Solve the system of equations by substitution. y = 2 x + y + z = 3 3x + 2y + 2z = 7 ____

Substitute y = 2: Substitute y = 2 into x + y + z = 3. x + 2 + z = 3 x + z = 1Simplify first equation: Substitute y = 2 into 3x + 2y + 2z = 7. 3x + 2*2 + 2z = 7 3x + 4 + 2z = 7 3x + 2z = 3Simplify second equation: Divide the equation 3x + 2z = 3 by 3 to simplify. x + (2/3)z = 1Divide to simplify: Now we have two equations with x and z: 1) x + z = 1 2) x + (2/3)z = 1 Subtract equation 2 from equation 1 to eliminate x. (x + z) - (x + (2/3)z) = 1 - 1 z - (2/3)z = 0 (1/3)z = 0Eliminate x: Solve for z. Multiply both sides by 3 to get rid of the fraction. z = 0Solve for z: Substitute z = 0 into x + z = 1 to find x. x + 0 = 1 x = 1Find x: We have found: x = 1 y = 2 z = 0 The solution is (1, 2, 0).

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Solve the system of equations by substitution. $\newline$ $y = 2$ $\newline$ $x + y + z = 3$ $\newline$ $3x + 2y + 2z = 7$

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Algebra 2

Solve a system of equations in three variables using substitution

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Q. Solve the system of equations by substitution.

\newline

y = 2

\newline

x + y + z = 3

\newline

3x + 2y + 2z = 7

Substitute $y = 2$ : Substitute $y = 2$ into $x + y + z = 3$ . $\newline$ $x + 2 + z = 3$ $\newline$ $x + z = 1$
Simplify first equation: Substitute $y = 2$ into $3x + 2y + 2z = 7$ . $\newline$ $3x + 2\cdot 2 + 2z = 7$ $\newline$ $3x + 4 + 2z = 7$ $\newline$ $3x + 2z = 3$
Simplify second equation: Divide the equation $3x + 2z = 3$ by $3$ to simplify. $\newline$ $x + \frac{2}{3}z = 1$
Divide to simplify: Now we have two equations with $x$ and $z$ :
$1$ ) $x + z = 1$
$2$ ) $x + \frac{2}{3}z = 1$
Subtract equation $2$ from equation $1$ to eliminate $x$ .
$(x + z) - (x + \frac{2}{3}z) = 1 - 1$
$z - \frac{2}{3}z = 0$
$\frac{1}{3}z = 0$
Eliminate $x$ : Solve for $z$ . $\newline$ Multiply both sides by $3$ to get rid of the fraction. $\newline$ $z = 0$
Solve for $z$ : Substitute $z = 0$ into $x + z = 1$ to find $x$ .
$x + 0 = 1$
$x = 1$
Find $x$ : We have found: $\newline$ $x = 1$ $\newline$ $y = 2$ $\newline$ $z = 0$ $\newline$ The solution is $(1, 2, 0)$ .