Solve each system of equations.4.x+2y=123y−4z=25x+6y+z=20

Q4. - Algebra 2 | StudyQuestionHub

Solve each system of equations.

$\begin{array}{l} x + 2 y = 12 \\ 3 y - 4 z = 25 \\ x + 6 y + z = 20 \end{array}$

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Answer

The solution is $x = 6, y = 3, z = - 4 .$

1Step-1 –Using elimination to obtain two equations in two variables

Given equations are

$\begin{array}{l} x + 2 y = 12 \\ 3 y - 4 z = 25 \\ x + 6 y + z = 20 \end{array}$

Subtracting the third equation from the first equation, we get

$\begin{array}{l} 2 y - 6 y - z = 12 - 20 \\ - 4 y - z = - 8 \\ 4 y + z = - 8 \end{array}$

2Step-2 –Solving the system of two equations in two variables

System of equations in two variables are

$\begin{array}{l} 3 y + 16 y = 25 + 32 \\ 19 y = 57 \\ y = 3 \end{array}$

Substituting the value of $y$ in first equation, we get

$\begin{array}{l} 3 y - 4 z = 25 \\ 3 (3) - 4 z = 25 \\ - 4 z = 25 - 9 \\ - 4 z = 16 \\ z = - 4 \end{array}$

3Step-3 –Putting the value of y and z in any of the equation with all three variables

$\begin{array}{l} x + 6 y + z = 20 \\ x + 6 (3) - 4 = 20 \\ x + 18 - 4 = 20 \\ x = 20 - 14 \\ x = 6. \end{array}$