Let starting salary = x
Increment every year = y
Therefore:
Salary after 4 years of service = x+4y
Salary after 10 years of service = x+10y
We have the equations:
[tex]\begin{gathered} x+4y=7500 \\ x+10y=9000 \end{gathered}[/tex]Substracting equation 1 from equation 2, we get:
[tex]x+10y-(x+4y)=9000-7500[/tex]Simplify:
[tex]\begin{gathered} x+10y-x-4y=1500 \\ 6y=1500 \\ Solve\text{ for y} \\ \frac{6y}{6}=\frac{1500}{6} \\ y=250 \end{gathered}[/tex]Next, substitute y = 250 in the equation 1:
[tex]x+4(250)=7500[/tex]And solve for x:
[tex]\begin{gathered} x+1000=7500 \\ x+1000-1000=7500-1000 \\ x=6500 \end{gathered}[/tex]Answer:
Starting salary = $6500
Annual increment = $250