Solve the system by substitution. {:[8x+8y=-40],[x=4y+5]:} (◻,◻)

Substitute x in first equation: Substitute the expression for x from the second equation into the first equation. Given the second equation x = 4y + 5, we can substitute x in the first equation 8x + 8y = -40. 8(4y + 5) + 8y = -40Distribute and combine like terms: Distribute 8 to the terms inside the parentheses and combine like terms. 8 * 4y + 8 * 5 + 8y = -40 32y + 40 + 8y = -40 40y + 40 = -40Isolate y term: Subtract 40 from both sides of the equation to isolate the term with y. 40y + 40 - 40 = -40 - 40 40y = -80Solve for y: Divide both sides of the equation by 40 to solve for y. 40y / 40 = -80 / 40 y = -2Substitute y back for x: Substitute the value of y back into the second equation to solve for x. x = 4y + 5 x = 4(-2) + 5 x = -8 + 5 x = -3Write solution as ordered pair: Write the solution as an ordered pair (x, y). The solution is (-3, -2).

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Solve the system by substitution.

{:[8x+8y=-40],[x=4y+5]:}

(◻,◻)

Solve the system by substitution. $\newline$ $\begin{aligned} 8 x+8 y & =-40 \\ x & =4 y+5 \end{aligned}$ $\newline$ $(\square, \square)$

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Algebra 2

Solve a system of equations in three variables using substitution

Full solution

Q. Solve the system by substitution.

\newline

\begin{aligned} 8 x+8 y & =-40 \\ x & =4 y+5 \end{aligned}

\newline

(\square, \square)

Substitute $x$ in first equation: Substitute the expression for $x$ from the second equation into the first equation. $\newline$ Given the second equation $x = 4y + 5$ , we can substitute $x$ in the first equation $8x + 8y = -40$ . $\newline$ $8(4y + 5) + 8y = -40$
Distribute and combine like terms: Distribute $8$ to the terms inside the parentheses and combine like terms. $\newline$ $8 \times 4y + 8 \times 5 + 8y = -40$ $\newline$ $32y + 40 + 8y = -40$ $\newline$ $40y + 40 = -40$
Isolate y term: Subtract $40$ from both sides of the equation to isolate the term with $y$ . $\newline$ $40y + 40 - 40 = -40 - 40$ $\newline$ $40y = -80$
Solve for y: Divide both sides of the equation by $40$ to solve for y. $\newline$ $\frac{40y}{40} = \frac{-80}{40}$ $\newline$ $y = -2$
Substitute $y$ back for $x$ : Substitute the value of $y$ back into the second equation to solve for $x$ .
$x = 4y + 5$
$x = 4(-2) + 5$
$x = -8 + 5$
$x = -3$
Write solution as ordered pair: Write the solution as an ordered pair $(x, y)$ . $\newline$ The solution is $(-3, -2)$ .