QuestionSolve the system of equations. $\begin{aligned} x-3 y-z & =6 \\ -x+8 y-4 z & =-16 \\ 2 x-15 y+7 z & =30 \end{aligned}$

Studdy Solution

STEP 1

1. We are given a system of three linear equations with three variables: $x$ , $y$ , and $z$ .
2. The goal is to find the values of $x$ , $y$ , and $z$ that satisfy all three equations simultaneously.
3. We will use the method of elimination to solve this system.

STEP 2

1. Eliminate one variable from two pairs of equations.
2. Solve the resulting system of two equations with two variables.
3. Substitute back to find the third variable.
4. Verify the solution by substituting back into the original equations.

STEP 3

Choose two pairs of equations to eliminate the same variable. Let's eliminate $x$ .
First, add the first equation to the second equation to eliminate $x$ :
$(x - 3y - z) + (-x + 8y - 4z) = 6 + (-16)$
Simplify:
$5y - 5z = -10$
Divide through by 5:
$y - z = -2$

STEP 4

Next, add the first equation to the third equation to eliminate $x$ :
$(x - 3y - z) + (2x - 15y + 7z) = 6 + 30$
Simplify:
$3x - 18y + 6z = 36$
Divide through by 3:
$x - 6y + 2z = 12$

STEP 5

Now, solve the system of two equations with two variables:
1. $y - z = -2$
2. $x - 6y + 2z = 12$
From equation 1, express $y$ in terms of $z$ :
$y = z - 2$
Substitute $y = z - 2$ into equation 2:
$x - 6(z - 2) + 2z = 12$
Simplify:
$x - 6z + 12 + 2z = 12$
Combine like terms:
$x - 4z + 12 = 12$
Subtract 12 from both sides:
$x - 4z = 0$

STEP 6

Solve for $x$ :
$x = 4z$
Now, substitute $x = 4z$ and $y = z - 2$ back into the original equations to find $z$ .

STEP 7

Substitute $x = 4z$ and $y = z - 2$ into the first original equation:
$4z - 3(z - 2) - z = 6$
Simplify:
$4z - 3z + 6 - z = 6$
Combine like terms:
$0z + 6 = 6$
This equation is always true, so we need to check the other equations to find $z$ .

STEP 8

Substitute $x = 4z$ and $y = z - 2$ into the second original equation:
$-4z + 8(z - 2) - 4z = -16$
Simplify:
$-4z + 8z - 16 - 4z = -16$
Combine like terms:
$0z - 16 = -16$
This equation is also always true. Therefore, we have infinite solutions along the line defined by $x = 4z$ and $y = z - 2$ .

STEP 9

Verify the solution by substituting $x = 4z$ and $y = z - 2$ into the third original equation:
$2(4z) - 15(z - 2) + 7z = 30$
Simplify:
$8z - 15z + 30 + 7z = 30$
Combine like terms:
$0z + 30 = 30$
This equation is also always true. Therefore, the solution is valid for any $z$ .
The solution is a line of infinite solutions given by: $x = 4z, \quad y = z - 2, \quad z = z$

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