Question

solving systems of linear elimination day 1

1. $-2x - 8y = 10$

$2x - 6y = 18$

1. $3x - y = -2$

$-2x + y = 3$

1. $2x + 5y = -24$

$3x - 5y = 14$

1. $-6x + 2y = 8$

$6x + 3y = -3$

1. $-2x + 3y = -23$

$x - 3y = 16$

1. $6x - 3y = 12$

$-5x + 3y = -5$

Explanation:

Problem 1: Solve \(

$$\begin{cases}-2x - 8y = 10\\2x - 6y = 18\end{cases}$$

\)

Step 1: Eliminate \(x\) by adding the two equations.

Adding \(-2x - 8y = 10\) and \(2x - 6y = 18\):
\[

$$\begin{align*} (-2x + 2x) + (-8y - 6y) &= 10 + 18\\ -14y &= 28 \end{align*}$$

\]

Step 2: Solve for \(y\).

Divide both sides by \(-14\):
\[
y = \frac{28}{-14} = -2
\]

Step 3: Substitute \(y = -2\) into one of the original equations (e.g., \(2x - 6y = 18\)) to solve for \(x\).

\[

$$\begin{align*} 2x - 6(-2) &= 18\\ 2x + 12 &= 18\\ 2x &= 18 - 12\\ 2x &= 6\\ x &= 3 \end{align*}$$

\]

Step 1: Eliminate \(y\) by adding the two equations.

Adding \(3x - y = -2\) and \(-2x + y = 3\):
\[

$$\begin{align*} (3x - 2x) + (-y + y) &= -2 + 3\\ x &= 1 \end{align*}$$

\]

Step 2: Substitute \(x = 1\) into one of the original equations (e.g., \(3x - y = -2\)) to solve for \(y\).

\[

$$\begin{align*} 3(1) - y &= -2\\ 3 - y &= -2\\ -y &= -2 - 3\\ -y &= -5\\ y &= 5 \end{align*}$$

\]

Step 1: Eliminate \(y\) by adding the two equations.

Adding \(2x + 5y = -24\) and \(3x - 5y = 14\):
\[

$$\begin{align*} (2x + 3x) + (5y - 5y) &= -24 + 14\\ 5x &= -10 \end{align*}$$

\]

Step 2: Solve for \(x\).

Divide both sides by \(5\):
\[
x = \frac{-10}{5} = -2
\]

Step 3: Substitute \(x = -2\) into one of the original equations (e.g., \(2x + 5y = -24\)) to solve for \(y\).

\[

$$\begin{align*} 2(-2) + 5y &= -24\\ -4 + 5y &= -24\\ 5y &= -24 + 4\\ 5y &= -20\\ y &= -4 \end{align*}$$

\]

Answer:

\((3, -2)\)

Problem 2: Solve \(

$$\begin{cases}3x - y = -2\\-2x + y = 3\end{cases}$$

\)