Systems of Equations

Grade 8 · algebra · 100 practice problems · read aloud

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Systems of Equations: Finding Where Two Lines Meet

🔍 What is a System of Equations?

A system of equations is a set of two or more equations with the same variables. In 8th grade, we work with two linear equations. The solution to the system is the point (x, y) where the two lines intersect on a graph. This is useful for solving real-world problems where two conditions must be true at the same time, like finding a break-even point in business or comparing two phone plans.

📝 How to Solve: The Substitution Method

  1. Step 1: Solve one equation for one variable (e.g., solve for y).
  2. Step 2: Substitute that expression into the other equation.
  3. Step 3: Solve the new equation for the remaining variable (e.g., x).
  4. Step 4: Substitute back to find the other variable (e.g., y).
  5. Step 5: Check your solution by plugging (x, y) into both original equations.

✨ Worked Example

Solve the system:
y = 2x + 1
3x + y = 11

Step 1: The first equation is already solved for y: y = 2x + 1

Step 2: Substitute (2x + 1) for y in the second equation:
3x + (2x + 1) = 11

Step 3: Solve for x:
5x + 1 = 11
5x = 10
x = 2

Step 4: Substitute x=2 back into y = 2x + 1:
y = 2(2) + 1
y = 5

Solution: The lines intersect at (2, 5).

⚠️ Common Mistakes to Avoid

  • Substitution Errors: When substituting an expression like (2x+1), always put it in parentheses! Forgetting this leads to sign errors.
  • Solving for the Wrong Variable: Choose to solve for the variable that has a coefficient of 1 (like y in 2x + y = 5) to make it easier.
  • Forgetting to Check: Always plug your final answer into both original equations to make sure it works for both.

💡 Tips & Tricks

  • Look for the "Lonely Variable": If one variable doesn't have a coefficient (like y in y = 3x - 2), that's the easiest one to solve for first.
  • Graph to Check: Quickly sketch both lines. Your solution (x, y) should be where they cross!
  • Stay Organized: Label your steps. A small mistake early on can throw off your entire answer.

🎯 How to Practice

Start with systems where one equation is already solved for y (like the example). Then, move to problems where you need to do that first step yourself. Try creating your own systems by choosing two lines and finding their intersection point. Many online math websites and your textbook will have practice problems with answer keys in the back to check your work.

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 2x + 3y = 12, x - y = 1

Hint: You can use substitution by solving one equation for a variable and plugging into the other equation, or use elimination by making coefficients match and adding/subtracting equations.

Show the answer

Answer: x = 3, y = 2

  1. Solve one equation for one variable. From equation (2), x - y = 1, we can solve for x: x = y + 1
  2. Substitute this expression for x into equation (1). Replace x in equation (1) with (y + 1): 2(y + 1) + 3y = 12
  3. Simplify and solve for y. 2y + 2 + 3y = 12 5y + 2 = 12 Subtract 2 from both sides: 5y = 10 Divide both sides by 5: y = 2
  4. Substitute y = 2 back into x = y + 1. x = 2 + 1 x = 3
  5. Check the solution in both original equations. For equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ For equation (2): 3 - 2 = 1 ✓ Final answer: x = 3, y = 2

We are given the system of equations: (1) 2x + 3y = 12 (2) x - y = 1

2 2x + 3y = 12; 3x - y = 7

Hint: You can solve this system by eliminating one variable. Try multiplying one equation to make the coefficients of y match or become opposites.

Show the answer

Answer: x = 3, y = 2

  1. Solve one equation for one variable. From equation (2): 3x - y = 7 Add y to both sides: 3x = 7 + y Subtract 7 from both sides: y = 3x - 7
  2. Substitute into the other equation. Substitute y = 3x - 7 into equation (1): 2x + 3(3x - 7) = 12
  3. Simplify and solve for x. 2x + 9x - 21 = 12 11x - 21 = 12 Add 21 to both sides: 11x = 33 Divide by 11: x = 3
  4. Substitute back to find y. y = 3x - 7 = 3(3) - 7 = 9 - 7 = 2
  5. Check the solution. In equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ In equation (2): 3(3) - 2 = 9 - 2 = 7 ✓ Final answer: x = 3, y = 2

