Solve the system by substitution. {:[3y=x],[-x+y=-12]:} (◻,◻)

Identify Equation: Identify the first equation where one variable is already isolated. The first equation is 3y = x. This equation can be used to substitute for x in the second equation.Substitute in Second: Substitute 3y for x in the second equation -x + y = -12. Replace x with 3y in the second equation to get -3y + y = -12.Simplify and Solve: Simplify the equation and solve for y. -3y + y = -12 simplifies to -2y = -12. Now, divide both sides by -2 to solve for y. y = -12 / -2 y = 6Substitute Back: Substitute y = 6 back into the first equation 3y = x to find x. 3y = x becomes 3*6 = x. x = 18Write Ordered Pair: Write the solution as an ordered pair (x, y). The solution is (18, 6).

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Solve the system by substitution.

{:[3y=x],[-x+y=-12]:}

(◻,◻)

Solve the system by substitution. $\newline$ $\begin{array}{c} 3 y=x \\ -x+y=-12 \end{array}$ $\newline$ $(\square, \square)$

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Algebra 2

Solve a system of equations in three variables using substitution

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Q. Solve the system by substitution.

\newline

\begin{array}{c} 3 y=x \\ -x+y=-12 \end{array}

\newline

(\square, \square)

Identify Equation: Identify the first equation where one variable is already isolated. $\newline$ The first equation is $3y = x$ . This equation can be used to substitute for $x$ in the second equation.
Substitute in Second: Substitute $3y$ for $x$ in the second equation $-x + y = -12$ . Replace $x$ with $3y$ in the second equation to get $-3y + y = -12$ .
Simplify and Solve: Simplify the equation and solve for $y$ . $-3y + y = -12$ simplifies to $-2y = -12$ . Now, divide both sides by $-2$ to solve for $y$ . $y = \frac{-12}{-2}$ $y = 6$
Substitute Back: Substitute $y = 6$ back into the first equation $3y = x$ to find $x$ . $\newline$ $3y = x$ becomes $3\times 6 = x$ . $\newline$ $x = 18$
Write Ordered Pair: Write the solution as an ordered pair $(x, y)$ . The solution is $(18, 6)$ .