Solve the system of equations by substitution. $\newline$ $x = 6$ $\newline$ $x + 2y - z = 18$ $\newline$ $-2x + 3y + 3z = 15$

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Precalculus

Solve a system of equations in three variables using substitution

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Q. Solve the system of equations by substitution.

\newline

x = 6

\newline

x + 2y - z = 18

\newline

-2x + 3y + 3z = 15

Substitute $x$ into equations: First, we need to substitute $x = 6$ into the second and third equations. $\newline$ $x + 2y - z = 18$ becomes $6 + 2y - z = 18$ . $\newline$ $-2x + 3y + 3z = 15$ becomes $-2(6) + 3y + 3z = 15$ .
Solve for z: Now, let's solve the first substituted equation for z. $\newline$ $6 + 2y - z = 18$ $\newline$ $2y - z = 18 - 6$ $\newline$ $2y - z = 12$ $\newline$ $z = 2y - 12$
Substitute $z$ and $x$ into third equation: Next, we'll substitute $z$ and $x$ into the third equation. $\newline$ $-2(6) + 3y + 3z = 15$ $\newline$ $-12 + 3y + 3(2y - 12) = 15$
Simplify the equation: Simplify the equation. $\newline$ $-12 + 3y + 6y - 36 = 15$ $\newline$ $9y - 48 = 15$
Add $48$ to both sides: Add $48$ to both sides to solve for $y$ . $9y - 48 + 48 = 15 + 48$ $9y = 63$
Divide by $9$ to find y: Divide both sides by $9$ to find y. $\newline$ $\frac{9y}{9} = \frac{63}{9}$ $\newline$ $y = 7$
Find z using equation: Now we have $y$ , let's find $z$ using the equation $z = 2y - 12$ .
$z = 2(7) - 12$
$z = 14 - 12$
$z = 2$
Check values in original equations: We have $x = 6$ , $y = 7$ , and $z = 2$ . Let's check these values in the original equations. $\newline$ First equation: $x = 6$ (This is given, so it's correct.) $\newline$ Second equation: $x + 2y - z = 18$ becomes $6 + 2(7) - 2 = 18$ , which simplifies to $6 + 14 - 2 = 18$ , and that's $18 = 18$ , so it's correct. $\newline$ Third equation: $-2x + 3y + 3z = 15$ becomes $-2(6) + 3(7) + 3(2) = 15$ , which simplifies to $y = 7$ $0$ , and that's $y = 7$ $1$ , so it's correct.