Answer :
we write an equation for each statement
Ten times the number of fast was 140 less than twice the number of slow
[tex]10f=2s-140[/tex]Also, one-half the number of slow exceeded 3 times the number of fast by 10
[tex]\begin{gathered} \frac{s}{2}=3f\times10 \\ \\ \frac{s}{2}=30f \end{gathered}[/tex]where f is the number of fast and s the numer of slow
Now we can solve one unknow from any equation and replace on the other equaiton
for example:
I will solve s from the second equation
[tex]\begin{gathered} \frac{s}{2}=30f \\ \\ s=2\times30f \\ s=60f \end{gathered}[/tex]and replace s on the other equation
[tex]\begin{gathered} 10f=2s-140 \\ 10f=2(60f)-140 \\ 10f=120f-140 \end{gathered}[/tex]now we place the terms with f on the same side
[tex]10f-120f=-140[/tex]simplify
[tex]\begin{gathered} (10-120)f=-140 \\ -110f=-140 \\ 110f=140 \\ \\ f=\frac{14}{11}\approx1.27 \end{gathered}[/tex]the number of fast was 1.27
now the number of slow we can check it if we replace f on any equation, for example on the Second
[tex]\begin{gathered} \frac{s}{2}=30f \\ \\ s=2\times30f \\ s=60f \\ s=60(\frac{14}{11}) \\ \\ s=\frac{840}{11}\approx76.36 \end{gathered}[/tex]