Question

which system of equations has the same solution as the system below?
$3x + 3y = 3$
$6x + 5y = 1$
answer
$3x - 6y = -6$
$6x + 5y = 1$
$-6x - 6y = -6$
$6x + 5y = 1$
$-6x - 6y = 3$
$6x + 5y = 1$
$-6x + 3y = -6$
$6x + 5y = 1$

Explanation:

Step1: Analyze the original equation

The original first equation is \(3x + 3y=3\). We can multiply both sides of this equation by a non - zero constant to get an equivalent equation. Let's recall that if we have an equation \(ax + by = c\), multiplying both sides by \(k\) (\(k
eq0\)) gives \(kax + kby=kc\), and the solution of the system remains the same.

Step2: Analyze the transformation of the first equation

For the equation \(3x + 3y = 3\), if we multiply both sides by \(- 2\), we have:
\[

$$\begin{align*} -2\times(3x + 3y)&=-2\times3\\ -6x-6y&=-6 \end{align*}$$

\]
The second equation of the original system is \(6x + 5y = 1\). So the system of equations \(

$$\begin{cases}-6x - 6y=-6\\6x + 5y = 1\end{cases}$$

\) has the same solution as the original system.

Answer:

\(-6x - 6y=-6\) and \(6x + 5y = 1\) (the second option in the given choices, i.e., the option with \(\boldsymbol{-6x - 6y=-6}\) and \(\boldsymbol{6x + 5y = 1}\))