Integrated Math 1: Elimination Expertise: Solving Systems Effectively

Elimination Method Explained 🧮

The elimination method, also known as the addition method, is a technique used to solve systems of linear equations. The goal is to eliminate one of the variables by adding or subtracting the equations. This leaves you with a single equation in one variable, which can be easily solved. Once you find the value of one variable, you can substitute it back into one of the original equations to find the value of the other variable.

Steps for Solving Systems Using Elimination 🪜

1. Align the Equations: Make sure the like terms (x's and y's) are aligned in columns.
2. Multiply (if needed): Multiply one or both equations by a constant so that the coefficients of one variable are opposites (e.g., 2x and -2x).
3. Add the Equations: Add the equations together. This should eliminate one of the variables.
4. Solve for the Remaining Variable: Solve the resulting equation for the remaining variable.
5. Substitute: Substitute the value found in step 4 back into one of the original equations to solve for the other variable.
6. Check Your Solution: Substitute both values into both original equations to verify the solution.

Example ✍️

Let's solve the following system of equations:

Equation 1: $2x + y = 7$

Equation 2: $x - y = 2$

Step 1: Align the Equations

The equations are already aligned.

Step 2: Multiply (if needed)

Notice that the coefficients of $y$ are already opposites ($+1$ and $-1$). No multiplication is needed.

Step 3: Add the Equations

Add Equation 1 and Equation 2:

  2x + y = 7
+ x - y = 2
----------
  3x + 0 = 9

Step 4: Solve for the Remaining Variable

Solve $3x = 9$ for $x$:

3x = 9
x = 9 / 3
x = 3

Step 5: Substitute

Substitute $x = 3$ into Equation 1:

2(3) + y = 7
6 + y = 7
y = 7 - 6
y = 1

Step 6: Check Your Solution

Check the solution $(x = 3, y = 1)$ in both original equations:

Equation 1: $2(3) + 1 = 6 + 1 = 7$ (Correct)

Equation 2: $3 - 1 = 2$ (Correct)

Therefore, the solution to the system of equations is $x = 3$ and $y = 1$.

Another Example with Multiplication ➕

Solve the following system:

Equation 1: $x + 2y = 5$

Equation 2: $3x - y = 1$

To eliminate $y$, multiply Equation 2 by 2:

2 * (3x - y) = 2 * 1
6x - 2y = 2

Now add the modified Equation 2 to Equation 1:

  x + 2y = 5
+ 6x - 2y = 2
----------
  7x + 0 = 7

Solve for $x$:

7x = 7
x = 1

Substitute $x = 1$ into Equation 1:

1 + 2y = 5
2y = 4
y = 2

The solution is $x = 1$ and $y = 2$.