Juan is buying flowers for his mother. He has $18 to spend and sees that roses are $3 each and each carnation are $1.50 each. He wants to buy 3 times as many carnations as roses and spends all his money on flowers. Write a system of equations that models this situation. Is there a viable solution that meets Juan’s conditions?

a. 3n₁ + 1.5n₂ = 18 and n₂ = 3n₁. b. there is no viable solution that meets Juan's conditions.

Step-by-step explanation:

a. Let n₁ be the number of roses and n₂ be the number of carnations. Now each rose costs $3. So the cost of n₁ roses will be 3n₁. Also, each carnation costs $1.5. So the cost of n₂ carnation will be 1.5n₂.

Since the total amount available is $18, then 3n₁ + 1.5n₂ = 18.

Also, he wants to buy 3 times as many carnations as roses. So, n₂ = 3n₁.

So, the system of equations are

3n₁ + 1.5n₂ = 18 and n₂ = 3n₁.

b. 3n₁ + 1.5n₂ = 18 and n₂ = 3n₁.

We solve the above system of equations by inserting the second equation into the first. So,

3n₁ + 1.5n₂ = 18

3n₁ + 1.5(3n₁.) = 18

3n₁ + 4.5n₁ = 18

7.5n₁ = 18

dividing both sides by 7.5, we have

n₁ = 18/7.5

= 2.4

So,  n₂ = 3n₁

= 3(2.4)

= 7.2

Since n₁ = 2.4 and n₂ = 7.2 are not whole numbers, there is no viable solution that meets Juan's conditions.

Approximating n₁ = 2 and n₂ = 7. Inserting them into the first equation, we have

3n₁ + 1.5n₂ = 18

3(2) + 1.5(7) = 18

6 + 10.5 = 16.5 ≠ 18

Since L.H.S ≠ R.H.S, there is no viable solution that meets Juan's conditions.

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