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Solve the system using multiplication for the linear combination method.

[tex]\[
\begin{array}{l}
6x - 3y = 3 \\
-2x + 6y = 14
\end{array}
\][/tex]

What is the solution to the system?

A. [tex]\((2, 1)\)[/tex]

B. [tex]\((2, -3)\)[/tex]

C. [tex]\((2, -1)\)[/tex]

D. [tex]\((2, 3)\)[/tex]


Sagot :

Let's solve the system of linear equations using the linear combination (or elimination) method.

Given the system of equations:

[tex]\[ \begin{array}{l} 6x - 3y = 3 \quad \text{(1)} \\ -2x + 6y = 14 \quad \text{(2)} \end{array} \][/tex]

Step 1: Align the coefficients

To eliminate [tex]\( y \)[/tex], we can align the coefficients of [tex]\( y \)[/tex] by manipulating the equations. We see that the coefficient of [tex]\( y \)[/tex] in equation [tex]\((1)\)[/tex] is [tex]\(-3y\)[/tex] and in [tex]\((2)\)[/tex] it is [tex]\(6y\)[/tex]. One way to align them is by multiplying equation [tex]\((1)\)[/tex] by 2, so the coefficients of [tex]\( y \)[/tex] become opposites:

[tex]\[ 2 \cdot (6x - 3y) = 2 \cdot 3 \implies 12x - 6y = 6 \quad \text{(3)} \][/tex]

Step 2: Add the equations

Now, we add equation [tex]\((3)\)[/tex] and equation [tex]\((2)\)[/tex]:

[tex]\[ \begin{array}{rcl} 12x - 6y & = & 6 \quad \text{(3)} \\ -2x + 6y & = & 14 \quad \text{(2)} \\ \hline (12x - 2x) + (-6y + 6y) & = & 6 + 14 \\ 10x & = & 20 \end{array} \][/tex]

Step 3: Solve for [tex]\( x \)[/tex]

[tex]\[ 10x = 20 \implies x = \frac{20}{10} = 2 \][/tex]

Step 4: Substitute [tex]\( x \)[/tex] back into one of the original equations

Let's substitute [tex]\( x = 2 \)[/tex] into the first equation [tex]\((1)\)[/tex]:

[tex]\[ 6(2) - 3y = 3 \implies 12 - 3y = 3 \][/tex]

Step 5: Solve for [tex]\( y \)[/tex]

[tex]\[ 12 - 3y = 3 \implies -3y = 3 - 12 \implies -3y = -9 \implies y = \frac{-9}{-3} = 3 \][/tex]

The solution to the system of equations is:

[tex]\[ (x, y) = (2, 3) \][/tex]

Thus, the correct solution is:

[tex]\((2, 3)\)[/tex]