Solving by Substitution Method

TL;DR

The substitution method helps you solve systems of equations by isolating one variable in one equation and then plugging that expression into another equation. This reduces the problem to a simpler one-variable equation, making it easier to find the values for all variables. It's a systematic way to break down complex systems into manageable steps.

1. The Mental Model

Imagine you have a riddle where you know "Alice is twice as old as Bob" and "Alice and Bob's ages add up to 30." Substitution lets you replace "Alice" with "twice Bob's age" in the second statement, so you're only dealing with Bob's age to solve the riddle.

2. The Core Material

The substitution method is great for solving systems of linear equations, especially when it's easy to get one variable by itself. The goal is to reduce a system of multiple equations with multiple variables into a single equation with just one variable, which you already know how to solve!

2.1 The Steps

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Here's how you do it:

1. Isolate a variable: Look at your equations and pick one where it's easiest to get one variable (like x or y or z) by itself. This means getting it alone on one side of the equals sign.
2. Substitute: Take the expression you just found for that isolated variable and plug it into another equation in your system. Make sure you use a different equation than the one you just used to isolate the variable!
3. Solve the new equation: Now you'll have an equation with only one variable. Solve it! This will give you the value of that variable.
4. Back-substitute: Take the value you just found and plug it back into any of your original equations (or the equation where you first isolated a variable) to find the value of the other variables.
5. Check your solution: Plug all the values you found back into all the original equations to make sure they all hold true.

2.2 Visualizing the Process

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graph TD
    A["Start with a system of equations"] --> B["Choose an equation and isolate one variable (e.g., x = ... or y = ...)"]
    B --> C{"Is the isolated variable expression ready?"}
    C -- "Yes" --> D["Substitute the expression into ANOTHER equation"]
    D --> E["Solve the resulting one-variable equation"]
    E --> F["Substitute the value found back into one of the original equations (or the isolated variable expression)"]
    F --> G["Solve for the second variable"]
    G --> H["(For 3x3 systems) Substitute both known values into a remaining equation"]
    H --> I["Solve for the third variable"]
    I --> J["Check your solution in ALL original equations"]
    J --> K["Solution found!"]

2.3 Solving 3x3 Systems with Substitution

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For a 3x3 system (three equations, three variables like x, y, z), the process extends:

1. Isolate: Pick one equation and isolate one variable (e.g., get x by itself). This will likely involve y and z.
2. Substitute (first time): Substitute this expression for x into the other two equations. Now you have a system of two equations with only y and z.
3. Solve 2x2 system: Use the substitution method again (or elimination) on this new 2x2 system to find the values for y and z.
4. Back-substitute (twice):
   • Plug the values of y and z into the expression you first found for x to get its value.
   • (Optional, for verification) Plug all three values back into the original equations.

3. Worked Example

Let's solve this 3x3 system:

1. x + y + z = 6
2. 2x - y + z = 3
3. 3x + 2y - z = 4

Step 1: Isolate a variable.
From equation (1), it's easy to get x by itself:
x = 6 - y - z (Let's call this equation (4))

Step 2: Substitute.
Substitute x from equation (4) into equations (2) and (3):

Into (2):
2(6 - y - z) - y + z = 3
12 - 2y - 2z - y + z = 3
12 - 3y - z = 3
-3y - z = 3 - 12
-3y - z = -9 (Let's call this equation (5))

Into (3):
3(6 - y - z) + 2y - z = 4
18 - 3y - 3z + 2y - z = 4
18 - y - 4z = 4
-y - 4z = 4 - 18
-y - 4z = -14 (Let's call this equation (6))

Now we have a 2x2 system with equations (5) and (6):
5. -3y - z = -9
6. -y - 4z = -14

Step 3: Solve the new 2x2 system.
From equation (5), let's isolate z:
-z = -9 + 3y
z = 9 - 3y (Let's call this equation (7))

Substitute z from (7) into equation (6):
-y - 4(9 - 3y) = -14
-y - 36 + 12y = -14
11y - 36 = -14
11y = -14 + 36
11y = 22
y = 2

Step 4: Back-substitute.
Now that we have y = 2, plug it back into equation (7) to find z:
z = 9 - 3(2)
z = 9 - 6
z = 3

Finally, we have y = 2 and z = 3. Plug both into equation (4) to find x:
x = 6 - y - z
x = 6 - 2 - 3
x = 1

So the solution is x = 1, y = 2, z = 3.

Step 5: Check your solution.
Plug x=1, y=2, z=3 into the original three equations:

1. 1 + 2 + 3 = 6 (True)
2. 2(1) - 2 + 3 = 2 - 2 + 3 = 3 (True)
3. 3(1) + 2(2) - 3 = 3 + 4 - 3 = 4 (True)

All equations hold true, so our solution is correct!

4. Key Takeaways

• Always isolate the easiest variable first, typically one with a coefficient of 1 or -1, to avoid fractions early on.
• When you substitute, always plug the expression into a different equation than the one you used to isolate the variable.
• Be super careful with signs, especially when distributing negative numbers.
• For 3x3 systems, you'll reduce it to a 2x2 system first, then solve that, and then back-substitute to find all three variables.
• Always check your final solution by plugging all values back into all original equations.

Common Mistakes to Avoid:

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• Substituting back into the same equation you used to isolate the variable – this will just give you 0=0.
• Making arithmetic errors, especially with negative numbers during distribution or combination of terms.
• Not checking your solution, which means you might not catch a mistake until it's too late.
• Getting overwhelmed by a 3x3 system; remember it's just two rounds of the 2x2 process.

5. Now Try It

Solve the following system of equations using the substitution method. You'll know you're successful if you find the values for x, y, and z that satisfy all three equations.

1. x + 2y - z = 4
2. 2x - y + 3z = -1
3. 3x + y - 2z = 3