Sketch the graph or use a graphing utility to determine which of the following solution scenarios is true for the system of equations.

⎧⎩⎨⎪⎪3x+7y−z=8x+y+z=10x−y+2z=3
{
3
𝑥
+
7
𝑦

𝑧
=
8
𝑥
+
𝑦
+
𝑧
=
10
𝑥

𝑦
+
2
𝑧
=
3


(1 point)
Responses

All three planes intersect at a single point, which means there is one solution.
All three planes intersect at a single point, which means there is one solution.

All three planes overlap, which means there are infinite solutions.
All three planes overlap, which means there are infinite solutions.

All three planes are parallel, which means there are no solutions.
All three planes are parallel, which means there are no solutions.

Two of the planes are parallel, which means there are no solutions.
Two of the planes are parallel, which means there are no solutions.
Graphing Calculator

1 answer

To determine the solution scenario for the given system of equations, we can analyze the equations step by step.

Let's rewrite the system:

  1. \( 3x + 7y - z = 3 \)
  2. \( 8x + y + z = 10 \)
  3. \( 10x - y + 2z = 3 \)

We can rearrange these equations to understand their relative positions (planes) in 3D space. The equations represent planes, and we can analyze whether they intersect, overlap, are parallel, or have some combination.

Analyzing the Planes Using Determinant

To determine the nature of these planes, we can check their coefficients and find the determinant of the coefficient matrix:

\[ A = \begin{bmatrix} 3 & 7 & -1 \ 8 & 1 & 1 \ 10 & -1 & 2 \end{bmatrix} \]

The determinant can indicate the types of solutions:

  1. If the determinant is non-zero, the planes intersect at a single point (one solution).
  2. If the determinant is zero, we check the rank of the matrix to determine if there are infinite solutions or no solutions.

Let's calculate the determinant of matrix \( A \):

\[ \text{det}(A) = 3 \begin{vmatrix} 1 & 1 \ -1 & 2 \end{vmatrix} - 7 \begin{vmatrix} 8 & 1 \ 10 & 2 \end{vmatrix} - 1 \begin{vmatrix} 8 & 1 \ 10 & -1 \end{vmatrix} \]

Calculating the 2x2 determinants:

  1. \( \begin{vmatrix} 1 & 1 \ -1 & 2 \end{vmatrix} = (1)(2) - (1)(-1) = 2 + 1 = 3 \)
  2. \( \begin{vmatrix} 8 & 1 \ 10 & 2 \end{vmatrix} = (8)(2) - (1)(10) = 16 - 10 = 6 \)
  3. \( \begin{vmatrix} 8 & 1 \ 10 & -1 \end{vmatrix} = (8)(-1) - (1)(10) = -8 - 10 = -18 \)

Now substituting back into the determinant calculation:

\[ \text{det}(A) = 3(3) - 7(6) - 1(-18) = 9 - 42 + 18 = -15 \]

The determinant of \( A \) is non-zero (i.e., \(-15\)), which indicates that the three planes intersect at a single point.

Conclusion

Given the nature of the determinant, we can conclude:

  • All three planes intersect at a single point, which means there is one solution.

This corresponds to the first response option provided in your question.