Full solution

Q. Solve the system of equations by elimination.

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3x + y - 3z = 2

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3x - 3y - z = -14

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x - 2y + 2z = 6

Eliminate z from Equations: First, let's eliminate z from the first two equations. $\newline$ $3x + y - 3z = 2$ (Equation $1$ ) $\newline$ $3x - 3y - z = -14$ (Equation $2$ ) $\newline$ Multiply Equation $2$ by $3$ to make the coefficients of z the same. $\newline$ $3*(3x - 3y - z) = 3*(-14)$ $\newline$ $9x - 9y - 3z = -42$ (Equation $2$ modified) $\newline$ Now, add Equation $1$ and Equation $2$ modified. $\newline$ $(3x + y - 3z) + (9x - 9y - 3z) = 2 + (-42)$ $\newline$ $12x - 8y = -40$
Solve for $x$ and $y$ : Next, let's eliminate $z$ from the second and third equations. $3x - 3y - z = -14$ (Equation $2$ ) $x - 2y + 2z = 6$ (Equation $3$ ) Multiply Equation $3$ by $-\frac{1}{2}$ to make the coefficients of $z$ the same. $-\frac{1}{2}\times(x - 2y + 2z) = -\frac{1}{2}\times6$ $-\frac{1}{2}x + y - z = -3$ (Equation $3$ modified) Now, add Equation $2$ and Equation $3$ modified. $(3x - 3y - z) + (-\frac{1}{2}x + y - z) = -14 + (-3)$ $\frac{5}{2}x - 2y = -17$
Substitute $x$ and $y$ into Equation: Now, let's solve the system formed by the two new equations we got. $12x - 8y = -40$ (From Step $1$ ) $\frac{5}{2}x - 2y = -17$ (From Step $2$ )Multiply the second equation by $4$ to make the coefficients of $y$ the same. $4*(\frac{5}{2}x - 2y) = 4*(-17)$ $10x - 8y = -68$ (Equation $4$ )Now, subtract Equation $4$ from the first new equation. $(12x - 8y) - (10x - 8y) = -40 - (-68)$ $2x = 28$ $y$ $0$
Find $z$ : Substitute $x = 14$ into Equation $4$ to find $y$ . $10x - 8y = -68$ $10(14) - 8y = -68$ $140 - 8y = -68$ $-8y = -68 - 140$ $-8y = -208$ $y = 26$
Find $z$ : Substitute $x = 14$ into Equation $4$ to find $y$ .
$10x - 8y = -68$
$10(14) - 8y = -68$
$140 - 8y = -68$
$-8y = -68 - 140$
$-8y = -208$
$y = 26$ Substitute $x = 14$ and $y = 26$ into one of the original equations to find $z$ .
$x = 14$ $2$ (Equation $3$ )
$x = 14$ $3$
$x = 14$ $4$
$x = 14$ $5$
$x = 14$ $6$
$x = 14$ $7$
$x = 14$ $8$