X+y = 10, y + z=20, x + z= 24 find x, y and z

We have three equations involving three variables \( x \), \( y \), and \( z \):

1. \( x + y = 10 \)
2. \( y + z = 20 \)
3. \( x + z = 24 \)

We can solve this system of equations by using substitution or elimination.

### Step 1: Solve for \( x \) in terms of \( y \) from Equation 1:
\[
x = 10 - y
\]

### Step 2: Substitute \( x = 10 - y \) into Equation 3:
\[
(10 - y) + z = 24
\]
Simplify:
\[
10 - y + z = 24
\]
\[
z = 24 - 10 + y = 14 + y
\]
Thus, \( z = 14 + y \).

### Step 3: Substitute \( z = 14 + y \) into Equation 2:
\[
y + (14 + y) = 20
\]
Simplify:
\[
2y + 14 = 20
\]
\[
2y = 6
\]
\[
y = 3
\]

### Step 4: Substitute \( y = 3 \) into the equations for \( x \) and \( z \):
- From \( x = 10 - y \):
  \[
  x = 10 - 3 = 7
  \]
- From \( z = 14 + y \):
  \[
  z = 14 + 3 = 17
  \]

### Final Solution:
- \( x = 7 \)
- \( y = 3 \)
- \( z = 17 \)

Would you like more details on this method or a different approach?

Here are 5 questions related to this:

1. How can the elimination method be applied to solve this system?
2. What are other ways to solve systems of equations?
3. What happens if one of the equations was slightly different?
4. Can we use matrices to solve this system of equations?
5. How would this system change if we introduced a fourth variable?

**Tip:** For solving linear systems, using substitution works well when equations are easily rearranged, while elimination is better for more complex systems.

X + y = 10, y + z = 20, x + z = 24. Find x, y, and z.

Learn how to solve the system of linear equations: X + y = 10, y + z = 20, x + z = 24. This solution uses the substitution method to find the values of x, y, and z. Suitable for students in Grades 7-9, covering key algebraic concepts.

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