Russ P. answered 10/09/14
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Patient MIT Grad For Math and Science Tutoring
Kahlee,
Y ou have a problem in two unknowns so you must have 2 independent equations to solve for them, which you do in this problem.
Let d = daytime rate in $ per hour
Let n = nighttime rate in $ per hour
Then 20d + 20n = 380 dividing thru by 2 to get 10d in first term: 10d + 10n = 190
and 30d + 12n = 276 dividing thru by -3 to get -10d in first term: -10d - 4n = -92
Adding the two equations eliminates the d unknown: (10d - 10d) + (10n -4n) = (190 - 92) = 98
So 6n = 98 and n = 98 / 6 = $ 16.33 per hour nighttime pay rate
and 10d = 190 - 10n = 190 -163.33 = 26.67, so d = $ 2.67 per hour daytime rate
Also, instead of adding the modified equations, you could instead have used the method of substitution: solve the first one for d interns of n, and then substitute that for d in the second equation and solve for n, then back to solve for d.
That solves the math problem, but you wouldn't see such disparities in the real world, where the night pay rate may be 50% more or double the day rate, not six times! And the day rate is well below the minimum wage at around $7.50
So the problem writer should have picked coefficients that would produce results closer to the real world.