You have a system of two equations with three variables \( x \), \( y \), and \( z \). The system is:

1. \( x + y + z = 100 \)
2. \( 1x + 5y + 0.05z = 100 \)

We can solve this system using substitution or elimination methods. Let's use substitution to express one variable in terms of the others.

### Step 1: Solve the first equation for \( x \)
From the first equation:
\[
x = 100 - y - z
\]

### Step 2: Substitute \( x = 100 - y - z \) into the second equation
Substitute into the second equation:
\[
1(100 - y - z) + 5y + 0.05z = 100
\]
Simplifying the equation:
\[
100 - y - z + 5y + 0.05z = 100
\]
Combine like terms:
\[
100 + 4y - 0.95z = 100
\]

### Step 3: Simplify the equation
Subtract 100 from both sides:
\[
4y - 0.95z = 0
\]
So,
\[
4y = 0.95z
\]
Now divide by 4:
\[
y = \frac{0.95z}{4}
\]
\[
y = 0.2375z
\]

### Step 4: Substitute \( y = 0.2375z \) into the first equation
Now substitute \( y = 0.2375z \) into the first equation:
\[
x + 0.2375z + z = 100
\]
Simplify:
\[
x + 1.2375z = 100
\]
Solve for \( x \):
\[
x = 100 - 1.2375z
\]

### Step 5: Parametric solutions
Now, you have expressions for both \( x \) and \( y \) in terms of \( z \):
\[
x = 100 - 1.2375z
\]
\[
y = 0.2375z
\]
Thus, the solution depends on the value of \( z \). You can assign any value to \( z \) and calculate corresponding values of \( x \) and \( y \).

Would you like to choose a specific value for \( z \), or should we explore this further? Let me know if you need more details!

### Relative questions:
1. How can we test specific values for \( z \) to find solutions?
2. What happens if we set \( z = 0 \)? How does it affect \( x \) and \( y \)?
3. Can we solve this system graphically, and what does the solution represent geometrically?
4. What is the effect of having more variables than equations in this system?
5. How does the substitution method work for larger systems of equations?

### Tip: When solving systems with more variables than equations, expect infinitely many solutions, typically expressed parametrically.

Explore how to solve a system of linear equations with three variables: x + y + z = 100 and 1x + 5y + 0.05z = 100. Learn the step-by-step process of using substitution to find parametric solutions. Suitable for students in Grades 9-12.

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