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For what value of a, if any, will the following system of equations in x, y and z have a solution?
2x + 3y = 4
x+y+z = 4
x + 2y - z = a
2x + 3y = 4
x+y+z = 4
x + 2y - z = a
- a)Any real number
- b)0
- c)1
- d)There is no such value
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
For what value of a, if any, will the following system of equations in...
Given:
System of equations:
2x + 3y = 4x
y + z = 4x
2y - z = a
To find:
The value of 'a' for which the given system of equations has a solution.
Solution:
We can solve the system of equations using the method of elimination or substitution.
Elimination Method:
Step 1: Multiply the first equation by (-1) and add it to the second equation to eliminate 'x'.
-2x - 3y = -4x
y + z = 4x
-----------------
-2y - z = 0
Step 2: Multiply the second equation by 2 and add it to the third equation to eliminate 'y'.
2y + 2z = 8x
2y - z = a
-----------------
3z = 8x + a
Now we have two equations:
-2y - z = 0
3z = 8x + a
We can solve these equations to find the values of 'x', 'y', and 'z'.
Substitution Method:
Step 1: Solve the first equation for 'x' in terms of 'y'.
2x = 3y
x = (3/2)y
Step 2: Substitute the value of 'x' in the second equation.
y + z = 4(3/2)y
y + z = 6y
Step 3: Simplify the equation.
z = 6y - y
z = 5y
Step 4: Substitute the values of 'x' and 'z' in the third equation.
2(3/2)y - (5y) = a
3y - 5y = a
-2y = a
From the above equation, we can see that the value of 'a' is equal to -2y.
Conclusion:
Since 'y' can take any real value, 'a' can also take any real value. Therefore, the correct answer is option 'A' - Any real number. The given system of equations will have a solution for any value of 'a'.
System of equations:
2x + 3y = 4x
y + z = 4x
2y - z = a
To find:
The value of 'a' for which the given system of equations has a solution.
Solution:
We can solve the system of equations using the method of elimination or substitution.
Elimination Method:
Step 1: Multiply the first equation by (-1) and add it to the second equation to eliminate 'x'.
-2x - 3y = -4x
y + z = 4x
-----------------
-2y - z = 0
Step 2: Multiply the second equation by 2 and add it to the third equation to eliminate 'y'.
2y + 2z = 8x
2y - z = a
-----------------
3z = 8x + a
Now we have two equations:
-2y - z = 0
3z = 8x + a
We can solve these equations to find the values of 'x', 'y', and 'z'.
Substitution Method:
Step 1: Solve the first equation for 'x' in terms of 'y'.
2x = 3y
x = (3/2)y
Step 2: Substitute the value of 'x' in the second equation.
y + z = 4(3/2)y
y + z = 6y
Step 3: Simplify the equation.
z = 6y - y
z = 5y
Step 4: Substitute the values of 'x' and 'z' in the third equation.
2(3/2)y - (5y) = a
3y - 5y = a
-2y = a
From the above equation, we can see that the value of 'a' is equal to -2y.
Conclusion:
Since 'y' can take any real value, 'a' can also take any real value. Therefore, the correct answer is option 'A' - Any real number. The given system of equations will have a solution for any value of 'a'.
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Community Answer
For what value of a, if any, will the following system of equations in...
The correct answer is d.
2x + 3y = 4 ....(1)
x + y + z = 4 ....(2)
x + 2y – z = a ....(3)
add (2) and (3),
x + y + z + x + 2y – z = 4 + a
2x + 3y = 4 + a ....(4)
from (1) and (4),
value of a must be 0 otherwise no solution
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