Prove that $F(x,y)=(x-4y,2x+3y)$ is bijective

For proving bijectiveness you have to show two things

1. For $(x_1,y_1)ne(x_2,y_2)$ we should have $F(x_1,y_1)ne F(x_2,y_2)$
2. For any $(a,b)$, there exists $(x,y)$, such that $F(x,y)=(a,b)$

For proving 1 we show that If $F(x_1,y_1)=F(x_2,y_2)$, then we must have $(x_1,y_1)=(x_2,y_2)$. Now begin{equation} x_1-4y_1=x_2-4y_2tag{1} end{equation}

begin{equation} 2x_1+3y_1=2x_2+3y_2tag{2} end{equation} Then $(2)-(1)times 2$ yields $11y_1=11y_2$, which implies $y_1=y_2$, which again implies that $x_1=x_2$.

For proving 2 we show that the system of equations

begin{equation} x-4y=atag{3} end{equation}

begin{equation} 2x+3y=btag{4} end{equation}

has a solution for any $a,binBbb{R}$. $(4)-(3)times2$ yields $11y=b-2aimplies y=frac{b-2a}{11}implies x=frac{4b+3a}{11}$. Thus there indeed is a solution.

Hence $F(x,y)$ is both one-to-one and onto, hence bijective.