Systems of Equations

Algebra is arithmetic with a blank space in it. A letter such as x stands for a number you have not found yet, and solving means finding the number that makes the sentence true.

Systems of Equations: the central idea

A solution to a system must satisfy every equation at the same time. Geometrically, it is an intersection shared by all represented lines or curves.

Words you need

System of equations

A system of equations is a set of two or more equations that use the same variables and must all be true at once.

Substitution

Substitution is the method where you solve one equation for a single variable and then put that expression in place of the variable in the other equation.

Elimination

Elimination is the method where you add or subtract multiples of the equations so that one variable cancels out.

Consistent system

A consistent system is one that has at least one solution, so its lines either cross or sit on top of each other.

Inconsistent system

An inconsistent system has no solution because its lines are parallel and never meet.

Dependent system

A dependent system has infinitely many solutions because both equations describe the very same line.

What to know before this lesson

Review linear equations, ordered pairs, substitution, and addition of signed expressions.

If one of those prerequisites is uncertain, use the Algebra subject guide to locate the earlier concept before memorizing a procedure.

Systems of equations: a worked example

Every step, with the arithmetic

1. Step 1 - Write the system$3x+2y=19$ and $4x-2y=16$
2. Step 2 - Notice the $y$ terms are already opposites$+2y$ and $-2y$ add to $0$, so elimination needs no extra multiplying
3. Step 3 - Add the two equations column by column$(3x+4x)+(2y-2y)=19+16$
4. Step 4 - Solve the one-variable equation left behind$7x=35$, so $x=5$
5. Step 5 - Back-substitute into the first equation$3(5)+2y=19$, so $2y=4$ and $y=2$
6. Step 6 - Check the pair in the other equation$4(5)-2(2)=20-4=16$, which matches
7. Step 7 - State the answer as a point$(5,2)$, the single point where the two lines cross

Adding $x+y=7$ and $x-y=1$ eliminates $y$ and gives $2x=8$. With $x=4$, either equation gives $y=3$.

How to tell which kind of system you have

The step-by-step method for systems of equations

1. Choose substitution when one variable is already isolated; choose elimination when coefficients can cancel efficiently.
2. Perform the chosen operation while retaining both original equations and their labels.
3. Solve for one variable, back-substitute for the other, and classify inconsistent or dependent systems separately.

How to check your answer

Check $(4,3)$ in both equations: $4+3=7$ and $4-3=1$. One successful equation is not enough.

Substitute a proposed value into the original statement—not only the last simplified line. For functions, also inspect domain, intercepts, and whether the graph agrees.

A mistake that changes the mathematics

Elimination permits adding equal multiples of equations; it does not permit multiplying only one term in an equation.

Where you will actually use this

Choosing between two plans

Gym A charges 40 dollars a month plus 2 dollars a class; gym B charges 10 dollars plus 5 dollars a class. Setting $40+2c=10+5c$ gives $30=3c$, so at $c=10$ classes both cost 60 dollars. Fewer than ten classes favors gym B; more than ten favors gym A.

Selling two ticket types

A show sells 200 tickets for 1,800 dollars, with adult tickets at 12 dollars and child tickets at 7 dollars. From $a+k=200$ and $12a+7k=1800$, substituting $k=200-a$ gives $5a+1400=1800$, so $a=80$ adults and $k=120$ children.

Blending two products

Mixing 20 pounds of coffee from an 8-dollar bean and a 12-dollar bean to sell at 9 dollars a pound means $x+y=20$ and $8x+12y=180$. Substituting $y=20-x$ gives $-4x=-60$, so 15 pounds of the cheaper bean and 5 pounds of the pricier one.

How systems of equations connects to the rest of algebra

• Solving linear equations — Both methods squeeze the system down to one equation in one variable, and from that line onward it is ordinary linear solving.
• Solving linear systems with matrices — With three or more variables the same elimination is written as row operations on an augmented matrix, which keeps the bookkeeping straight.
• The quadratic formula — If one equation in the system is a parabola instead of a line, substitution produces a quadratic and the formula supplies the intersection points.

Try a transfer problem

Solve $2x+3y=13$ and $4x-y=5$ twice—once by substitution and once by elimination—and compare the intermediate work.

Show the worked answer

Substitution: the second equation gives $y=4x-5$. Putting that into the first gives $2x+3(4x-5)=13$, so $2x+12x-15=13$, then $14x=28$ and $x=2$. Then $y=4(2)-5=3$. Elimination: multiply the second equation by 3 to get $12x-3y=15$, and add it to $2x+3y=13$. The $y$ terms cancel and give $14x=28$, so $x=2$, and $4(2)-y=5$ gives $y=3$. Both routes give $(2,3)$, and both pass through the same line $14x=28$. Check in both originals: $2(2)+3(3)=4+9=13$ and $4(2)-3=5$. The difference is only bookkeeping. Substitution needed one distribution step because $y$ was already alone; elimination needed one multiplication to line the $y$ terms up as opposites.

Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to algebra practice questions.

Questions about systems of equations

How do I choose between substitution and elimination?

Use substitution when a variable is already alone or has a coefficient of 1, as in $4x-y=5$. Use elimination when the variable terms are already opposites, or when one small multiplication makes them opposites, as in $3x+2y=19$ paired with $4x-2y=16$.

What does it mean when both variables disappear?

The sentence left behind tells you the answer. If it is false, like $1=8$, the lines are parallel and the system has no solution. If it is true, like $0=0$, the two equations are the same line and every point on it is a solution.

Can a system have exactly two solutions?

Not with two straight lines, since lines cross at most once. But a system with a curve can. A line and a circle, or a line and a parabola, can meet at two points, and substitution then produces a quadratic with two roots.

Do I really have to check in both equations?

Yes. A wrong pair can easily satisfy one equation and fail the other. That is why the check for $(5,2)$ is done in the equation that was not used for back-substitution.