Question

solve the system.
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$$\begin{cases} x + 9y + z = 20 x + 10y - 2z = 18 3x + 27y + 2z = 58 \\end{cases}$$

enter your answer as an ordered triple (x, y, z).
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Explanation:

Step1: Eliminate x from first two equations

Subtract the first equation \(x + 9y + z = 20\) from the second equation \(x + 10y - 2z = 18\).
\((x + 10y - 2z)-(x + 9y + z)=18 - 20\)
Simplify: \(y - 3z=-2\) (Equation 4)

Step2: Eliminate x from first and third equations

Multiply the first equation by 3: \(3x + 27y + 3z = 60\) (Equation 5)
Subtract the third equation \(3x + 27y + 2z = 58\) from Equation 5:
\((3x + 27y + 3z)-(3x + 27y + 2z)=60 - 58\)
Simplify: \(z = 2\)

Step3: Substitute z into Equation 4

Substitute \(z = 2\) into \(y - 3z=-2\):
\(y-3\times2=-2\)
\(y - 6=-2\)
Add 6 to both sides: \(y = 4\)

Step4: Substitute y and z into first equation

Substitute \(y = 4\) and \(z = 2\) into \(x + 9y + z = 20\):
\(x+9\times4 + 2=20\)
\(x + 36+2=20\)
\(x+38 = 20\)
Subtract 38: \(x=-18\)

Answer:

\((-18, 4, 2)\)