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Solve the following system.
2x + 4y + 3z = 2
x + 2y - z = 0
4x + y - z = 6
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2x + 4y + 3z = 2 (1)
x + 2y - z = 0 (2)
4x + y - z = 6 (3)
Multiply equation (2) by 2. You will get
2x + 4y - 2z = 0. (2')
Now distract eqn.(2') from eqn.(1). You will get
5z = 2. Hence, z = 0.4.
Thus we just found the unknown z. It is known now.
Next, substitute this value of z into equations 2) and (3). You will get
x + 2y = 0.4, (4)
4x + y = 6.4. (5)
So, we reduced the original 3x3 system to 2x2-system.
At this point, I think that if you got an assignment to solve the system in 3 unknowns, you easily can solve the system in 2 unknowns.
Can you complete the solution on your own?
My lessons in this site on solving systems of linear equations in three unknowns by the Substitution and the Elimination methods are
- Solving systems of linear equations in 3 unknowns by the Substitution method,
- BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method,
- Solving systems of linear equations in 3 unknowns by the Elimination method and
- BRIEFLY on solving systems of linear equations in 3 unknowns by the Elimination method
My lessons in this site on solving systems of two linear equations in two unknowns are
- Solution of a linear system of two equations in two unknowns by the Substitution method
- Solution of a linear system of two equations in two unknowns by the Elimination method
- Solution of a linear system of two equations in two unknowns using determinant
- Geometric interpretation of a linear system of two equations in two unknowns
- Solving word problems using linear systems of two equations in two unknowns