Question

1. choose the solution for each linear system.

\

$$\begin{cases}x - 2y = 4 \\\\ 4x + 2y = 6\\end{cases}$$

\bigcirc\\ (2, -1)
\bigcirc\\ (3, 0)
\bigcirc\\ (-3, -1)
\bigcirc\\ (-2, 1)

1. choose the solution for each linear system.

\

$$\begin{cases}6x - 2y = -4 \\\\ -4x + 3y = -4\\end{cases}$$

\bigcirc\\ (2, 4)
\bigcirc\\ (-4, -2)
\bigcirc\\ (-2, -4)
\bigcirc\\ (4, 2)

Explanation:

Question 10

Step1: Add the two equations

We have the system:

$$\begin{cases}x - 2y = 4\\4x + 2y = 6\end{cases}$$

Adding the two equations to eliminate \(y\):
\((x - 2y)+(4x + 2y)=4 + 6\)
Simplify: \(x+4x-2y + 2y=10\)
Which gives: \(5x=10\)

Step2: Solve for \(x\)

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

Step3: Substitute \(x = 2\) into the first equation

Substitute \(x = 2\) into \(x-2y = 4\):
\(2-2y=4\)
Subtract 2 from both sides: \(-2y=4 - 2=2\)
Divide by -2: \(y=\frac{2}{-2}=-1\)

So the solution is \((2,-1)\)

Step1: Multiply equations to eliminate a variable

We have the system:

$$\begin{cases}6x - 2y=-4\\-4x + 3y=-4\end{cases}$$

Multiply the first equation by 3 and the second equation by 2 to make the coefficients of \(y\) opposites:
First equation: \(3\times(6x - 2y)=3\times(-4)\Rightarrow18x-6y=-12\)
Second equation: \(2\times(-4x + 3y)=2\times(-4)\Rightarrow - 8x+6y=-8\)

Step2: Add the new equations

Add the two new equations:
\((18x-6y)+(-8x + 6y)=-12+(-8)\)
Simplify: \(18x-8x-6y + 6y=-20\)
Which gives: \(10x=-20\)

Step3: Solve for \(x\)

Divide both sides by 10:
\(x=\frac{-20}{10}=-2\)

Step4: Substitute \(x=-2\) into the first equation

Substitute \(x = - 2\) into \(6x-2y=-4\):
\(6\times(-2)-2y=-4\)
Simplify: \(-12-2y=-4\)
Add 12 to both sides: \(-2y=-4 + 12=8\)
Divide by -2: \(y=\frac{8}{-2}=-4\)

So the solution is \((-2,-4)\)

Answer:

(2, -1)

Question 11