Lesson 59: Solving Systems of Linear Equations by Substitution

New Concept

To be a solution to a system of equations, an ordered pair must satisfy both equations.

What’s next

Next, you'll learn the substitution method, a powerful algebraic technique for finding the exact point where two different relationships intersect.

Property

1. Rearrange one equation to be in the form $y = mx + b$ or $x = my + b$.
2. Substitute this expression into the other equation.
3. Solve the new equation for the single variable.
4. Substitute this value back into one of the original equations to find the second variable.
5. Write the solution as an ordered pair $(x, y)$.

Explanation

Think of it as a secret agent mission! You find a secret identity for one variable, like $y = 2x - 5$, and then substitute this disguise into the other equation to unmask the first value. With that clue, you can plug it back into any original equation to find the partner variable. Mission accomplished, case closed!

Examples

• For the system $y = 2x + 1$ and $y = -x + 10$, substitute to get $2x + 1 = -x + 10$.
• Solving $2x + 1 = -x + 10$ gives $3x = 9$, so $x = 3$. Then, find y: $y = 2(3) + 1 = 7$. The solution is $(3, 7)$.
• For $x = 4y - 2$ and $2x + y = 10$, substitute for $x$ to get the new equation: $2(4y - 2) + y = 10$.

Property

When substituting an expression for a variable with a coefficient, the coefficient must be distributed to every term inside the parentheses. For example, in $5x - 2y = -1$, if $y = (-2x-4)$, the equation becomes $5x - 2(-2x - 4) = -1$.

Explanation

Think of the number outside the parentheses as a party host! It must greet every single guest inside. When you substitute an expression, the coefficient outside must be multiplied by every term inside—no exceptions. Don't let anyone get left out, not even the negative signs who are just a bit shy and hiding in the back!

Examples

• In the expression $12(-2y + 11)$, distribute the $12$ to get $12(-2y) + 12(11) = -24y + 132$.
• Be careful with negatives! In $5x - 2(-3x + 4) = 1$, distributing the $-2$ correctly gives $5x + 6x - 8 = 1$.
• After substituting $x = -3y + 9$ into $10x - 5y = 10$, you get $10(-3y + 9) - 5y = 10$.

Property

If neither equation is in a form like $y = mx + b$, you must first rearrange one equation to isolate a single variable. Whenever possible, choose a variable with a coefficient of 1, as it is the easiest to solve for.

Explanation

Sometimes your equations are messy and not ready for substitution. It’s like trying to fit a puzzle piece in the wrong spot! Your first job is to rearrange one equation to isolate a variable, like solving for $y$. Pro tip: always look for a variable with a coefficient of 1—it's the easiest one to get by itself.

Examples

• Given the equation $3x + y = 10$, it is easy to rearrange for $y$. Just subtract $3x$ from both sides to get $y = -3x + 10$.
• In the system $3x + y = 10$ and $6x - 2y = 4$, first rearrange to $y = -3x + 10$, then substitute to get $6x - 2(-3x + 10) = 4$.
• For $4a - 2b = 8$, solve for $b$ by first getting $-2b = -4a + 8$, then dividing by $-2$ to find $b = 2a - 4$.

Sometimes neither equation is ready for substitution. Let's see how to cleverly rearrange one to unlock the solution.

Example Problem

Solve the system of equations by substitution: $3x - y = 9$ and $x + 2y = -4$.

Step-by-Step

1. First, we need to rearrange one equation so a variable is isolated. The first equation, $3x - y = 9$, can be easily solved for $y$.
2. Start with the first equation:

3. Subtract $3x$ from both sides:

4. Multiply the entire equation by $-1$ to solve for $y$:

5. Now, substitute this expression for $y$ into the second equation.

6. Distribute the $2$ to the terms in the parentheses:

7. Combine the like terms ($x$ and $6x$):

8. Add $18$ to both sides to isolate the term with $x$:

9. Divide by $7$ to solve for $x$:

10. Substitute $x = 2$ back into one of the original equations to find $y$. Let's use the first one:

11. The solution is the ordered pair $(2, -3)$.