x+y+z=100 and x/20+2y+5z=100 then find the value of x, y and z
Answers
Answer:
simplify the equation i.e x+2y+16z=80(dividing first eqn by 25)
get relation of x in terms of y and z with help of second equation. x=20-y-z
substitute x in simplified equation. ( 20-y-z)+2y+16z=80, ==>y+15z=60==>z=4-(y/15)
now we got the relation between y and z ,remember z is always less than 4
Now go for trial and error method to make y as a whole number substituting y=15 ,then z =4-(15/15) ==>z=4–1=3 also we know that x+y+z=20 substitute y and z values we get x=2
check x=2,y=15 and z=3 in first equation ,since LHS= RHS this will be the correct solution.
Answer:
Mr. Smith has $100.00 to buy 100 animals with. Cows cost $10.00 each, pigs cost $3.00 each and chickens cost $.50 each. How many can he buy with exactly $100.00. I have this equation so far, but where do I go from here? 10x + 3y + .5z = 100 Multiply through by 10 to clear decimal: 100x + 30y + 5z = 1000 Divide by 5 20x + 6y + z = 200 x + y + z = 100 19x + 5y = 100 19x = 100 - 5y 19x = 5(20 - y) x = 5(20 - y)/19 Since x is a positive integer, the 19 must divide evenly into 5(20 - y). the 19 will not divide evenly into the 5, so 19 must divide into 20 - y. 20 - y is at most 19, since y must be at least 1. The only integer at least 19 which is divisible by 19 is 19 itself. Therefore 20 - y = 19 -y = -1 y = 1 x = 5(20 - y)/19 x = 5(20 - 1)/19 x = 5(19)/19 x = 5 x + y + z = 100 5 + 1 + z = 100 6 + z = 100 z = 94 Therefore he bought 5 cows, 1 pig, and 94 chickens. Edwin