Solve the system of equations by elimination. $\newline$ $2x - 3y - z = 1$ $\newline$ $3x + 3y - 2z = -12$ $\newline$ $-2x - y - 2z = 16$

View step-by-step help

Home

Math Problems

Algebra 2

Solve a system of equations in three variables using elimination

Full solution

Q. Solve the system of equations by elimination.

\newline

2x - 3y - z = 1

\newline

3x + 3y - 2z = -12

\newline

-2x - y - 2z = 16

Eliminate $x$ by adding equations: Add the first and third equations to eliminate $x$ .
$(2x - 3y - z) + (-2x - y - 2z) = 1 + 16$
$2x - 3y - z - 2x - y - 2z = 17$
$-4y - 3z = 17$
Eliminate $x$ by multiplying and adding equations: Multiply the second equation by $2$ and add it to the third equation to eliminate $x$ .
$(2 \times (3x + 3y - 2z)) + (-2x - y - 2z) = 2 \times (-12) + 16$
$6x + 6y - 4z - 2x - y - 2z = -24 + 16$
$4x + 5y - 6z = -8$
Solve for y from new equations: Now we have two new equations: $\newline$ $-4y - 3z = 17$ $\newline$ $4x + 5y - 6z = -8$ $\newline$ Let's solve for y from the first new equation. $\newline$ $-4y = 17 + 3z$ $\newline$ $y = -\frac{17 + 3z}{4}$
Substitute y in second new equation: Substitute y in the second new equation. $\newline$ $4x + 5\left(-\frac{17 + 3z}{4}\right) - 6z = -8$ $\newline$ $4x - \left(\frac{85}{4} + \frac{15z}{4}\right) - 6z = -8$ $\newline$ $4x - \frac{85}{4} - \frac{15z}{4} - \frac{24z}{4} = -8$ $\newline$ $4x - \frac{85}{4} - \frac{39z}{4} = -8$
Clear fractions by multiplying: Multiply everything by $4$ to clear the fractions. $\newline$ $4 \times (4x - \frac{85}{4} - \frac{39z}{4}) = 4 \times (-8)$ $\newline$ $16x - 85 - 39z = -32$
Solve for $x$ from first equation: Add $85$ to both sides. $\newline$ $16x - 39z = -32 + 85$ $\newline$ $16x - 39z = 53$
Substitute $x$ in original equation: Now we have two equations with two variables: $\newline$ $16x - 39z = 53$ $\newline$ $-4y - 3z = 17$ $\newline$ Let's solve for $x$ from the first equation. $\newline$ $16x = 53 + 39z$ $\newline$ $x = \frac{53 + 39z}{16}$
Clear fractions by multiplying: Substitute $x$ in the first original equation. $2\left(\frac{53 + 39z}{16}\right) - 3y - z = 1$ $\frac{106 + 78z}{16} - 3y - z = 1$
Combine like terms: Multiply everything by $16$ to clear the fractions. $\newline$ $16 \times \left(\frac{106 + 78z}{16}\right) - 16 \times (3y) - 16 \times (z) = 16 \times (1)$ $\newline$ $106 + 78z - 48y - 16z = 16$
Solve for $z$ from second equation: Combine like terms. $106 + 62z - 48y = 16$
Substitute $z$ in equation: Subtract $106$ from both sides. $\newline$ $62z - 48y = 16 - 106$ $\newline$ $62z - 48y = -90$
Clear fractions by multiplying: Now we have two equations with two variables: $\newline$ $62z - 48y = -90$ $\newline$ $-4y - 3z = 17$ $\newline$ Let's solve for $z$ from the second equation. $\newline$ $-3z = 17 + 4y$ $\newline$ $z = -\frac{17 + 4y}{3}$
Distribute $-62$ : Substitute $z$ in the equation $62z - 48y = -90$ . $62\left(-\frac{17 + 4y}{3}\right) - 48y = -90$ $-62\left(17 + 4y\right) / 3 - 48y = -90$
Combine like terms: Multiply everything by $3$ to clear the fractions. $3 \times (-62(17 + 4y) / 3) - 3 \times (48y) = 3 \times (-90)$ $-62(17 + 4y) - 144y = -270$
Solve for $y$ : Distribute $-62$ .
$-62 \times 17 - 62 \times 4y - 144y = -270$
$-1054 - 248y - 144y = -270$
Substitute $y$ in equation: Combine like terms. $-1054 - 392y = -270$
Solve for $z$ : Add $1054$ to both sides. $\newline$ $-392y = -270 + 1054$ $\newline$ $-392y = 784$
Substitute $z$ in equation: Divide by $-392$ to solve for $y$ . $y = \frac{784}{-392}$ $y = -2$
Substitute $z$ in equation: Substitute $y$ back into the equation $z = -\frac{17 + 4y}{3}$ . $z = -\frac{17 + 4(-2)}{3}$ $z = -\frac{17 - 8}{3}$ $z = -\frac{9}{3}$ $z = -3$
Substitute $z$ in equation: Substitute $y$ back into the equation $z = -\frac{17 + 4y}{3}$ .
$z = -\frac{17 + 4(-2)}{3}$
$z = -\frac{17 - 8}{3}$
$z = -\frac{9}{3}$
$z = -3$ Substitute $z$ back into the equation $x = \frac{53 + 39z}{16}$ .
$x = \frac{53 + 39(-3)}{16}$
$y$ $0$
$y$ $1$
$y$ $2$