For the following system of equations in echelon​ form, tell how many solutions there are in nonnegative integers.

4 x plus 3 y plus 2 z4x+3y+2z

equals=9090

y minus 2 zy−2z

equals=1010

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Part 1

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

There is at least one​ solution,

left parenthesis nothing comma nothing comma z right parenthesisenter your response here,enter your response here,z.

​(Use integers or fractions for any numbers in the​ expression.)

B.

There is no solution.

We are tasked with determining how many solutions exist in nonnegative integers for the given system of equations:

1. \( 4x + 3y + 2z = 90 \)
2. \( y - 2z = 10 \)

### Step 1: Solve the second equation for \( y \)

From the second equation, we can express \( y \) in terms of \( z \):

\[
y = 2z + 10
\]

### Step 2: Substitute into the first equation

Now, substitute \( y = 2z + 10 \) into the first equation \( 4x + 3y + 2z = 90 \):

\[
4x + 3(2z + 10) + 2z = 90
\]

Simplify the expression:

\[
4x + 6z + 30 + 2z = 90
\]
\[
4x + 8z + 30 = 90
\]
\[
4x + 8z = 60
\]
\[
x + 2z = 15
\]

### Step 3: Solve for \( x \) in terms of \( z \)

\[
x = 15 - 2z
\]

### Step 4: Analyze the constraints

We now have the following conditions:

- \( x = 15 - 2z \)
- \( y = 2z + 10 \)

For \( x \) and \( y \) to be nonnegative integers, we need to ensure that:

1. \( x \geq 0 \), meaning \( 15 - 2z \geq 0 \), or \( z \leq 7.5 \). Since \( z \) must be a nonnegative integer, the possible values for \( z \) are \( z = 0, 1, 2, 3, 4, 5, 6, 7 \).

2. \( y \geq 0 \), meaning \( 2z + 10 \geq 0 \), which is always true since \( z \) is nonnegative.

### Step 5: Find the integer solutions

Now, let's verify the integer solutions for each \( z \):

- \( z = 0 \): \( x = 15 - 2(0) = 15 \), \( y = 2(0) + 10 = 10 \)
- \( z = 1 \): \( x = 15 - 2(1) = 13 \), \( y = 2(1) + 10 = 12 \)
- \( z = 2 \): \( x = 15 - 2(2) = 11 \), \( y = 2(2) + 10 = 14 \)
- \( z = 3 \): \( x = 15 - 2(3) = 9 \), \( y = 2(3) + 10 = 16 \)
- \( z = 4 \): \( x = 15 - 2(4) = 7 \), \( y = 2(4) + 10 = 18 \)
- \( z = 5 \): \( x = 15 - 2(5) = 5 \), \( y = 2(5) + 10 = 20 \)
- \( z = 6 \): \( x = 15 - 2(6) = 3 \), \( y = 2(6) + 10 = 22 \)
- \( z = 7 \): \( x = 15 - 2(7) = 1 \), \( y = 2(7) + 10 = 24 \)

### Step 6: Conclusion

There are 8 solutions for \( (x, y, z) \) in nonnegative integers.

Thus, the correct answer is:

**A. There is at least one solution, \( (x, y, z) = (15 - 2z, 2z + 10, z) \), where \( z = 0, 1, 2, 3, 4, 5, 6, 7 \).**

Would you like more details on the steps or have any questions?

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**Relative Questions:**

1. What is the significance of transforming equations into echelon form?
2. Can a system of equations have an infinite number of nonnegative integer solutions?
3. How does the nature of coefficients affect the existence of integer solutions?
4. What methods can be used to solve systems of linear Diophantine equations?
5. How do constraints like nonnegative integers influence solution sets?

**Tip:** Always simplify one equation first when solving systems—it reduces complexity in substitution steps.

For the following system of equations in echelon form, tell how many solutions there are in nonnegative integers.

4x + 3y + 2z = 90

y - 2z = 10

A. There is at least one solution, (x, y, z).
B. There is no solution.

Discover how to solve the system of linear equations 4x + 3y + 2z = 90 and y - 2z = 10 for nonnegative integer solutions. Learn how to apply the substitution method step-by-step and find all possible solutions for x, y, and z. Suitable for students in Grades 9-12.

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