Question

solve the system of equations below.
\\\

$$\begin{aligned} 3x + 4y &= 10 \\\\ 6x - 2y &= 40 \\end{aligned}$$

\\

(6, -2)
(-2, -6)
(2, 6)
(2, -6)

Explanation:

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<reasoning>

Set up the system of equations

\[

$$\begin{cases} 3x + 4y = 10 \quad \text{--- (Equation 1)} \\ 6x - 2y = 40 \quad \text{--- (Equation 2)} \end{cases}$$

\]

Eliminate the variable y

Multiply Equation 2 by 2 and add it to Equation 1:
\[

$$\begin{aligned} 2(6x - 2y) &= 2(40) \\ 12x - 4y &= 80 \\ \\ (3x + 4y) + (12x - 4y) &= 10 + 80 \\ 15x &= 90 \\ x &= 6 \end{aligned}$$

\]

Solve for y

Substitute \(x = 6\) into Equation 1:
\[

$$\begin{aligned} 3(6) + 4y &= 10 \\ 18 + 4y &= 10 \\ 4y &= -8 \\ y &= -2 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) (6,-2)</mcq-correct>
<mcq-option>(B) (-2,-6)</mcq-option>
<mcq-option>(C) (2,6)</mcq-option>
<mcq-option>(D) (2,-6)</mcq-option>
</answer>

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Answer:

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<reasoning>

Set up the system of equations

\[

$$\begin{cases} 3x + 4y = 10 \quad \text{--- (Equation 1)} \\ 6x - 2y = 40 \quad \text{--- (Equation 2)} \end{cases}$$

\]

Eliminate the variable y

Multiply Equation 2 by 2 and add it to Equation 1:
\[

$$\begin{aligned} 2(6x - 2y) &= 2(40) \\ 12x - 4y &= 80 \\ \\ (3x + 4y) + (12x - 4y) &= 10 + 80 \\ 15x &= 90 \\ x &= 6 \end{aligned}$$

\]

Solve for y

Substitute \(x = 6\) into Equation 1:
\[

$$\begin{aligned} 3(6) + 4y &= 10 \\ 18 + 4y &= 10 \\ 4y &= -8 \\ y &= -2 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) (6,-2)</mcq-correct>
<mcq-option>(B) (-2,-6)</mcq-option>
<mcq-option>(C) (2,6)</mcq-option>
<mcq-option>(D) (2,-6)</mcq-option>
</answer>

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"knowledge_point": [
"Mathematics",
"Algebra",
"Systems of Equations Elimination"
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