Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned. it will not count toward your grade.
solve the following system of equations for x, y, and z using the substitution method. what is the value of y?
$4x + 4y + 4z = 12$
$2x + y + z = 5$
$3x + 6y - 3z = 12$
(1 point)
○ 1
○ 0
○ 2
○ 3

Explanation:

Step1: Simplify first equation

Divide \(4x + 4y + 4z = 12\) by 4: \(x + y + z = 3\). So \(x + y + z = 3\) (Equation 1).

Step2: Express \(y + z\) from Equation 1

From \(x + y + z = 3\), we get \(y + z = 3 - x\) (Equation 1a).

Step3: Substitute into second equation

Second equation: \(2x + y + z = 5\). Substitute \(y + z = 3 - x\) (from Equation 1a): \(2x + (3 - x) = 5\). Simplify: \(2x + 3 - x = 5\) → \(x + 3 = 5\) → \(x = 2\).

Step4: Substitute \(x = 2\) into Equation 1a

\(y + z = 3 - 2 = 1\) → \(z = 1 - y\) (Equation 1b).

Step5: Substitute \(x = 2\) and \(z = 1 - y\) into third equation

Third equation: \(3x + 6y - 3z = 12\). Substitute \(x = 2\) and \(z = 1 - y\): \(3(2) + 6y - 3(1 - y) = 12\). Simplify: \(6 + 6y - 3 + 3y = 12\) → \(3 + 9y = 12\) → \(9y = 9\) → \(y = 1\).

Answer:

\(y = 1\) (corresponding to the option "1")