Resolver 4(2y-5)^2=f(3y-1)^2

$$4 ( 2 y - 5 ) ^ { 2 } = f ( 3 y - 1 ) ^ { 2 }$$

Ebatzi: f

$$f=4\times \left(\frac{2y-5}{3y-1}\right)^{2}$$

$$y\neq \frac{1}{3}$$

Ekuazio lineala ebazteko urratsak

\left(2y-5\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$4\left(4y^{2}-20y+25\right)=f\left(3y-1\right)^{2}$$

Erabili banaketa-propietatea 4 eta 4y^{2}-20y+25 biderkatzeko.

$$16y^{2}-80y+100=f\left(3y-1\right)^{2}$$

\left(3y-1\right)^{2} zabaltzeko, erabili Newton-en binomioa \left(a-b\right)^{2}=a^{2}-2ab+b^{2}.

$$16y^{2}-80y+100=f\left(9y^{2}-6y+1\right)$$

Erabili banaketa-propietatea f eta 9y^{2}-6y+1 biderkatzeko.

$$16y^{2}-80y+100=9fy^{2}-6fy+f$$

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$9fy^{2}-6fy+f=16y^{2}-80y+100$$

Konbinatu f duten gai guztiak.

$$\left(9y^{2}-6y+1\right)f=16y^{2}-80y+100$$

Zatitu ekuazioaren bi aldeak 9y^{2}-6y+1 balioarekin.

$$\frac{\left(9y^{2}-6y+1\right)f}{9y^{2}-6y+1}=\frac{4\left(2y-5\right)^{2}}{9y^{2}-6y+1}$$

9y^{2}-6y+1 balioarekin zatituz gero, 9y^{2}-6y+1 balioarekiko biderketa desegiten da.

$$f=\frac{4\left(2y-5\right)^{2}}{9y^{2}-6y+1}$$

Zatitu 4\left(2y-5\right)^{2} balioa 9y^{2}-6y+1 balioarekin.

$$f=\frac{4\left(2y-5\right)^{2}}{\left(3y-1\right)^{2}}$$

Ebatzi: y (complex solution)

$$\left\{\begin{matrix}y=-\frac{-\sqrt{f}+10}{3\sqrt{f}-4}\text{; }y=\frac{\sqrt{f}+10}{3\sqrt{f}+4}\text{, }&f\neq \frac{16}{9}\\y=\frac{17}{12}\text{, }&f=\frac{16}{9}\end{matrix}\right.$$

Ebatzi: y

$$\left\{\begin{matrix}y=-\frac{-\sqrt{f}+10}{3\sqrt{f}-4}\text{; }y=\frac{\sqrt{f}+10}{3\sqrt{f}+4}\text{, }&f\neq \frac{16}{9}\text{ and }f\geq 0\\y=\frac{17}{12}\text{, }&f=\frac{16}{9}\end{matrix}\right.$$