Question:
If the solution of the system of simultaneous linear equations: $$ x + y - z = 6, $$ $$ 3x + 2y - z = 5, $$ $$ 2x - y - 2z + 3 = 0 $$ is \( x = \alpha, y = \beta, z = \gamma \), then \( \alpha + \beta = ?\)
If the solution of the system of simultaneous linear equations: $$ x + y - z = 6, $$ $$ 3x + 2y - z = 5, $$ $$ 2x - y - 2z + 3 = 0 $$ is \( x = \alpha, y = \beta, z = \gamma \), then \( \alpha + \beta = ?\)
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For solving a system of linear equations, use substitution or elimination to express one variable in terms of others and simplify step by step.
Updated On: May 22, 2025
- \( -7 \)
- \( 2 \)
- \( 1 \)
- \( -2 \)
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The Correct Option is B
Approach Solution - 1
To solve the system of simultaneous linear equations, we have the following equations:
- \( x + y - z = 6 \)
- \( 3x + 2y - z = 5 \)
- \( 2x - y - 2z + 3 = 0 \)
First, simplify the third equation:
\( 2x - y - 2z + 3 = 0 \)
\( \Rightarrow 2x - y - 2z = -3 \)
We can write the system of equations as:
- (1) \( x + y - z = 6 \)
- (2) \( 3x + 2y - z = 5 \)
- (3) \( 2x - y - 2z = -3 \)
To eliminate \( z \), subtract equation (1) from equation (2):
\( (3x + 2y - z) - (x + y - z) = 5 - 6 \)
\( 2x + y = -1 \) ...(4)
Next, eliminate \( z \) between equations (1) and (3):
\( (2x - y - 2z) - 2(x + y - z) = -3 - 2(6) \)
\( 2x - y - 2z - 2x - 2y + 2z = -15 \)
\( -3y = -15 \)
\( y = 5 \)
Substitute \( y = 5 \) into equation (4):
\( 2x + 5 = -1 \)
\( 2x = -6 \)
\( x = -3 \)
Now substitute \( x = -3 \) and \( y = 5 \) into equation (1) to find \( z \):
\( -3 + 5 - z = 6 \)
\( 2 - z = 6 \)
\( z = -4 \)
Now we have \( x = -3 \), \( y = 5 \), \( z = -4 \). Hence, \( \alpha = -3 \), \( \beta = 5 \), and \( \gamma = -4 \).
Finally, \(\alpha + \beta = -3 + 5 = 2\).
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Approach Solution -2
We are given the system of equations:
\[
x + y - z = 6,
\]
\[
3x + 2y - z = 5,
\]
\[
2x - y - 2z + 3 = 0.
\]
Step 1: Convert to standard form
Rearrange the third equation:
\[
2x - y - 2z = -3.
\]
Thus, the system of equations is:
\[
x + y - z = 6,
\]
\[
3x + 2y - z = 5,
\]
\[
2x - y - 2z = -3.
\]
Step 2: Solve for variables
Using the first equation:
\[
x + y = z + 6 \Rightarrow z = x + y - 6.
\]
Substituting \( z = x + y - 6 \) into the second equation:
\[
3x + 2y - (x + y - 6) = 5.
\]
Simplify:
\[
3x + 2y - x - y + 6 = 5.
\]
\[
2x + y + 6 = 5 \Rightarrow 2x + y = -1.
\]
Substituting \( z = x + y - 6 \) into the third equation:
\[
2x - y - 2(x + y - 6) = -3.
\]
Expanding:
\[
2x - y - 2x - 2y + 12 = -3.
\]
\[
-3y + 12 = -3.
\]
\[
-3y = -15 \Rightarrow y = 5.
\]
Substituting \( y = 5 \) into \( 2x + y = -1 \):
\[
2x + 5 = -1.
\]
\[
2x = -6 \Rightarrow x = -3.
\]
Using \( z = x + y - 6 \):
\[
z = -3 + 5 - 6 = -4.
\]
Step 3: Compute \( \alpha + \beta \)
\[
\alpha + \beta = x + y = -3 + 5 = 2.
\]
Thus, the correct answer is:
\[
\boxed{2}
\]
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