Question

what is the value of z in this linear system?
$2x + 3y + 7z = 13$
$-2x - 2y + 4z = -12$
$3x + 4y + 4z = 19$
a. 1
b. 0
c. 5
d. 3

Explanation:

Step1: Eliminate \(x\) from first two equations

Add the first equation \(2x + 3y + 7z = 13\) and the second equation \(-2x - 2y + 4z = -12\).
\((2x - 2x)+(3y - 2y)+(7z + 4z)=13 + (-12)\)
Simplify to get: \(y + 11z = 1\) --- Equation (4)

Step2: Manipulate equations to eliminate \(x\) again

Multiply the first equation \(2x + 3y + 7z = 13\) by 3: \(6x + 9y + 21z = 39\) --- Equation (5)
Multiply the third equation \(3x + 4y + 4z = 19\) by 2: \(6x + 8y + 8z = 38\) --- Equation (6)
Subtract Equation (6) from Equation (5):
\((6x - 6x)+(9y - 8y)+(21z - 8z)=39 - 38\)
Simplify to get: \(y + 13z = 1\) --- Equation (7)

Step3: Solve for \(z\)

Subtract Equation (4) from Equation (7):
\((y - y)+(13z - 11z)=1 - 1\)
\(2z = 0\)
Divide both sides by 2: \(z = 0\)

Answer:

B. 0