For this case we have a system of these equations with two unknowns:
[tex]a + b = 50\\a + 5 = \frac {2} {3} (b-5)[/tex]
According to the first equation we have:
[tex]a + 5 = \frac {2} {3} b- \frac {10} {3}\\a = \frac {2} {3} b- \frac {10} {3} -5\\a = \frac {2} {3} b - (\frac {25} {3})\\a = \frac {2} {3} b- \frac {25} {3}[/tex]
According to the second equation we have:
[tex]a = 50-b[/tex]
We match:
[tex]50-b = \frac {2} {3} b- \frac {25} {3}\\-b- \frac {2} {3} b = - \frac {25} {3} -50\\- \frac {5} {3} b = - \frac {175} {3}\\\frac {5} {3} b = \frac {175} {3}\\b = \frac {175 * 3} {3 * 5}\\b = \frac {525} {15}\\b = 35[/tex]
Now, we find the value of the variable "a":
[tex]a = 50-35\\a = 15[/tex]
Finally we have to:
[tex]3 (15) -35 = 45-35 = 10[/tex]
Answer:
[tex]3a-b = 10[/tex]