We have the system of equations: (1) 2x + 3y = 12 (2) 3x - y = 7

3 3x + 2y = 16; 2x - y = 1

Hint: Try using the elimination method by making the coefficients of one variable match in both equations

Show the answer

Answer: x = 2, y = 3

  1. Start with the system: 3x + 2y = 16 and 2x - y = 1
  2. Multiply the second equation by 2 to match the y-coefficient: 2(2x - y) = 2(1) → 4x - 2y = 2
  3. Add this to the first equation: (3x + 2y) + (4x - 2y) = 16 + 2 → 7x = 18
  4. Solve for x: x = 18/7 = 2
  5. Substitute x = 2 into the second equation: 2(2) - y = 1 → 4 - y = 1
  6. Solve for y: -y = 1 - 4 → -y = -3 → y = 3
  7. Check in first equation: 3(2) + 2(3) = 6 + 6 = 12 The solution is x = 2, y = 3.

4 3x + 2y = 16; 2x - y = 6

Hint: Consider using the elimination method by making the coefficients of one variable match in both equations.

Show the answer

Answer: x = 4, y = 2

  1. Start with the system: 3x + 2y = 16 and 2x - y = 6
  2. Multiply the second equation by 2 to eliminate y: 2(2x - y) = 2(6) → 4x - 2y = 12
  3. Add this to the first equation: (3x + 2y) + (4x - 2y) = 16 + 12 → 7x = 28
  4. Solve for x: x = 28 ÷ 7 = 4
  5. Substitute x = 4 into the second equation: 2(4) - y = 6 → 8 - y = 6
  6. Solve for y: -y = 6 - 8 → -y = -2 → y = 2
  7. Check in first equation: 3(4) + 2(2) = 12 + 4 = 16 ✓ The solution is x = 4, y = 2.

5 2x + 3y = 12; 3x - 2y = 5

Hint: You can use elimination by making the coefficients of one variable opposites, or substitution by solving one equation for a variable and plugging into the other.

Show the answer

Answer: x = 3, y = 2

  1. Choose a method** We can use elimination or substitution. Let's use elimination. We want the coefficients of either x or y to be the same in both equations so we can eliminate one variable. --- **
  2. Make coefficients of y match** Multiply Equation (1) by 2: 2*(2x + 3y) = 2*12 4x + 6y = 24 ...(1a) Multiply Equation (2) by 3: 3*(3x - 2y) = 3*5 9x - 6y = 15 ...(2a) --- **
  3. Add the two new equations** (1a) + (2a): (4x + 6y) + (9x - 6y) = 24 + 15 4x + 9x + 6y - 6y = 39 13x = 39 --- **
  4. Solve for x** x = 39 / 13 x = 3 --- **
  5. Substitute x = 3 into Equation (1)** 2*(3) + 3y = 12 6 + 3y = 12 3y = 12 - 6 3y = 6 y = 6 / 3 y = 2 --- **
  6. Check in Equation (2)** 3*(3) - 2*(2) = 9 - 4 = 5, which matches. --- **Final Answer:** x = 3, y = 2

We are solving the system of equations: Equation (1): 2x + 3y = 12 Equation (2): 3x - 2y = 5 --- **

6 5x + 2y = 24; 3x - 2y = 8

Hint: Notice that both equations have a 2y term but with opposite signs. Consider what happens when you add the two equations together.

Show the answer

Answer: x = 4, y = 2

  1. Add the two equations to eliminate y (5x + 2y) + (3x - 2y) = 24 + 8 5x + 2y + 3x - 2y = 32 8x = 32
  2. Solve for x x = 32 ÷ 8 x = 4
  3. Substitute x = 4 into the first equation 5(4) + 2y = 24 20 + 2y = 24
  4. Solve for y 2y = 24 - 20 2y = 4 y = 4 ÷ 2 y = 2
  5. Verify with the second equation 3(4) - 2(2) = 12 - 4 = 8 ✓ The solution is x = 4, y = 2.
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