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

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

Solve for f

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

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

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2y-5\right)^{2}.

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

Use the distributive property to multiply 4 by 4y^{2}-20y+25.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3y-1\right)^{2}.

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

Use the distributive property to multiply f by 9y^{2}-6y+1.

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

Swap sides so that all variable terms are on the left hand side.

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

Combine all terms containing f.

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

Divide both sides by 9y^{2}-6y+1.

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

Dividing by 9y^{2}-6y+1 undoes the multiplication by 9y^{2}-6y+1.

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

Divide 4\left(2y-5\right)^{2} by 9y^{2}-6y+1.

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

Solve for y

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