\text{ and }p=1\right)\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

Kendu a^{2}d bi aldeetatik.

$$2ap+p^{2}d-pd-a^{2}d=2a^{2}-ad$$

Gehitu ad bi aldeetan.

$$2ap+p^{2}d-pd-a^{2}d+ad=2a^{2}$$

Kendu 2ap bi aldeetatik.

$$p^{2}d-pd-a^{2}d+ad=2a^{2}-2ap$$

Konbinatu d duten gai guztiak.

$$\left(p^{2}-p-a^{2}+a\right)d=2a^{2}-2ap$$

Zatitu ekuazioaren bi aldeak p^{2}-p-a^{2}+a balioarekin.

$$\frac{\left(p^{2}-p-a^{2}+a\right)d}{p^{2}-p-a^{2}+a}=\frac{2a\left(a-p\right)}{p^{2}-p-a^{2}+a}$$

p^{2}-p-a^{2}+a balioarekin zatituz gero, p^{2}-p-a^{2}+a balioarekiko biderketa desegiten da.

$$d=\frac{2a\left(a-p\right)}{p^{2}-p-a^{2}+a}$$

Zatitu 2a\left(a-p\right) balioa p^{2}-p-a^{2}+a balioarekin.

$$d=-\frac{2a}{p+a-1}$$

Ebatzi: a

$$\left\{\begin{matrix}\\a=p\text{, }&\text{unconditionally}\\a=\frac{d\left(1-p\right)}{d+2}\text{, }&d\neq -2\\a\in \mathrm{R}\text{, }&p=1\text{ and }d=-2\end{matrix}\right.$$