Solve each system of equations.145x+2y=43x+4y+2z=67x+3y+4z=29

Q14. - Algebra 2 | StudyQuestionHub

Solve each system of equations.

$\begin{array}{l} 5 x + 2 y = 4 \\ 3 x + 4 y + 2 z = 6 \\ 7 x + 3 y + 4 z = 29 \end{array}$

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Answer

The solution is $x = 2, y = - 3, z = 6 .$ .

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

Given system of equations are

$\begin{array}{l} 5 x + 2 y = 4 \\ 3 x + 4 y + 2 z = 6 \\ 7 x + 3 y + 4 z = 29 \end{array}$

Multiplying the second equation by $2$ and subtracting from the third equation, we get

$\begin{array}{l} 7 x - 6 x + 3 y - 8 y = 29 - 12 \\ x - 5 y = 17 \end{array}$

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

$\begin{array}{l} 5 x + 2 y = 4 \\ x - 5 y = 17 \end{array}$ System of two equations with two variables are

Multiplying the second equation by $5$ and subtracting from the first equation, we get

$\begin{array}{l} 2 y + 25 y = 4 - 85 \\ 27 = - 81 \\ y = - 3 \end{array}$

Putting the value of $y$ in the second equation, we get

$\begin{array}{l} x - 5 y = 17 \\ x - 5 (- 3) = 17 \\ x + 15 = 17 \\ x = 17 - 15 \\ x = 2 \end{array}$

3Step-3 –Substituting the value of x and y in any equation with three variables

$\begin{array}{l} 3 x + 4 y + 2 z = 6 \\ 3 (2) + 4 (- 3) + 2 z = 6 \\ 6 - 12 + 2 z = 6 \\ 2 z = 6 + 6 \\ 2 z = 12 \\ z = 6. \end{array}$