Question

consider the system of equations
3x + 4y = 0
-3x + -4y = 2
adding the two equations side - by - side and simplifying yields 0 = 2.
which of the following can be concluded about the system of equations?

Explanation:

Step1: Analyze the system of equations

We have the system:
\[

$$\begin{cases} 3x + 4y = 0 \\ -3x - 4y = 2 \end{cases}$$

\]
When we add the two equations side - by - side, we add the left - hand sides and the right - hand sides separately.

The left - hand side of the first equation is \(3x + 4y\) and the left - hand side of the second equation is \(-3x-4y\). Adding them together: \((3x + 4y)+(-3x - 4y)=3x+4y - 3x - 4y\).

The right - hand side of the first equation is \(0\) and the right - hand side of the second equation is \(2\). Adding them together: \(0 + 2=2\).

Step2: Simplify the left - hand side

Simplify \(3x+4y - 3x - 4y\). The \(3x\) and \(-3x\) cancel out (\(3x-3x = 0\)), and the \(4y\) and \(-4y\) cancel out (\(4y-4y = 0\)). So the left - hand side simplifies to \(0\).

After adding and simplifying, we get \(0=2\), which is a false statement.

In a system of linear equations \(a_1x + b_1y=c_1\) and \(a_2x + b_2y=c_2\), if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}
eq\frac{c_1}{c_2}\) (in the case of two - variable linear equations), the system of equations is inconsistent, meaning it has no solution.

Here, \(\frac{3}{-3}=\frac{4}{-4}=- 1\), and \(\frac{0}{2}=0
eq - 1\). So the system of equations has no solution.

Answer:

The system of equations \(

$$\begin{cases}3x + 4y = 0\\-3x - 4y = 2\end{cases}$$

\) has no solution because adding the two equations results in the false statement \(0 = 2\), indicating that the two lines represented by the equations are parallel and do not intersect.