How many solutions does the system of equations below have? -3x + y - 3z = 4 x - y - z = 16 3x - y + 3z = -4 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Combine Equations: Combine the first and third equations to eliminate y and z. (-3x + y - 3z) + (3x - y + 3z) = 4 + (-4) 0x + 0y + 0z = 0 This simplifies to 0 = 0, which is always true.Combine Second and Third: Now, let's combine the second and third equations. (x - y - z) + (3x - y + 3z) = 16 + (-4) 4x - 2y + 2z = 12 Divide the entire equation by 2 to simplify. 2x - y + z = 6Simplified Equations: Look at the simplified equations: 0 = 0 (from combining the first and third equations) 2x - y + z = 6 (from combining the second and third equations) The first equation is always true, and the second equation represents a plane in three-dimensional space.Infinite Solutions: Since the first equation is always true, it doesn't restrict the solutions. The second equation represents a plane, and there's no third independent equation to further restrict the solutions. Therefore, the system has infinitely many solutions because any point on the plane represented by the second equation will satisfy the system.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

How many solutions does the system of equations below have? $\newline$ $-3x + y - 3z = 4$ $\newline$ $x - y - z = 16$ $\newline$ $3x - y + 3z = -4$ $\newline$ Choices: $\newline$ (A)no solution $\newline$ (B)one solution $\newline$ (C)infinitely many solutions

View step-by-step help

Home

Math Problems

Algebra 2

Determine the number of solutions to a system of equations in three variables

Full solution

Q. How many solutions does the system of equations below have?

\newline

-3x + y - 3z = 4

\newline

x - y - z = 16

\newline

3x - y + 3z = -4

\newline

Choices:

\newline

(A)no solution

\newline

(B)one solution

\newline

(C)infinitely many solutions

Combine Equations: Combine the first and third equations to eliminate $y$ and $z$ . $(-3x + y - 3z) + (3x - y + 3z) = 4 + (-4)$ $0x + 0y + 0z = 0$ This simplifies to $0 = 0$ , which is always true.
Combine Second and Third: Now, let's combine the second and third equations. $\newline$ $(x - y - z) + (3x - y + 3z) = 16 + (-4)$ $\newline$ $4x - 2y + 2z = 12$ $\newline$ Divide the entire equation by $2$ to simplify. $\newline$ $2x - y + z = 6$
Simplified Equations: Look at the simplified equations: $\newline$ $0 = 0$ (from combining the first and third equations) $\newline$ $2x - y + z = 6$ (from combining the second and third equations) $\newline$ The first equation is always true, and the second equation represents a plane in three-dimensional space.
Infinite Solutions: Since the first equation is always true, it doesn't restrict the solutions. The second equation represents a plane, and there's no third independent equation to further restrict the solutions. $\newline$ Therefore, the system has infinitely many solutions because any point on the plane represented by the second equation will satisfy the system.