01 / 21

Solving Systems by
Substitution

Learn how to solve a system of two equations by replacing one variable with an expression, then find the point where both lines meet.

Pre-Algebra

What is a System of Equations?

A system of equations is a set of two or more equations that share the same variables. A solution to the system is the pair \((x, y)\) that makes both equations true at the same time.

The Solution is an Intersection Point

1 2 3 4 1 2 3 4 5 6 7 8 x y
\( y = 2x + 1 \)
\( y = -x + 7 \)
 (2, 5)

Why Use Substitution?

Graphing works, but reading exact coordinates from a graph can be tricky. Substitution is an algebraic method that gives you the exact answer every time: no guessing, no rounding.

Graphing
Visual, but can be imprecise when the intersection falls between grid lines.
Substitution
Algebraic and exact: replace one variable with an expression to solve step-by-step.

The Big Idea of Substitution

This is the substitution principle. We take one equation that is already solved for a variable and plug it into the other equation. This eliminates one variable, leaving a single equation with one unknown.

When Does Substitution Work Best?

Substitution is the ideal method when at least one equation is already solved for a variable, or can be solved easily. Look for equations in these forms:

\( y = \dots \)
One variable is already isolated on one side of the equation.
\( x = \dots \)
Either variable works: just substitute the expression into the other equation.
Coefficient of \(1\)
If a variable has a coefficient of \(1\) (like \( x + 3y = 7 \)), it is easy to isolate.

Substitution vs. Elimination

Substitution Elimination
Strategy Replace a variable with an expression Add or subtract equations to cancel a variable
Best when A variable is already isolated Coefficients line up nicely
Key move Plug in Add/subtract
Result One equation, one unknown One equation, one unknown

The 4-Step Method

1
Isolate one variable in one equation (choose the easiest one).
2
Substitute that expression into the other equation.
3
Solve the resulting one-variable equation.
4
Back-substitute to find the other variable, then check in both original equations.

Example 1: Basic Substitution

1
Isolate: \( y \) is already isolated in Equation 1: \( y = 3x - 1 \).
2
Substitute: Replace \( y \) in Equation 2: \( 2x + (3x - 1) = 9 \).
3
Solve: \( 5x - 1 = 9 \) → \( 5x = 10 \) → \( x = 2 \).
4
Back-substitute: \( y = 3(2) - 1 = 5 \). Solution: \((2, 5)\).

Check Your Answer

Always verify by plugging the solution \((2, 5)\) back into both original equations:

Equation 1
\( y = 3x - 1 \)
\( 5 = 3(2) - 1 \)
\( 5 = 5 \) ✓
Equation 2
\( 2x + y = 9 \)
\( 2(2) + 5 = 9 \)
\( 9 = 9 \) ✓

Example 2: Isolate First

1
Isolate: From Eq. 1, \( x = 8 - 2y \) (coefficient of \(x\) is \(1\), so it is easy).
2
Substitute into Eq. 2: \( 3(8 - 2y) - y = 3 \).
3
Solve: \( 24 - 6y - y = 3 \) → \( -7y = -21 \) → \( y = 3 \).
4
Back-substitute: \( x = 8 - 2(3) = 2 \). Solution: \((2, 3)\).

Example 3: Negative Coefficients

1
Isolate: \( y \) is already isolated: \( y = -2x + 10 \).
2
Substitute: \( 4x + 3(-2x + 10) = 20 \).
3
Distribute and solve: \( 4x - 6x + 30 = 20 \) → \( -2x = -10 \) → \( x = 5 \).
4
Back-substitute: \( y = -2(5) + 10 = 0 \). Solution: \((5, 0)\).

What Does the Solution Mean?

The ordered pair you find is the coordinates of the point where the two lines intersect on a graph. It represents the one set of values that makes both equations true.

1 2 3 4 5 2 4 6 8 10
\( y = -2x + 10 \)
\( 4x + 3y = 20 \)
 (5, 0)

Real-World Application

1
Define variables: Let \(n\) = notebooks, \(p\) = pens.
2
Write equations: \( n + p = 10 \) and \( 3n + p = 18 \).
3
Isolate: \( p = 10 - n \). Substitute: \( 3n + (10 - n) = 18 \).
4
Solve: \( 2n + 10 = 18 \) → \( n = 4 \), \( p = 6 \). You bought 4 notebooks and 6 pens.

Common Mistakes to Avoid

✕ Forgetting to Distribute
When you substitute \( 3(8 - 2y) \), you must multiply 3 by every term inside: \( 24 - 6y \), not \( 24 - 2y \).
✕ Substituting into the Same Equation
You must plug the expression into the other equation: substituting back into the same one just gives \( 0 = 0 \).
✕ Dropping the Negative Sign
Watch signs carefully. If \( y = -2x + 10 \), then \( 3y = 3(-2x + 10) = -6x + 30 \), not \( 6x + 30 \).

Substitution Summary

Step Action What You Get
1. Isolate Solve one equation for \(x\) or \(y\) An expression like \( y = \dots \)
2. Substitute Replace the variable in the other equation One equation with one variable
3. Solve Simplify and solve for the remaining variable A numerical value
4. Back-substitute Plug the value back into the isolated expression The second value and solution \((x, y)\)

Question 1 of 5

Solve the system using substitution:
\( y = x + 3 \)
\( 2x + y = 12 \)

What is the solution \((x, y)\)?

Question 2 of 5

What is the first step in the substitution method?

Question 3 of 5

Given the system:
\( x = 5 - y \)
\( 3x + 2y = 11 \)

After substitution into the second equation, which equation do you get?

Question 4 of 5

Solve the system:
\( y = -x + 8 \)
\( x - y = 2 \)

What is \( x \)?

Question 5 of 5

A student solves the system below and gets \((4, 1)\).
\( y = 2x - 7 \)
\( x + y = 5 \)

Is the student correct?