1. The first equation gives you an equivalent for y. Use that in the second equation.
.. 4x + (x+5) = 20
.. 5x + 5 = 20 . . . . collect terms
.. 5x = 15 . . . . . . . . subtract 5
.. x = 3 . . . . . . . . . . divide by 5
The first equation tells you how to find y.
.. y = x + 5
.. y = 3 + 5 = 8
The solution is (x, y) = (3, 8).
2. Add 2x to the first equation to get an expression for y.
.. y = 3 + 2x
Use this in the second equation.
.. 6x - 3(3 +2x) = 21
.. 6x - 9 - 6x = 21 . . . eliminate parentheses
.. -9 = 21 . . . . . . . . . . false. There is no solution to this set of equations.
3. Subtract 2y from the first equation to get an expression for x.
.. x = -1 - 2y
Use this in the second equation.
.. 4(-1 -2y) -4y = 20 . . . . . substitute for x
.. -4 -8y -4y = 20 . . . . . . . eliminate parentheses
.. -12y = 24 . . . . . . . . . . . . collect terms, add 4
.. y = -2 . . . . . . . . . . . . . . .divide by -12
.. x = -1 -2*(-2) . . . . . . . . . use the equation for x to find x
.. x = 3
The solution is (x, y) = (3, -2).
Word Problem
a) Let f and n represent the total dollar cost of membership in the "fee" and "no-fee" gyms. Let m represent the number of months of membership.
.. f = 150 + 35m . . . . $150 plus $35 for each month
.. n = 60m . . . . . . . . . $60 each month
b) The costs will be the same when f = n.
.. f = n
.. 150 +35m = 60m
.. 150 = 25m . . . . . . . . . subtract 35m
.. 6 = m . . . . . . . . . . . . . divide by 25
The cost of membership will be the same after 6 months.
The cost will be $60*6 = $150 +$35*6 = $360.
c) If Cathy cancels in 5 months, the no-fee gym will cost less.
.. n = 60*5 = 300
.. f = 150 +35*5 = 325
