Solve the system by substitution. {:[-2y+10=x],[-x-3y=-8]:} (◻,◻)

Isolate x: Isolate x in the first equation. The first equation is -2y + 10 = x. We can rewrite this as x = -2y + 10.Substitute x: Substitute the expression for x into the second equation. The second equation is -x - 3y = -8. Substituting x from the first equation, we get -( -2y + 10) - 3y = -8.Simplify and solve: Simplify the equation and solve for y. 2y - 10 - 3y = -8 -y - 10 = -8 Add 10 to both sides: -y = 2 Multiply both sides by -1: y = -2Substitute y into x: Substitute the value of y back into the first equation to find x. Using the first equation x = -2y + 10 and substituting y = -2, we get: x = -2(-2) + 10 x = 4 + 10 x = 14

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Solve the system by substitution.

{:[-2y+10=x],[-x-3y=-8]:}

(◻,◻)

Solve the system by substitution. $\newline$ $\begin{array}{l} -2 y+10=x \\ -x-3 y=-8 \end{array}$ $\newline$ $(\square, \square)$

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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}{l} -2 y+10=x \\ -x-3 y=-8 \end{array}

\newline

(\square, \square)

Isolate x: Isolate $x$ in the first equation. $\newline$ The first equation is $-2y + 10 = x$ . We can rewrite this as $x = -2y + 10$ .
Substitute $x$ : Substitute the expression for $x$ into the second equation. $\newline$ The second equation is $-x - 3y = -8$ . Substituting $x$ from the first equation, we get $-( -2y + 10) - 3y = -8$ .
Simplify and solve: Simplify the equation and solve for $y$ . $2y - 10 - 3y = -8$ $-y - 10 = -8$ Add $10$ to both sides: $-y = 2$ Multiply both sides by $-1$ : $y = -2$
Substitute $y$ into $x$ : Substitute the value of $y$ back into the first equation to find $x$ . Using the first equation $x = -2y + 10$ and substituting $y = -2$ , we get: $x = -2(-2) + 10$ $x = 4 + 10$ $x = 14$