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Book Free DemoLet us learn how to find the solution to the system of linear equations in three variables.
Example:
Solve the system of linear equations in three variables \(6x + 4y - 2z = 12\), \(-2x + 2y + z = 3\), \(2x + 2y + 2z = 8\).
Solution:
Let us name the equations.
\(6x + 4y - 2z = 12\) ---- (\(1\))
\(-2x + 2y + z = 3\) ---- (\(2\))
\(2x + 2y + 2z = 8\) ---- (\(3\))
Step 1: Solving equations (\(2\)) and (\(3\)).
\(-2x + 2y + z = 3\)
\(2x + 2y + 2z = 8\)
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\(4y + 3z = 11\) ---- (\(4\))
Step 2: Similarly, let us eliminate the variable \(x\) from equations \(1\) and \(2\).
\(6x + 4y -2z = 12\)
\(-6x + 6y + 3z = 9\)
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\(10y + z = 21\) ---- (\(5\))
Step 3: Solve equations (\(4\)) and (\(5\)).
\(4y + 3z = 11\)
\(30y + 3z = 63\)
(\(-\)) (\(-\)) (\(-\))
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\(- 26y = - 52\)
\(y = 2\)
Substitute the value of \(y\) in equation (\(5\)), we get:
\(20 + z = 21\)
\(z = 1\)
Step 4: Substitute the value of \(y\) and \(z\) in equation (\(1\)), we get:
\(6x + 4(2) - 2(1) = 12\)
\(6x + 8 - 2 = 12\)
\(6x + 6 = 12\)
\(6x = 6\)
\(x = 1\)
Therefore, the solution is \(x = 1\), \(y = 2\) and \(z = 1\).
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