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1 pound of cereal a contains 10g of potassium, 2g of calcium, and 4g of magnesium. 1 pound of cereal B contains 6g of potassium, 0g of calcium and 3g of magnesium. 1 pound of cereal C contains 12g of potassium, 6g of calcium and 2g of magnesium. how many pounds of each cereal is needed to have a mixture containing 98g of potassium, 28g of calcium and 32g of magnesium

1 Answer

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Answer:

5 pounds of Cereal A

2 pounds of Cereal B

3 pounds of Cereal C

Step-by-step explanation:

Let x represent grams of cereal A

Let y represent grams of cereal B

Let z represent grams of cereal C

For potassium needed,

10g of cereal A, 6g of cereal B and 12g of cereal C which totals to 98g

Mathematically, we have 10x + 6y + 12z = 98 --------- Equation 1

For calcium,

2g of cereal A, 0g of cereal B and 6g of cereal C which totals to 28g.

Mathematically, we have 2x + 0y + 6z = 28

2x + 6z = 28 ---------- Equation 2

For magnesium

4g of cereal A, 3g of cereal B and 2g of cereal C, which totals to 32g

Mathematically, we have 4x + 3y + 2z = 32 ----------- Equation 3

Write out the three equations

10x + 6y + 12z = 98

2x + 6z = 28

4x + 3y + 2z = 32

We solve the above simultaneous equations using substitution method

In equation 2, we make x the subject of the formula

From 2x + 6z = 28

We have; 2x = 28 - 6z --------- Divide through by 2

2x/2 = (28-6z)/2

x = 28/2 - 6z/2

x = 14 - 3z ------------ Equation 4

Substitute (14-3z) for x in equation 1 and 2

In equation 1,

From 10x + 6y + 12z = 98

We have 10(14-3z) + 6y + 12z = 98 ---------- Open the bracket

140 - 30z + 6y + 12z = 98 -------------- Collect like terms

6y + 12z - 30z= 98 - 140

6y - 18z = -42 --------------- Divide through by 6

(6y - 18z)/6 = -42/6

y - 3z = -7 ---------------- Make y the subject of formula

y = 3z - 7 ------------- Equation 5

In equation 3,

From 4x + 3y + 2z = 32

We have; 4(14-3z) + 3y+ 2z = 32 -------------- Open the bracket

56 - 12z + 3y + 2z = 32 ------------------- Collect like terms

3y + 2z - 12z = 32 - 56

3y - 10z = -24 ---------------- Equation 6

Substitute (3z - 7) for y in equation 6

In equation 6;

From 3y - 10z = -24

We have 3(3z - 7) - 10z= -24 ------------ Open the bracket

9z - 21 - 10z = -24 ------- Collect like terms

9z- 10z = -24 +21

-z = -3

z = 3

Substitute 3 for z in equation 5

In equation 5;

From y = 3z - 7

We have; y = 3(3) - 7

y = 9 - 7

y = 2

Substitute 3 for z in equation 4

In equation 4;

From x= 14- 3z

We have; x = 14 - 3(3)

x = 14 - 9

x = 5

So, x = 5 y = 2 and z= 3

User Nathan Werry
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