Solve the system by substitution. {:[-2x+5=y],[10 x-4y=-38]:} (◻,◻)

Isolate y: Isolate y in the first equation. The first equation is -2x + 5 = y. We can rewrite this as y = -2x + 5.Substitute expression for y: Substitute the expression for y into the second equation. The second equation is 10x - 4y = -38. We substitute y with -2x + 5 to get 10x - 4(-2x + 5) = -38.Simplify and solve for x: Simplify and solve for x. 10x + 8x - 20 = -38 18x - 20 = -38 18x = -38 + 20 18x = -18 x = -18 / 18 x = -1Substitute x back: Substitute x back into the first equation to solve for y. We have y = -2(-1) + 5. y = 2 + 5 y = 7Write solution as ordered pair: Write the solution as an ordered pair (x, y). The solution is (-1, 7).

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Solve the system by substitution.

{:[-2x+5=y],[10 x-4y=-38]:}

(◻,◻)

Solve the system by substitution. $\newline$ $\begin{aligned} -2 x+5 & =y \\ 10 x-4 y & =-38 \end{aligned}$ $\newline$ $(\square, \square)$

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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{aligned} -2 x+5 & =y \\ 10 x-4 y & =-38 \end{aligned}

\newline

(\square, \square)

Isolate y: Isolate $y$ in the first equation. $\newline$ The first equation is $-2x + 5 = y$ . We can rewrite this as $y = -2x + 5$ .
Substitute expression for y: Substitute the expression for y into the second equation. $\newline$ The second equation is $10x - 4y = -38$ . We substitute $y$ with $-2x + 5$ to get $10x - 4(-2x + 5) = -38$ .
Simplify and solve for x: Simplify and solve for x. $\newline$ $10x + 8x - 20 = -38$ $\newline$ $18x - 20 = -38$ $\newline$ $18x = -38 + 20$ $\newline$ $18x = -18$ $\newline$ $x = \frac{-18}{18}$ $\newline$ $x = -1$
Substitute $x$ back: Substitute $x$ back into the first equation to solve for $y$ . We have $y = -2(-1) + 5$ . $y = 2 + 5$ $y = 7$
Write solution as ordered pair: Write the solution as an ordered pair $(x, y)$ . The solution is $(-1, 7)$ .