Answer :
To solve the system of linear equations using the linear combination (or elimination) method, we need to eliminate one of the variables by combining the two equations. The given system is:
[tex]\[ \begin{array}{l} 6x - 3y = 3 \\ -2x + 6y = 14 \end{array} \][/tex]
Step 1: Multiply the equations to make the coefficients of one of the variables (either [tex]\(x\)[/tex] or [tex]\(y\)[/tex]) the same (with the same or opposite sign).
We can multiply the first equation by 2 and the second equation by 1 (leaving it unchanged) to make the coefficients of [tex]\(x\)[/tex] compatible for elimination:
[tex]\[ \begin{array}{l} 2(6x - 3y) = 2(3) \\ 1(-2x + 6y) = 1(14) \end{array} \][/tex]
This simplifies to:
[tex]\[ \begin{array}{l} 12x - 6y = 6 \quad \Rightarrow \quad (1) \\ -2x + 6y = 14 \quad \Rightarrow \quad (2) \end{array} \][/tex]
Step 2: Add the two equations to eliminate [tex]\(y\)[/tex]:
[tex]\[ (12x - 6y) + (-2x + 6y) = 6 + 14 \][/tex]
This simplifies to:
[tex]\[ 12x - 6y - 2x + 6y = 20 \][/tex]
Combining like terms gives us:
[tex]\[ 10x = 20 \][/tex]
Step 3: Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{20}{10} \][/tex]
[tex]\[ x = 2 \][/tex]
Step 4: Substitute [tex]\(x = 2\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. We'll use the first equation:
[tex]\[ 6x - 3y = 3 \][/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\[ 6(2) - 3y = 3 \][/tex]
[tex]\[ 12 - 3y = 3 \][/tex]
Step 5: Solve for [tex]\(y\)[/tex]:
[tex]\[ 12 - 3 = 3y \][/tex]
[tex]\[ 9 = 3y \][/tex]
[tex]\[ y = \frac{9}{3} \][/tex]
[tex]\[ y = 3 \][/tex]
Solution:
The solution to the system of equations is [tex]\((x, y) = (2, 3)\)[/tex].
Answer:
[tex]\((2, 3)\)[/tex]
[tex]\[ \begin{array}{l} 6x - 3y = 3 \\ -2x + 6y = 14 \end{array} \][/tex]
Step 1: Multiply the equations to make the coefficients of one of the variables (either [tex]\(x\)[/tex] or [tex]\(y\)[/tex]) the same (with the same or opposite sign).
We can multiply the first equation by 2 and the second equation by 1 (leaving it unchanged) to make the coefficients of [tex]\(x\)[/tex] compatible for elimination:
[tex]\[ \begin{array}{l} 2(6x - 3y) = 2(3) \\ 1(-2x + 6y) = 1(14) \end{array} \][/tex]
This simplifies to:
[tex]\[ \begin{array}{l} 12x - 6y = 6 \quad \Rightarrow \quad (1) \\ -2x + 6y = 14 \quad \Rightarrow \quad (2) \end{array} \][/tex]
Step 2: Add the two equations to eliminate [tex]\(y\)[/tex]:
[tex]\[ (12x - 6y) + (-2x + 6y) = 6 + 14 \][/tex]
This simplifies to:
[tex]\[ 12x - 6y - 2x + 6y = 20 \][/tex]
Combining like terms gives us:
[tex]\[ 10x = 20 \][/tex]
Step 3: Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{20}{10} \][/tex]
[tex]\[ x = 2 \][/tex]
Step 4: Substitute [tex]\(x = 2\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. We'll use the first equation:
[tex]\[ 6x - 3y = 3 \][/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\[ 6(2) - 3y = 3 \][/tex]
[tex]\[ 12 - 3y = 3 \][/tex]
Step 5: Solve for [tex]\(y\)[/tex]:
[tex]\[ 12 - 3 = 3y \][/tex]
[tex]\[ 9 = 3y \][/tex]
[tex]\[ y = \frac{9}{3} \][/tex]
[tex]\[ y = 3 \][/tex]
Solution:
The solution to the system of equations is [tex]\((x, y) = (2, 3)\)[/tex].
Answer:
[tex]\((2, 3)\)[/tex]
