System of Equations Problems- Practice and Solutions

What Is a System of Equations?

A system of equations is two or more equations solved together. The solution is the point where all equations in the system are true at the same time.

In algebra, you're usually working with linear equations in two variables, written as:

ax + by = c

The goal is to find values for x and y that satisfy every equation simultaneously. If the lines intersect at one point, you have one solution. If they're parallel, no solution exists. If they're the same line, you have infinitely many solutions.

Three Methods to Solve Systems of Equations

You have three main tools. Each works, but some are faster depending on the problem.

Method 1: Substitution

Substitution works by expressing one variable in terms of the other, then plugging that expression into the second equation.

Step-by-Step

• Step 1: Solve one equation for one variable
• Step 2: Substitute that expression into the other equation
• Step 3: Solve for the remaining variable
• Step 4: Plug back to find the first variable

Example

Solve:

y = 2x + 3
x + y = 9

Equation 1 already has y isolated. Plug it into equation 2:

x + (2x + 3) = 9
3x + 3 = 9
3x = 6
x = 2

Now find y:

y = 2(2) + 3 = 7

Solution: (2, 7)

Method 2: Elimination

Elimination works by adding or subtracting equations to cancel one variable. If coefficients don't match, multiply one or both equations first.

Step-by-Step

• Step 1: Align equations in standard form (ax + by = c)
• Step 2: Multiply one or both equations so a variable has matching or opposite coefficients
• Step 3: Add or subtract equations to eliminate one variable
• Step 4: Solve for the remaining variable
• Step 5: Substitute back to find the first variable

Example

Solve:

2x + y = 10
x - y = 2

The y coefficients are already opposites (+y and -y). Add the equations:

2x + y + x - y = 10 + 2
3x = 12
x = 4

Plug into x - y = 2:

4 - y = 2
-y = -2
y = 2

Solution: (4, 2)

Method 3: Graphing

Graph both equations on the same coordinate plane. The intersection point is your solution.

This method is less precise unless the intersection falls exactly on grid lines. It's useful for understanding what systems represent visually.

Example

Graph y = x + 1 and y = -2x + 4.

The first line crosses the y-axis at (0,1) and rises 1 unit for every unit right. The second crosses at (0,4) and falls 2 units for every unit right.

They intersect at approximately (1, 2). Plugging in confirms: 2 = 1 + 1 ✓ and 2 = -2(1) + 4 ✓

Practice Problems with Solutions

Work through these. Cover the solutions, try them yourself, then check.

Problem 1

Solve using substitution:

3x + 2y = 12
x = 4

Solution: Plug x = 4 into the first equation: 3(4) + 2y = 12 → 12 + 2y = 12 → 2y = 0 → y = 0. Answer: (4, 0)

Problem 2

Solve using elimination:

4x + 3y = 25
2x - 3y = 5

Solution: Add equations: 4x + 3y + 2x - 3y = 25 + 5 → 6x = 30 → x = 5. Plug back: 2(5) - 3y = 5 → 10 - 3y = 5 → -3y = -5 → y = 5/3. Answer: (5, 5/3)

Problem 3

Solve:

x + y = 8
2x - y = 4

Solution: Add equations: x + y + 2x - y = 8 + 4 → 3x = 12 → x = 4. Then 4 + y = 8 → y = 4. Answer: (4, 4)

Word Problems: Setting Up the System

Most people struggle before they even solve anything. They can't set up the equations. Here's how to fix that.

Ticket Problem

"A theater sells 120 tickets total. Adult tickets cost $15, child tickets cost $8. Total revenue is $1,485. How many of each were sold?"

Define variables: Let a = adult tickets, c = child tickets.

Set up equations:

a + c = 120 (total tickets)
15a + 8c = 1485 (total revenue)

Solve by substitution or elimination. Using substitution: c = 120 - a. Plug into second equation: 15a + 8(120 - a) = 1485 → 15a + 960 - 8a = 1485 → 7a = 525 → a = 75. Then c = 45.

Mixture Problem

"How many liters of a 30% acid solution must be mixed with a 50% acid solution to get 20 liters of 40% acid?"

Let x = liters of 30% solution, y = liters of 50% solution.

x + y = 20 (total volume)
0.30x + 0.50y = 0.40(20) (acid content)

Solve: From first equation, y = 20 - x. Plug in: 0.30x + 0.50(20 - x) = 8 → 0.30x + 10 - 0.50x = 8 → -0.20x = -2 → x = 10. Then y = 10. You need 10 liters of each.

Common Mistakes to Avoid

• Forgetting to distribute when multiplying equations. Check every term.
• Adding unlike terms. You can only add coefficients of the same variable.
• Arithmetic errors. Most mistakes happen in the arithmetic, not the method.
• Not checking your answer. Plug both values into both original equations.
• Using the wrong method. If substitution looks messy, try elimination instead.

Getting Started: Choosing the Right Method

Look at your system. Answer these questions:

• Is one variable already isolated, or easy to isolate? → Use substitution
• Do coefficients match or cancel easily? → Use elimination
• Are the numbers small and friendly? → Either method works
• Do you need to see the answer visually? → Use graphing

With practice, you'll see which method fits before you even write anything down. The first problem below will test whether you're ready to move on.

Practice Set

Solve these systems. Show your work.

1. x + 2y = 10 and 3x - 2y = 6

2. y = 3x - 5 and 2x + y = 10

3. 5x + 2y = 16 and 3x - 4y = 8

4. Two numbers add to 30. One number is three times the other. Find both numbers.

5. A farmer has chickens and rabbits. Total heads = 50, total legs = 140. How many of each?

Solutions

1. Add equations: 4x = 16 → x = 4. Then 4 + 2y = 10 → y = 3. Answer: (4, 3)

2. Substitute: 2x + (3x - 5) = 10 → 5x = 15 → x = 3. Then y = 3(3) - 5 = 4. Answer: (3, 4)

3. Multiply first equation by 2: 10x + 4y = 32. Add to second: 10x + 4y + 3x - 4y = 32 + 8 → 13x = 40 → x = 40/13. Then 5(40/13) + 2y = 16 → 200/13 + 2y = 208/13 → 2y = 8/13 → y = 4/13. Answer: (40/13, 4/13)

4. Let x + y = 30 and x = 3y. Substitute: 3y + y = 30 → y = 7.5, x = 22.5. Answer: 22.5 and 7.5

5. Chickens (c) have 2 legs, rabbits (r) have 4 legs. c + r = 50, 2c + 4r = 140. Divide second equation by 2: c + 2r = 70. Subtract first from this: (c + 2r) - (c + r) = 70 - 50 → r = 20. Then c = 30. Answer: 30 chickens, 20 rabbits