Solving Systems of Linear Equations Algebraically

Solving Systems of Linear Equations Algebraically

Question 3

Solve the following systems of linear equations for $x$ and $y$ algebraically. You can use either the substitution method or the elimination method.

System of Equations:

1. $x + 2y = 6$
2. $x - y = 3$

Methods to Solve:

• Substitution Method
• Elimination Method

Solution Using Substitution Method:

1. Solve the second equation for $x$:
   $x - y = 3$ can be rearranged to $x = y + 3$.

2. Substitute $x$ in the first equation:
   Replace $x$ in the first equation $x + 2y = 6$ with $(y + 3)$.
   $(y + 3) + 2y = 6$

3. Simplify and solve for $y$:
   $y + 3 + 2y = 6$
   $3y + 3 = 6$
   $3y = 6 - 3$
   $3y = 3$
   $y = 1$

4. Substitute $y$ back into the equation $x = y + 3$ to find $x$:
   $x = 1 + 3$
   $x = 4$

Solution Using Elimination Method:

1. Align the equations:
   Write the equations one above the other:
   $x + 2y = 6$
   $x - y = 3$

2. Subtract the second equation from the first equation to eliminate $x$:
   $(x + 2y) - (x - y) = 6 - 3$
   $x + 2y - x + y = 3$
   $3y = 3$

3. Solve for $y$:
   $y = \frac{3}{3}$
   $y = 1$

4. Substitute $y$ back into one of the original equations to find $x$. Using the first equation:
   $x + 2(1) = 6$
   $x + 2 = 6$
   $x = 6 - 2$
   $x = 4$

Final Solution:

The solution to the system of equations is:

$x = 4$
$y = 1$

This solution satisfies both equations in the system.