cost of 1 rose = $ 3
cost of 1 carnation = $ 1
cost of 1 tulip = $ 3
Solution:
Let "r" be the cost of 1 rose
Let "c" be the cost of 1 carnation
Let "t" be the cost of 1 tulip
3 roses, 2 carnations, and 1 tulip cost $14
So we can frame a equation as:
3 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 14
[tex]3 \times r + 2 \times c + 1 \times t = 14[/tex]
3r + 2c + 1t = 14 ----- eqn 1
6 roses, 2 carnations, and 6 tulips cost $38
So we can frame a equation as:
6 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 6 tulip x cost of 1 tulip = $ 38
[tex]6 \times r + 2 \times c + 6 \times t = 38[/tex]
6r + 2c + 6t = 38 ------ eqn 2
1 rose, 12 carnations, and 1 tulip cost $18
So we can frame a equation as:
1 rose x cost of 1 rose + 12 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 18
[tex]1 \times r + 12 \times c + 1 \times t = 18[/tex]
r + 12c + t = 18 ----- eqn 3
Let us solve eqn 1 and eqn 2 and eqn 3 to find values of "r" "c" "t"
3r + 2c + 1t = 14 ----- eqn 1
6r + 2c + 6t = 38 ------ eqn 2
r + 12c + t = 18 ----- eqn 3
From eqn 1,
3r = 14 - 2c - t
[tex]r = \frac{14 - 2c - t}{3}[/tex]
Substitute the above value of r in eqn 2
[tex]6(\frac{14 - 2c - t}{3})+ 2c + 6t = 38\\\\28 - 4c - 2t + 2c + 6t = 38\\\\-2c +4t = 10\\\\-2c = 10 - 4t\\\\2c = 4t - 10\\\\c = 2t - 5[/tex]
Substitute c = 2t - 5 and [tex]r = \frac{14 - 2c - t}{3}[/tex] in eqn 3
[tex]12(2t - 5) + \frac{14 - 2c - t}{3} + t = 18\\\\24t - 60 + \frac{14-2(2t - 5) - t}{3} + t = 18\\\\72t - 180 + 14 - 4t +10 - t + 3t = 54\\\\70t = 54 + 180 - 14 -10\\\\70t = 210\\\\t = 3[/tex]
t = 3
Substitute t = 3 in c = 2t - 5
c = 2(3) - 5
c = 1
Substitute t = 3 and c = 1 in [tex]r = \frac{14 - 2c - t}{3}[/tex]
[tex]r = \frac{14 - 2(1) - 3}{3}\\\\r = \frac{14 - 2 - 3}{3}\\\\r = \frac{9}{3} = 3[/tex]
r = 3
Summarizing the results:
cost of 1 rose = $ 3
cost of 1 carnation = $ 1
cost of 1 tulip = $ 3
