Question

Solve the equations:A+C=1,B-2A=0,C-2B=3

Answer 1

To solve the system of equations given by:

1. \( A + C = 1 \)
2. \( B - 2A = 0 \)
3. \( C - 2B = 3 \)

let's express the variables in terms of one another.

From equation (2), we can express \( B \) in terms of \( A \):
\[
B = 2A
\]

Now we can substitute \( B \) into equation (3):

Substituting \( B = 2A \) into equation (3):
\[
C - 2(2A) = 3
\]
This simplifies to:
\[
C - 4A = 3 \quad \text{(4)}
\]

Now we have two equations:
1. \( A + C = 1 \)
2. \( C - 4A = 3 \)

We can express \( C \) from equation (1):
\[
C = 1 - A
\]

Now, substitute \( C = 1 - A \) into equation (4):
\[
(1 - A) - 4A = 3
\]
This simplifies to:
\[
1 - 5A = 3
\]
Solving for \( A \):
\[
-5A = 3 - 1
\]
\[
-5A = 2
\]
\[
A = -\frac{2}{5}
\]

Now substituting \( A = -\frac{2}{5} \) back into the equation for \( B \):
\[
B = 2A = 2\left(-\frac{2}{5}\right) = -\frac{4}{5}
\]

Now substituting \( A = -\frac{2}{5} \) into the equation for \( C \):
\[
C = 1 - A = 1 - \left(-\frac{2}{5}\right) = 1 + \frac{2}{5} = \frac{5}{5} + \frac{2}{5} = \frac{7}{5}
\]

Thus, the solutions are:
\[
A = -\frac{2}{5}, \quad B = -\frac{4}{5}, \quad C = \frac{7}{5}
\]

Therefore, the final answer is:
\[
(A, B, C) = \left(-\frac{2}{5}, -\frac{4}{5}, \frac{7}{5}\right)
\]