Question

Hazel has an assortment of red, blue, and green balls. The number of red balls is 2/3 the number of blue balls. The number of green balls is 1 more than 1/3 the number of blue balls. In total, she has 15 balls.

An equation created to find the number of blue balls will have

- no solution
- one solution
- infinitely many solutions

Answer 1

x = 4
`(2)/(3)` y = 7 and z =3
`(1)/(3)`

This implies the equation has just one solution

Explanation:

To create the equation, we need to be able to write the information or interpret the question mathematically.

Let x equal to the number of red balls.

Let y equal to the number of blue balls.

Let z equal to the number of green balls.

From the question; "The number of red balls is 2/3 the number of blue balls" can be mathematically written as : x =
`(2)/(3)` y ---------------------------(1)

The next statement; "The number of green balls is 1 more than 1/3 the number of blue balls" can be written mathematically as: z = 1+
`(1)/(3)` y ----------------------------(2)

The next statement; "she has 15 balls." can be mathematically written as: x + y + z = 15 ----------------------------------------(3)

Substitute equation (1) and equation (2) into equation (3)

`(2)/(3)` y + y +1 +
`(1)/(3)` y = 15

We can rearrange this equation

`(2)/(3)` y +
`(1)/(3)` y + y +1 = 15

`(3)/(3)` y + y + 1 = 15

y + y + 1 = 15

2y + 1 = 15

subtract 1 from both-side of the equation

2y + 1 -1 = 15 -1

2y = 14

Divide both-side of the equation by 2

2y/2 = 14/2

y = 7

Substitute y = 7 into equation (1)

x =
`(2)/(3)` y

x =
`(2)/(3)` (7)

x = 14/3

x = 4
`(2)/(3)`

Substitute y= 7 in equation (2)

z = 1+
`(1)/(3)` y

z = 1+
`(1)/(3)` (7)

z = 1+ 7/3

z = 10/3

z =3
`(1)/(3)`

Therefore;

x = 4
`(2)/(3)` y = 7 and z =3
`(1)/(3)`

This implies the equation has just one solution.

Answer 2

it would have one solution since you know the total number of balls