Question 849811
let x = cost to feed coyotes and y = cost to feed lions.


on wednesday, 21x + 21y = 315


on tuesday, 28x + 36y = 484


solve these 2 equations simultaneously to find the value of x and y.


equations are:


21x + 21y = 315
28x + 36y = 484


you can solve by substitution, or by elimination, or by graphing.


i'll solve by elimination.


least common multiple of 21 and 28 is 84.


multiply first equation by 4 and second equation by 3 to get:


84x + 84y = 1260
84x + 108y = 1452


subtract first equation from the second to get:


24y = 192


divide 192 by 24 to get:


y = 8


now that you know the value of y, substitute for y in either equation to find the value of x.


substituting in 21x + 21y = 315 gets:


21x + 21(8) = 315
simplify to get:
21x + 168  = 315
subtract 168 from both sides of the equation to get:
21x = 147
divide both sides of the equation by 21 to get:
x = 7


your answer should be:


x = 7 = cost to feed one coyote.
y = 8 = cost to feed one lion.


confirm this solution is correct by substituting for x and y in both equations.


both original equations are:


21x + 21y = 315
28x + 36y = 484


after substituting, these equations become:


21(7) = 21(8) = 315
28(7) + 36(8) = 484


evaluate both equations to get:


315 = 315
484 = 484


the solution is confirmed to be good.


x = 7
y = 8


cost to feed one coyote = 7.
cost to feed one lion = 8.


if the price is increased by 5%, then:


cost to feed one coyote becomes 7 + .05*7 = 7.35
cost to feed one lion becomes 8 + .05*8 = 8.4.


the cost to feed 35 coyotes and 40 lions then becomes:


35(7.35) + 40(8.4) = 593.25