Question

1. -10x - 8y = -12

x + 16y = -14

1. 20x + 4y = 12

-10x + y = 3

Explanation:

Problem 5: Solve the system \(

$$\begin{cases}-10x - 8y=-12\\x + 16y=-14\end{cases}$$

\)

Step 1: Solve the second equation for \(x\)

From \(x + 16y=-14\), we get \(x=-14 - 16y\).

Step 2: Substitute \(x\) into the first equation

Substitute \(x = - 14-16y\) into \(-10x - 8y=-12\):
\[

$$\begin{align*} -10(-14 - 16y)-8y&=-12\\ 140 + 160y-8y&=-12\\ 140+152y&=-12 \end{align*}$$

\]

Step 3: Solve for \(y\)

Subtract 140 from both sides:
\[

$$\begin{align*} 152y&=-12 - 140\\ 152y&=-152\\ y&=-1 \end{align*}$$

\]

Step 4: Substitute \(y = - 1\) back into \(x=-14 - 16y\)

\[

$$\begin{align*} x&=-14-16\times(-1)\\ x&=-14 + 16\\ x&=2 \end{align*}$$

\]

Step 1: Solve the second equation for \(y\)

From \(-10x + y=3\), we get \(y = 10x+3\).

Step 2: Substitute \(y\) into the first equation

Substitute \(y = 10x + 3\) into \(20x + 4y=12\):
\[

$$\begin{align*} 20x+4(10x + 3)&=12\\ 20x+40x+12&=12\\ 60x+12&=12 \end{align*}$$

\]

Step 3: Solve for \(x\)

Subtract 12 from both sides:
\[

$$\begin{align*} 60x&=12 - 12\\ 60x&=0\\ x&=0 \end{align*}$$

\]

Step 4: Substitute \(x = 0\) back into \(y = 10x+3\)

\[

$$\begin{align*} y&=10\times0 + 3\\ y&=3 \end{align*}$$

\]

Answer:

\(x = 2,y=-1\)

Problem 7: Solve the system \(

$$\begin{cases}20x + 4y=12\\-10x + y=3\end{cases}$$

\)