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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]



Answer :

To solve the system of equations using the linear combination (or elimination) method, we'll follow these steps:

1. Write down the system of equations:

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

2. Find a common coefficient for either [tex]\( x \)[/tex] or [tex]\( y \)[/tex] to facilitate elimination.

Notice that the coefficients of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the equations are not directly the same, but they can be manipulated to be. Let's focus on eliminating [tex]\( x \)[/tex]:

Equation 1: [tex]\( 6x - 3y = 3 \)[/tex]

Equation 2: [tex]\( -2x + 6y = 14 \)[/tex]

If we multiply Equation 2 by 3, we get:

[tex]\[ 3(-2x + 6y) = 3(14) \][/tex]

Simplifying, we obtain:

[tex]\[ -6x + 18y = 42 \][/tex]

3. Rewrite the system with the new equation:

[tex]\[ \begin{cases} 6x - 3y = 3 \\ -6x + 18y = 42 \end{cases} \][/tex]

4. Add the two equations to eliminate [tex]\( x \)[/tex]:

[tex]\[ (6x - 3y) + (-6x + 18y) = 3 + 42 \][/tex]

This simplifies to:

[tex]\[ 15y = 45 \][/tex]

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

[tex]\[ y = \frac{45}{15} \][/tex]

[tex]\[ y = 3 \][/tex]

6. Substitute [tex]\( y \)[/tex] back into one of the original equations to solve for [tex]\( x \)[/tex]:

Let's substitute [tex]\( y = 3 \)[/tex] into the first equation:

[tex]\[ 6x - 3(3) = 3 \][/tex]

Simplifying,

[tex]\[ 6x - 9 = 3 \][/tex]

[tex]\[ 6x = 3 + 9 \][/tex]

[tex]\[ 6x = 12 \][/tex]

[tex]\[ x = \frac{12}{6} \][/tex]

[tex]\[ x = 2 \][/tex]

The solution to the system of equations is:

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

Therefore, the correct answer is:

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

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