Answer:
11 years
Step-by-step explanation:
<u>Understanding the situation</u>
School A has graduated 1,700 students and graduates 50 students every year. The total amount of graduates at school A can be represented by the equation g = 50Y + 1700
School B has graduated 600 students, and graduates 150 every year. The total amount of graduated students at school B can be represented by the equation g = 150Y + 600
Using these two equations we want to find how many years it will take before school B has as many graduates as school A
<u>Determining the number of years</u>
To do so we must set the two equations equal to each other as both represent equations that represent the total number of graduates at the school after "y" years.
Doing so we would get 50Y + 1700 = 150Y + 600
Now we would just solve for Y algebraically
50Y + 1700 = 150Y + 600
==> subtract 600 from both sides
50Y + 1100 = 150Y
==> subtract 50Y from both sides
1100 = 100Y
==> divide both sides by 100
11 = y
So it will take 11 years for school B to have as many graduates as school A
<u>Checking our work</u>
We can also check our work by plugging in y = 11 and evaluating , if both equations have the same outcome we are correct
for school A : g = 50Y + 1700
==> plug in y = 11
g = 50(11) + 1700
==> multiply 50 and 11
g = 550 + 1700
==> add 1700 and 550
g = 2250
for school B : g = 150y + 600
==> plug in y = 11
g = 150(11) + 600
==> multiply 150 and 11
g = 1650 + 600
==> add 1650 and 600
g = 2250
both had an outcome of 2250 graduates so our answer is correct!
