Romeo bought a mixture of 20-cent, 35-cent,
and 50-cent valentines. The number of 20-cent
valentines was 1 more than twice the number of
35-cent valentines, and the number of 50-cent
valentines was 2 less than the number of
35-cent ones. If he spent $4.20 all together,
how many valentines of each kind did he buy?

Question

contestada

Respuesta :

sqdancefan sqdancefan · Answer 1 · 2021-08-22T22:24:01+02:00

9514 1404 393

Answer:

Step-by-step explanation:

Let x, y, z represent the numbers of 20-cent, 35-cent, and 50-cent valentines. Then the given relations can be written as ...

x = 1 +2y

z = -2 +y

0.20x +0.35y +0.50z = 4.20

__

Using the first two equations to substitute for x and y in the third equation, we get ...

0.20(1 +2y) +0.35y +0.50(-2 +y) = 4.20

1.25y -0.80 = 4.20 . . . . . simplify

1.25y = 5.00 . . . . . . . . . . add 0.80

y = 4 . . . . . . . . . . . . . . . divide by 1.25

x = 1 +2(4) = 9

z = -2 +4 = 2

Romeo bought 9 20-cent, 4 35-cent, and 2 50-cent valentines.