Systems of Linear Equations

Grade 10 · mathematics · 85 practice problems · read aloud

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Systems of Linear Equations

What Is It & Why Learn It?

A system of linear equations is two or more equations with the same variables. For example, two lines on a graph. We solve the system to find the point (x, y) where the lines intersect. This is useful for real-world problems like comparing phone plans or finding break-even points in business.

How to Solve: Step-by-Step

We'll focus on the Substitution Method and Elimination Method.

Example 1: Substitution Method

Solve: y = 2x - 1 and 3x + y = 9

  1. Step 1: The first equation is already solved for y: y = 2x - 1.
  2. Step 2: Substitute (2x - 1) for y in the second equation: 3x + (2x - 1) = 9.
  3. Step 3: Solve for x: 5x - 1 = 9 → 5x = 10 → x = 2.
  4. Step 4: Substitute x=2 back into y = 2x - 1: y = 2(2) - 1 → y = 3.
  5. Step 5: The solution is the point (2, 3). ✅

Example 2: Elimination Method

Solve: 2x + 3y = 12 and 5x - 3y = 9

  1. Step 1: Notice the y-terms: +3y and -3y. They already add to zero.
  2. Step 2: Add the equations vertically: (2x+3y) + (5x-3y) = 12+9 → 7x = 21.
  3. Step 3: Solve for x: x = 3.
  4. Step 4: Substitute x=3 into 2x+3y=12: 2(3)+3y=12 → 6+3y=12 → 3y=6 → y=2.
  5. Step 5: The solution is (3, 2). ✅

Common Mistakes to Avoid 🚫

  • Sign Errors: The most common mistake! When substituting or distributing, be careful with negative signs.
  • Not Checking Your Answer: Always plug your (x, y) solution back into both original equations to verify.
  • Inconsistent Systems: If all variables cancel and you get a false statement (like 0=5), it means no solution (the lines are parallel).

Tips & Tricks

  • Choosing a Method: Use Substitution if one variable is already isolated. Use Elimination if the coefficients of a variable are opposites or can easily be made opposites.
  • Check Your Point: The intersection point must satisfy both equations.
  • Graphical Insight: Sketching a quick graph can help you visualize the solution and catch errors.

How to Practice

Start with simple systems where one variable is already isolated. Then, move to problems that require one extra step to isolate a variable. Finally, practice with elimination problems where you need to multiply one or both equations by a number first. Always check your final answer!

Practice problems

6 of the 85, worked through step by step — try them before opening the answer.

1 2x + 3y = 12; 4x - y = 10

Hint: Consider using the elimination method by making the coefficients of one variable opposites, or use substitution by solving one equation for a variable and substituting into the other equation.

Show the answer

Answer: x = 3, y = 2

  1. Solve one equation for one variable** From equation (2): 4x - y = 10 Add y to both sides: 4x = 10 + y Subtract 10 from both sides: y = 4x - 10 --- **
  2. Substitute into the other equation** Substitute y = 4x - 10 into equation (1): 2x + 3(4x - 10) = 12 --- **
  3. Simplify and solve for x** 2x + 12x - 30 = 12 14x - 30 = 12 Add 30 to both sides: 14x = 42 Divide by 14: x = 42/14 = 3 --- **
  4. Solve for y** Substitute x = 3 into y = 4x - 10: y = 4(3) - 10 = 12 - 10 = 2 --- **
  5. Check the solution** In (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ In (2): 4(3) - 2 = 12 - 2 = 10 ✓ --- **Final Answer:** x = 3, y = 2

We are given the system of equations: (1) 2x + 3y = 12 (2) 4x - y = 10 --- **

2 2x + 3y = 7; 5x - 2y = 8; x = ?

Hint: Consider using elimination by making the coefficients of one variable opposites

Show the answer

Answer: 2

  1. Multiply the first equation by 2: 2(2x + 3y) = 2(7) → 4x + 6y = 14
  2. Multiply the second equation by 3: 3(5x - 2y) = 3(8) → 15x - 6y = 24
  3. Add the two equations: (4x + 6y) + (15x - 6y) = 14 + 24 → 19x = 38
  4. Solve for x: x = 38 ÷ 19 = 2

The answer is x = 2.

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

Hint: You can use substitution or elimination to solve this system. Try isolating one variable in one equation and substituting into the other equation.

Show the answer

Answer: 3

  1. Solve one equation for one variable** From equation (2): 4x - y = 10 Subtract 4x from both sides: -y = 10 - 4x Multiply both sides by -1: y = 4x - 10 --- **
  2. Substitute into the other equation** Substitute y = 4x - 10 into equation (1): 2x + 3(4x - 10) = 12 --- **
  3. Simplify and solve for x** 2x + 12x - 30 = 12 14x - 30 = 12 Add 30 to both sides: 14x = 42 Divide both sides by 14: x = 42/14 x = 3 --- **
  4. Check (optional but good practice)** If x = 3, then from y = 4x - 10: y = 4(3) - 10 = 12 - 10 = 2 Check equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✅ Check equation (2): 4(3) - 2 = 12 - 2 = 10 ✅ --- **Final answer:** x = 3

We are given the system of equations: 1) 2x + 3y = 12 2) 4x - y = 10 We want to solve for x. --- **

4 2x + 3y = 12; 3x - 2y = 5; x = ?

Hint: To solve a system of equations, you can use elimination by making coefficients of one variable opposites, or substitution by expressing one variable in terms of the other.

Show the answer

Answer: 3

  1. Choose a method** We can use elimination or substitution. Let's use elimination to eliminate y. --- **
  2. Make coefficients of y equal in magnitude** Multiply equation (1) by 2: 2 * (2x + 3y) = 2 * 12 4x + 6y = 24 ...(3) Multiply equation (2) by 3: 3 * (3x - 2y) = 3 * 5 9x - 6y = 15 ...(4) --- **
  3. Add equations (3) and (4) to eliminate y** (4x + 6y) + (9x - 6y) = 24 + 15 4x + 9x + 6y - 6y = 39 13x = 39 --- **
  4. Solve for x** x = 39 / 13 x = 3 --- **
  5. Conclusion** The value of x is 3. We can check by substituting x = 3 into equation (1): 2(3) + 3y = 12 → 6 + 3y = 12 → 3y = 6 → y = 2. Then check equation (2): 3(3) - 2(2) = 9 - 4 = 5, correct. **Final answer:** x = 3

We are given the system of equations: 1) 2x + 3y = 12 2) 3x - 2y = 5 We need to solve for x. --- **

5 2x + 3y = 16; 5x - 2y = 21; x = ?

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

Show the answer

Answer: 5

  1. Multiply the first equation by 2: 2(2x + 3y) = 2(16) → 4x + 6y = 32
  2. Multiply the second equation by 3: 3(5x - 2y) = 3(21) → 15x - 6y = 63
  3. Add the two new equations: (4x + 6y) + (15x - 6y) = 32 + 63 → 19x = 95
  4. Solve for x: x = 95 ÷ 19 = 5

The answer is 5.

6 3x + 2y = 16; 2x - 3y = -5; x = ?

Hint: Consider using the elimination method by making coefficients of one variable opposites

Show the answer

Answer: 2

  1. Multiply the first equation by 3: 3(3x + 2y) = 3(16) → 9x + 6y = 48
  2. Multiply the second equation by 2: 2(2x - 3y) = 2(-5) → 4x - 6y = -10
  3. Add the two equations: (9x + 6y) + (4x - 6y) = 48 + (-10) → 13x = 38
  4. Solve for x: x = 38/13 → x = 2

The answer is x = 2.

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