Solve each system of equations 6x+2y+4z=23x+4y−8z=−3−3x−6y+12z=5

The solution set of the given system of equations is 13,-12,14.

Q21. - Algebra 2 | StudyQuestionHub

Q21.

Question

Solve each system of equations

$\begin{array}{l} 6 x + 2 y + 4 z = 2 \\ 3 x + 4 y - 8 z = - 3 \\ - 3 x - 6 y + 12 z = 5 \end{array}$

Step-by-Step Solution

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Answer

The solution set of the given system of equations is $(\frac{1}{3}, - \frac{1}{2}, \frac{1}{4})$ .

1Step 1 – Use the elimination method to get the system of equations in two variables.

Multiply the equation $6 x + 2 y + 4 z = 2$ by $2$ and add the new resultant equation to the equation $3 x + 4 y - 8 z = - 3$ .

$\begin{array}{l} \underset{}{\begin{array}{l} 6 x + 2 y + 4 z = 2 \\ 3 x + 4 y - 8 z = - 3 \end{array}} \begin{matrix} multiply by 2 \end{matrix} \underset{}{\begin{array}{l} 12 x + 4 y + 8 z = 4 \\ 3 x + 4 y - 8 z = - 3 \end{array}} \\ 15 x + 8 y + 0 = 1 \end{array}$

So, the resultant equation is $15 x + 8 y = 1$

Multiply the equation $3 x + 4 y - 8 z = - 3$ by $3$ and multiply the equation $- 3 x - 6 y + 12 z = 5$ by $2$ .

Then add the two new resultant equations.

$\begin{array}{l} \underline{\begin{array}{l} 3 x + 4 y - 8 z = - 3 \\ - 3 x - 6 y + 12 z = 5 \end{array}} \begin{matrix} multiply by 3 \to \\ multiply by 2 \to \end{matrix} \underline{\begin{array}{l} 9 x + 12 y - 24 z = - 9 \\ - 6 x - 12 y + 24 z = 10 \end{array}} \\ 3 x - 0 + 0 = 1 \end{array}$

So, the resultant equation is $3 x = 1$ .

2Step 2 – Use the elimination method to solve the system of two equations.

Solve $3 x = 1$ for $x$ :

$\begin{array}{l} 3 x = 1 \\ \frac{3 x}{3} = \frac{1}{3} Divide both sides by 3 \\ x = \frac{1}{3} \end{array}$

3Step 3 – Find the values of   y and z .

Substitute $x = \ f r a c {1} {3} $ i n 15 x + 8 y = 1$ and find the value of $y$ .

$\begin{array}{l} 15 x + 8 y = 1 \\ 15 (\frac{1}{3}) + 8 y = 1 Substitute \frac{1}{3} for x \\ 5 + 8 y = 1 Simplify \\ 8 y = - 4 Subtract 5 from both sides \\ y = - \frac{1}{2} Divide both sides by 8 \end{array}$

Substitute $x = \frac{1}{3}$ , $y = - \frac{1}{2} i n 3 x + 4 y - 8 z = - 3$ and find the value of $z$

$\begin{array}{l} 3 x + 4 y - 8 z = - 3 \\ 3 (\frac{1}{3}) + 4 (- \frac{1}{2}) - 8 z = - 3 substitute \frac{1}{3} for x, - \frac{1}{2} for y \\ 1 - 2 - 8 z = - 3 simplify \\ - 1 - 8 z = - 3 simplify \\ - 8 z = - 2 add 1 on both sides \\ z = \frac{1}{4} Divide both sides by - 8 \end{array}$

Hence, the solution of the given system of equations is $(x, y, z) = (\frac{1}{3}, - \frac{1}{2}, \frac{1}{4})$ .

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