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Alik [6]
3 years ago
11

Three bouquets of flowers are ordered at a florist. 3 roses, 2 carnations, and 1 tulip cost $14. 6 roses, 2 carnations, and 6 tu

lips cost $38. 1 rose, 12 carnations, and 1 tulip cost $18. How much does each item cost?
Mathematics
1 answer:
STALIN [3.7K]3 years ago
6 0

cost of 1 rose = $ 3

cost of 1 carnation = $ 1

cost of 1 tulip = $ 3

<em><u>Solution:</u></em>

Let "r" be the cost of 1 rose

Let "c" be the cost of 1 carnation

Let "t" be the cost of 1 tulip

<em><u>3 roses, 2 carnations, and 1 tulip cost $14</u></em>

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

3 \times r + 2 \times c + 1 \times t = 14

3r + 2c + 1t = 14 ----- eqn 1

<em><u>6 roses, 2 carnations, and 6 tulips cost $38</u></em>

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

6 \times r + 2 \times c + 6 \times t = 38

6r + 2c + 6t = 38 ------ eqn 2

<u><em>1 rose, 12 carnations, and 1 tulip cost $18</em></u>

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

1 \times r + 12 \times c + 1 \times t = 18

r + 12c + t = 18 ----- eqn 3

<em><u>Let us solve eqn 1 and eqn 2 and eqn 3 to find values of "r" "c" "t"</u></em>

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

r = \frac{14 - 2c - t}{3}

Substitute the above value of r in eqn 2

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

Substitute c = 2t - 5 and r = \frac{14 - 2c - t}{3} in eqn 3

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

<h3>t = 3</h3>

Substitute t = 3 in c = 2t - 5

c = 2(3) - 5

<h3>c = 1</h3>

Substitute t = 3 and c = 1 in r = \frac{14 - 2c - t}{3}

r = \frac{14 - 2(1) - 3}{3}\\\\r = \frac{14 - 2 - 3}{3}\\\\r = \frac{9}{3} = 3

<h3>r = 3</h3>

<em><u>Summarizing the results:</u></em>

cost of 1 rose = $ 3

cost of 1 carnation = $ 1

cost of 1 tulip = $ 3

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