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#8)One type of lawn fertilizer consists of a mixture of nitrogen, phosphorus and potassium. An 80lb sample contains 8 move lbs of nitrogen and phosphorus than potassium. There is 16lb more potasssium than 5 times the amount of phosphorus. Let x be the amount of nitrogen, let y be the amount of phosphorus and let z be the amount potassium in the sample. Find the amount of each of the chemicals in the sample.

X = _____
y = _____
z = _____
(Hint: you must set up and solve a 3 by 3 systems of equations.)

To solve this problem, we need to set up a system of equations using the given information and then solve for the unknowns, x, y, and z.

Step 1: Assign variables to the unknowns. Let x represent the amount of nitrogen, y represent the amount of phosphorus, and z represent the amount of potassium in the sample.

Step 2: Write down the given information as equations.

From the problem, we have the following information:

- The total weight of the sample is 80 lbs: x + y + z = 80

- The sample contains 8 more pounds of nitrogen and phosphorus compared to potassium: x + y = z + 8

- There is 16 lbs more potassium than 5 times the amount of phosphorus: z = 5y + 16

Step 3: Solve the system of equations using substitution or elimination.

By substituting the third equation into the second equation, we get:

x + y = 5y + 16 + 8

Simplifying the equation:

x - 4y = 24 (Equation 4)

Now, we have two equations with two variables:

x + y + z = 80 (Equation 1)
x - 4y = 24 (Equation 4)

We can solve this system of equations by substitution or elimination. For simplicity, let's use the substitution method.

From Equation 4, we have:

x = 4y + 24

By substituting this value of x into Equation 1, we get:

(4y + 24) + y + z = 80

Combining like terms:

5y + z = 56 (Equation 5)

Now, we have two equations:

5y + z = 56 (Equation 5)
x - 4y = 24 (Equation 4)

To eliminate z, we can subtract Equation 5 from Equation 1:

(x + y + z) - (5y + z) = 80 - 56

Simplifying the equation:

x - 4y = 24 (Equation 4)

Now, we have:

x - 4y = 24 (Equation 6)

We can solve Equation 6 for x:

x = 4y + 24 (Equation 7)

Step 4: Substitute the value of x from Equation 7 into Equation 5 to solve for y.

5y + z = 56

4y + 24 - 4y + z = 56 (substituting x = 4y + 24)

Simplifying the equation:

z = 32

Step 5: Substitute the value of z into Equation 5 to solve for y.

5y + 32 = 56

5y = 24

y = 4.8

Step 6: Substitute the value of y into Equation 7 to solve for x.

x = 4(4.8) + 24

x = 43.2

Step 7: Verify the solution by checking all three equations with the obtained values of x, y, and z.

x + y + z = 43.2 + 4.8 + 32 = 80

x - 4y = 43.2 - 4(4.8) = 24

5y + z = 5(4.8) + 32 = 56

The obtained values of x = 43.2, y = 4.8, and z = 32 satisfy all three equations.

Therefore, the amount of each chemical in the sample is as follows:

X = 43.2 lbs
y = 4.8 lbs
z = 32 lbs