2. Quan plans to spend less than $80 for buying groceries. He plans to spend $68.25 on food and spend the rest on juice. Each juice carton costs $3. He is curious how many juice cartons he can purchase before he runs out of money.
(a) Use x to represent the number of juice cartons Quan can purchase and write an inequality that can be used to solve for x.
(b) Solve the inequality. Use the solution to determine the number of juice cartons Quan can purchase.

Respuesta :

Y=3X+68.25

80=3X+68.25
Subtract 68.25 from both sides
11.75=3X

Divide

11.75/3=3.91
He can't buy a portion of a juice but he doesn't have enough to buy 4

So he can afford 3 juice cartons

P.S. I do k12 also if that is what you are doing.

Answer:

a)[tex]3x+68.25<80[/tex]

b)3

Step-by-step explanation:

Let x be the number of juice cartons.

Cost of 1 carton = $3

Cost of x cartons = 3x

He plans to spend $68.25 on food .

His total cost = 3x+68.25

He also plans He plans to spend $68.25 on food .

So, Equation becomes:

[tex]3x+68.25<80[/tex]

[tex]3x<80-68.25[/tex]

[tex]3x<11.75[/tex]

[tex]x<\frac{11.75}{3}[/tex]

[tex]x<3.9166[/tex]

So, he can purchase 3 juice cartons.

Hence an inequality that can be used to solve for x is [tex]3x+68.25<80[/tex]