How many solutions does the system of equations below have? -x + 2y + 3z = -10 -3x - 2y - 2z = -4 -3x + 2y - z = 10 Choices: (A)no solution (B)one solution (C)infinitely many solutions

Write Equations: First, let's write down the system of equations to see if we can spot anything obvious. -x + 2y + 3z = -10 -3x - 2y - 2z = -4 -3x + 2y - z = 10Eliminate Variables: We can try to add the first and second equations to eliminate y. (-x + 2y + 3z) + (-3x - 2y - 2z) = -10 + (-4) This simplifies to -4x + z = -14Simplify Equations: Now, let's add the second and third equations to eliminate y again. (-3x - 2y - 2z) + (-3x + 2y - z) = -4 + 10 This simplifies to -6x - 3z = 6Match Coefficients: We can multiply the equation from step 3 by 3 to match the z coefficient in the equation from step 4. (-4x + z) * 3 = -14 * 3 This gives us -12x + 3z = -42Eliminate Variable z: Now we can add the equation from step 4 and the new equation from step 5 to eliminate z. (-12x + 3z) + (-6x - 3z) = -42 + 6 This simplifies to -18x = -36Solve for x: Divide both sides by -18 to solve for x. -18x / -18 = -36 / -18 This gives us x = 2Substitute x into Equation: Now that we have a value for x, we can substitute it back into the equations to solve for y and z. Let's substitute x = 2 into the equation from step 3. -4(2) + z = -14 This gives us -8 + z = -14

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? $\newline$ $-x + 2y + 3z = -10$ $\newline$ $-3x - 2y - 2z = -4$ $\newline$ $-3x + 2y - z = 10$ $\newline$ Choices: $\newline$ (A)no solution $\newline$ (B)one solution $\newline$ (C)infinitely many solutions

View step-by-step help

Home

Math Problems

Precalculus

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

-x + 2y + 3z = -10

\newline

-3x - 2y - 2z = -4

\newline

-3x + 2y - z = 10

\newline

Choices:

\newline

(A)no solution

\newline

(B)one solution

\newline

(C)infinitely many solutions

Write Equations: First, let's write down the system of equations to see if we can spot anything obvious. $\newline$ $-x + 2y + 3z = -10$ $\newline$ $-3x - 2y - 2z = -4$ $\newline$ $-3x + 2y - z = 10$
Eliminate Variables: We can try to add the first and second equations to eliminate $y$ . $(-x + 2y + 3z) + (-3x - 2y - 2z) = -10 + (-4)$ This simplifies to $-4x + z = -14$
Simplify Equations: Now, let's add the second and third equations to eliminate $y$ again. $(-3x - 2y - 2z) + (-3x + 2y - z) = -4 + 10$ This simplifies to $-6x - 3z = 6$
Match Coefficients: We can multiply the equation from step $3$ by $3$ to match the $z$ coefficient in the equation from step $4$ . $\newline$ $(-4x + z) \times 3 = -14 \times 3$ $\newline$ This gives us $-12x + 3z = -42$
Eliminate Variable $z$ : Now we can add the equation from step $4$ and the new equation from step $5$ to eliminate $z$ . $(-12x + 3z) + (-6x - 3z) = -42 + 6$ This simplifies to $-18x = -36$
Solve for x: Divide both sides by $-18$ to solve for x. $\newline$ $-18x / -18 = -36 / -18$ $\newline$ This gives us $x = 2$
Substitute $x$ into Equation: Now that we have a value for $x$ , we can substitute it back into the equations to solve for $y$ and $z$ . Let's substitute $x = 2$ into the equation from step $3$ . $-4(2) + z = -14$ This gives us $-8 + z = -14$