Define the equations that describe the situation
[tex]\begin{gathered} x=4y+3 \\ x\cdot y=45 \end{gathered}[/tex]Clear x from the second equation
[tex]x=\frac{45}{y}[/tex]Make both equations equal
[tex]\begin{gathered} 4y+3=\frac{45}{y} \\ y(4y+3)=45 \\ 4y^2+3y=45 \\ 4y^2+3y-45=0 \end{gathered}[/tex]Solve the cuadratic equation
[tex]\begin{gathered} y=\frac{-3\pm\sqrt[]{3^2-4\cdot4\cdot(-45)}}{2\cdot4} \\ y=\frac{-3\pm\sqrt[]{9+720}}{8} \\ y=\frac{-3\pm27}{8} \\ y1=-\frac{30}{8}=-\frac{15}{4} \\ y2=\frac{24}{8}=3 \end{gathered}[/tex]As we are dealing with integers, the value of y is 3, now use one of the first equations to find x
[tex]\begin{gathered} x=\frac{45}{y} \\ x=\frac{45}{3} \\ x=15 \end{gathered}[/tex]The integers are 3 and 15
