Question

which system has the same solution as the system of equations below?
$3x + 2y = -5$
$2x + 3y = 5$

a. $\

$$\begin{cases}6x + 4y = -5\\\\6x + 9y = 5\\end{cases}$$

$

b. $\

$$\begin{cases}9x + 6y = -15\\\\4x + 6y = 10\\end{cases}$$

$

c. $\

$$\begin{cases}6x + 4y = -10\\\\-6x + 9y = 15\\end{cases}$$

$

d. $\

$$\begin{cases}6x + 4y = -10\\\\6x + 9y = 5\\end{cases}$$

$

Explanation:

Step1: Recall the property of linear systems

To find a system with the same solution, we can multiply each equation in the original system by a non - zero constant (this is based on the property that if we multiply both sides of an equation \(ax + by=c\) by a non - zero number \(k\), we get \(kax + kby = kc\), and the solution of the new equation is the same as the original equation). The original system is \(

$$\begin{cases}3x + 2y=-5\\2x+3y = 5\end{cases}$$

\)

Step2: Analyze option B

• For the first equation of the original system \(3x + 2y=-5\), if we multiply both sides by \(3\), we have:

\(3\times(3x + 2y)=3\times(-5)\)
Using the distributive property \(a(b + c)=ab+ac\), we get \(9x+6y=-15\)

• For the second equation of the original system \(2x + 3y = 5\), if we multiply both sides by \(2\), we have:

\(2\times(2x + 3y)=2\times5\)
Using the distributive property, we get \(4x + 6y = 10\)

So the system \(

$$\begin{cases}9x + 6y=-15\\4x+6y = 10\end{cases}$$

\) (which is option B) is obtained by multiplying the original equations by non - zero constants, and thus has the same solution as the original system.

We can also check other options:

• Option A: For the first equation \(3x + 2y=-5\), multiplying by \(2\) gives \(6x + 4y=-10

eq - 5\), so this system is not equivalent.

• Option C: The second equation \(2x + 3y = 5\) multiplied by \(3\) gives \(6x+9y = 15\), but the first equation transformation and the sign of the second equation in option C are incorrect.
• Option D: For the first equation \(3x + 2y=-5\), multiplying by \(2\) gives \(6x + 4y=-10\) (correct for the first part), but for the second equation \(2x + 3y = 5\), multiplying by \(3\) gives \(6x+9y = 15

eq5\), so this system is not equivalent.

Answer:

B. \(

$$\begin{cases}9x + 6y=-15\\4x+6y = 10\end{cases}$$

\)