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
