Which system has the same solution as the system of equations shown? $$ \begin{array}{l} 8 x+3 y=5 \ 4 x+2 y=3 \end{array} $$ A. $$ 8 x+3 y=5 $$ $$ -8 x+4 y=6 $$ B. $$ 8 x+3 y=5 $$ $$ -8 x-4 y=-6 $$ C. $$ 16 x+6 y=5 $$ $$ 12 x-6 y=0 $$ D. $$ 8 x+3 y=5 $$ $$ 8 x+4 y=3 $$

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We start with the original system:
$$
\begin{aligned}
8x + 3y &= 5 \quad (1)\[1mm]
4x + 2y &= 3 \quad (2)
\end{aligned}
$$

A common strategy is to eliminate one variable. Notice that if we multiply equation (2) by 2, we get:
$$
2(4x + 2y) = 2\cdot3 \quad\Longrightarrow\quad 8x + 4y = 6. \quad (3)
$$

Now we have:
$$
\begin{aligned}
8x + 3y &= 5 \quad (1)\[1mm]
8x + 4y &= 6 \quad (3)
\end{aligned}
$$
Subtract equation (1) from equation (3):
$$
(8x+4y) - (8x+3y) = 6-5 \quad\Longrightarrow\quad y = 1.
$$

Substitute $$ y = 1 $$ into equation (1):
$$
8x + 3(1) = 5 \quad\Longrightarrow\quad 8x = 5 - 3 = 2, \quad \text{so} \quad x = \frac{2}{8} = \frac{1}{4}.
$$

Thus, the solution of the original system is:
$$
\left(x,y\right)=\left(\frac{1}{4}, 1\right).
$$

Now, we compare this with the given options. Look at option B:
$$
\begin{aligned}
8x + 3y &= 5,\[1mm]
-8x - 4y &= -6.
\end{aligned}
$$

Multiply the second equation by $$-1$$ to obtain:
$$
8x + 4y = 6,
$$
which is exactly the equation (3) we derived. Therefore, the solution to system B is also $$ \left(\frac{1}{4},1\right) $$.

Thus, the system in option B has the same solution as the given system.
Option B has the same solution as the original system.

Which system has the same solution as the system of equations shown?

A.

B.

C.

D.

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