Question

algebra 2a semester online practice
complete this assessment to review what you’ve learned. it will not count toward your grade.
clothilde is using the elimination method to solve the following system of equations for x, y, and z. what is the value of y?
$x - 2y + 3z = 4$
$4x - 2y + 2z = -2$
$8x + 2y + 2z = 2$
(1 point)
○ 2
○ -1
○ 3
○ 4

Explanation:

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

Let the equations be:

1. \(x - 2y + 3z = 4\)
2. \(4x - 2y + 2z = -2\)

Multiply equation 1 by 4: \(4x - 8y + 12z = 16\)
Subtract equation 2 from this new equation: \((4x - 8y + 12z)-(4x - 2y + 2z)=16 - (-2)\)
Simplify: \(-6y + 10z = 18\) (Equation 4)

Step2: Eliminate \(x\) from first and third equations

Third equation: \(8x + 2y + 2z = 2\)
Multiply equation 1 by 8: \(8x - 16y + 24z = 32\)
Subtract third equation from this: \((8x - 16y + 24z)-(8x + 2y + 2z)=32 - 2\)
Simplify: \(-18y + 22z = 30\) (Equation 5)

Step3: Solve Equation 4 and Equation 5 for \(y\) and \(z\)

From Equation 4: \(-6y + 10z = 18\) can be simplified to \(-3y + 5z = 9\) (divide by 2), so \(5z = 3y + 9\), \(z=\frac{3y + 9}{5}\)
Substitute \(z\) into Equation 5: \(-18y + 22\times\frac{3y + 9}{5}=30\)
Multiply through by 5: \(-90y + 22(3y + 9)=150\)
Expand: \(-90y + 66y + 198 = 150\)
Simplify: \(-24y = 150 - 198=-48\)
Solve for \(y\): \(y=\frac{-48}{-24}=2\)? Wait, no, wait, let's check again. Wait, maybe better to use elimination between Equation 4 and Equation 5.
Multiply Equation 4 by 3: \(-18y + 30z = 54\) (Equation 6)
Subtract Equation 5 from Equation 6: \((-18y + 30z)-(-18y + 22z)=54 - 30\)
Simplify: \(8z = 24\), so \(z = 3\)
Substitute \(z = 3\) into Equation 4: \(-6y + 10\times3 = 18\)
\(-6y + 30 = 18\), \(-6y = -12\), \(y = 2\)? Wait, but let's check with original equations. Wait, maybe I made a mistake. Wait, let's check the second and third equations. Wait, the second equation: \(4x - 2y + 2z = -2\), third: \(8x + 2y + 2z = 2\). Let's add second and third equations: \(12x + 4z = 0\), so \(3x + z = 0\), \(z = -3x\). Let's use first equation: \(x - 2y + 3(-3x)=4\), \(x - 2y -9x = 4\), \(-8x -2y =4\), \(4x + y = -2\) (Equation A). From second equation: \(4x - 2y + 2(-3x)=-2\), \(4x -2y -6x=-2\), \(-2x -2y=-2\), \(x + y = 1\) (Equation B). Now subtract Equation B from Equation A: \(3x = -3\), so \(x = -1\). Then from Equation B: \(-1 + y = 1\), so \(y = 2\)? Wait, but the options have 2 as an option. Wait, but let's check with \(x=-1\), \(y=2\), \(z=3\) in first equation: \(-1 -4 +9=4\), which is correct. Second equation: \(4(-1)-4 +6=-4 -4 +6=-2\), correct. Third equation: \(8(-1)+4 +6=-8 +4 +6=2\), correct. So \(y=2\)? Wait, but the options have 2 as the first option. Wait, but let me check again. Wait, maybe I messed up earlier steps, but the correct value of \(y\) is 2? Wait, no, wait the options: 2, -1, 3, 4. Wait, when I did the last method, adding second and third equations: \(4x -2y +2z +8x +2y +2z=-2 +2\), so \(12x +4z=0\), \(3x + z=0\), \(z=-3x\). Then first equation: \(x -2y +3(-3x)=4\), \(x -2y -9x=4\), \(-8x -2y=4\), divide by -2: \(4x + y=-2\) (Equation 1a). Second equation: \(4x -2y +2(-3x)=-2\), \(4x -2y -6x=-2\), \(-2x -2y=-2\), divide by -2: \(x + y=1\) (Equation 2a). Now solve Equation 1a and 2a: Equation 1a: \(4x + y=-2\), Equation 2a: \(x + y=1\). Subtract Equation 2a from Equation 1a: \(3x=-3\), \(x=-1\). Then from Equation 2a: \(-1 + y=1\), so \(y=2\). Yes, that's correct. So \(y=2\).

Answer:

\(2\) (corresponding to the option "2")