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Solving Systems of
Linear Equations

Master three powerful methods to find where two lines intersect: graphing, substitution, and elimination.

Algebra 2

What Is a System of Equations?

A system of linear equations is a set of two or more equations with the same variables. A solution is any ordered pair \((x, y)\) that makes every equation in the system true at the same time.

x y 0 1 2 3 4 0 1 2 3 (2, 3) \(y = x + 1\) \(y = -0.5x + 4\)

Three Possible Outcomes

One Solution
Lines intersect at exactly one point. The system is consistent and independent.
No Solution
Lines are parallel: same slope, different intercepts. The system is inconsistent.
Infinitely Many
Lines are identical: same slope, same intercept. The system is consistent and dependent.

Why Do Systems Matter?

Whenever a real-world problem involves two unknowns and two constraints, you are solving a system. Examples include mixing solutions in chemistry, balancing budgets, or finding break-even points in business.

Break-Even Analysis
Revenue line meets cost line: the intersection is the break-even quantity.
Mixture Problems
Two ingredients, two conditions on concentration and volume.

Three Methods to Solve

1
Graphing: Plot both lines and read the intersection point. Best for visualizing and estimating.
2
Substitution: Solve one equation for a variable, then plug into the other. Best when one variable is already isolated.
3
Elimination: Add or subtract equations to cancel a variable. Best when coefficients are easy to match.

Method 1: Graphing

Rewrite each equation in slope-intercept form \(y = mx + b\), plot both lines, and identify the point where they cross.

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

Method 2: Substitution

Isolate one variable in one equation, then substitute that expression into the other equation to solve for the remaining variable.

1
Isolate a variable: pick the equation where a variable has coefficient \(1\) or \(-1\).
2
Substitute the expression into the other equation. You now have one equation in one unknown.
3
Solve for that unknown, then back-substitute to find the other variable.
4
Check your answer in both original equations.

Substitution: Worked Example

1
Equation 1 already gives \(y = 3x - 2\).
2
Substitute into Equation 2: \(2x + (3x - 2) = 8\).
3
Simplify: \(5x - 2 = 8 \Rightarrow 5x = 10 \Rightarrow\) \(x = 2\).
4
Back-substitute: \(y = 3(2) - 2 = 4\). Solution: \((2, 4)\).

Method 3: Elimination

Multiply one or both equations by constants so that adding (or subtracting) them eliminates one variable entirely.

1
Align both equations in standard form: \(ax + by = c\).
2
Multiply one or both equations so the coefficients of one variable are opposites.
3
Add the equations to eliminate that variable. Solve for the remaining one.
4
Back-substitute and check in both original equations.

Elimination: Worked Example

1
The \(y\)-coefficients are already opposites: \(+2y\) and \(-2y\).
2
Add the equations: \((3x + 2y) + (x - 2y) = 16 + (-4)\).
3
Result: \(4x = 12 \Rightarrow\) \(x = 3\).
4
Substitute: \(3 - 2y = -4 \Rightarrow -2y = -7 \Rightarrow\) \(y = 3.5\). Solution: \((3, 3.5)\).

Substitution vs. Elimination

Substitution
Best when one equation already has a variable isolated, like \(y = \ldots\) or \(x = \ldots\). Avoids extra multiplication steps.
Elimination
Best when both equations are in standard form \(ax + by = c\) and coefficients are easy to match or already opposites.

How to Choose Your Method

1
Read both equations. Are they in slope-intercept form, standard form, or a mix?
2
Check for an isolated variable. If \(y = \ldots\) or \(x = \ldots\) appears, lean toward substitution.
3
Check for matching or opposite coefficients. If you see them, lean toward elimination.
4
Solve, back-substitute, and always verify in both original equations.

Harder Example: Scaling First

1
Eliminate \(y\): multiply Eq 1 by 2 and Eq 2 by \(-3\).
2
\(4x + 6y = 24\) and \(-15x - 6y = -33\). Add: \(-11x = -9\).
3
\(x = \frac{9}{11}\). Substitute back: \(2\!\left(\frac{9}{11}\right) + 3y = 12\).
4
\(3y = 12 - \frac{18}{11} = \frac{114}{11}\), so \(y = \frac{38}{11}\). Solution: \(\left(\frac{9}{11},\, \frac{38}{11}\right)\).

Special Cases: No Solution and Infinite Solutions

No Solution
Elimination produces a false statement like \(0 = 5\). The lines are parallel.
Infinitely Many
Elimination produces a true identity like \(0 = 0\). The lines are the same line.
Parallel (No Solution) Same slope, different intercept Identical (Infinite Solutions) Same slope, same intercept

Common Mistakes

✕ Parentheses in Substitution
When substituting an expression, especially with a minus sign, use parentheses. E.g., for \(2x - y = 1\) with \(y = 3x - 2\), write \(2x - (3x - 2) = 1\), not \(2x - 3x - 2 = 1\).
✕ Sign Errors in Elimination
When multiplying an equation by a negative, every term changes sign: constants too. Double-check each term.
✕ Not Checking the Answer
Always plug your \((x, y)\) back into both original equations. Checking only one can hide errors.

Method Comparison

Graphing Substitution Elimination
Best When Estimating or visualizing A variable is isolated Coefficients match easily
Precision Approximate Exact Exact
Speed Slow (plotting) Medium Fast (often)
Drawback Hard to read non-integer solutions Fractions can get messy Requires careful sign work

Question 1 of 5

Solve the system using substitution:
\( y = 2x + 1 \)
\( 3x + y = 11 \)
What is the solution?

Question 2 of 5

Solve by elimination:
\( 4x + y = 10 \)
\( 4x - y = 6 \)
What is \( x \)?

Question 3 of 5

What does it mean if solving a system gives you \( 0 = 7 \)?

Question 4 of 5

Which method is most efficient for this system?
\( x = 5y - 3 \)
\( 2x + 7y = 20 \)

Question 5 of 5

Solve the system:
\( 2x - y = 1 \)
\( 4x - 2y = 2 \)
How many solutions does it have?