Each salad contains red beans, Lima beans, and black eyed beans. Use the information below to determine how many of each of the three types of beans are needed

A.) Black eyed beans = at least 3 ; Red beans = atleast 4 ; Lima beans = atleast 5

B.) Red and Lima beans = 5 each ; Black eyed beans = 8

C.) Lima beans = 4 ; Black eyed beans = 2 ; Red beans = 2

Step-by-step explanation:

Let :

Black eyed beans = x

Red beans = x + 1

Lima beans = x + 1 +1 = x + 2

Total beans is atleast 12

x + (x +1) + (x + 2) ≥ 12

x + x + 1 + x + 2 ≥ 12

3x + 3 ≥ 12

3x ≥ 9

x ≥ 3

Black eyed beans = at least 3

Red beans = atleast 4

Lima beans = atleast 5

B.)

Same number of Red and Lima beans

Black eyed beans is 3 more than red beans

Total beans = 18

Red and Lima = x + x

Black eyed beans = x + 3

x + x + x + 3 = 18

3x + 3 = 18

3x = 18 - 3

3x = 15

x = 5

Red and Lima beans = 5 each

Black eyed beans = 3 + 5 = 8

C.)

Total beans in salad = x

Number of red beans = 2

Lima beans = 2 * 2 = 4

Lima beans = 1/2 of x ; x/2

4 = x /2

8 = x = total beans in salad

Red + lima + black eyed = 8

2 + 4 + black-eyed = 8

Black eyed beans = 8 - 6 = 2

Lima beans = 4

Black eyed beans = 2

Red beans = 2

question:

could we have options?

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step-by-step explanation: