Solve Systems Exactly

Grade 9 · algebra · 101 practice problems · read aloud

🔊 Listen to this explanation

Solving Systems of Equations Exactly

What is it and Why is it Useful? 🤔

A system of equations is two or more equations with the same variables. "Solving exactly" means finding the precise point (x, y) where the lines cross. This is crucial for real-world problems like calculating when two businesses will make the same profit or finding the best phone plan.

How to Solve: The Substitution Method

  1. Isolate a Variable: Solve one equation for one variable.
  2. Substitute: Plug that expression into the other equation.
  3. Solve: Solve the new one-variable equation.
  4. Back-Substitute: Plug the solution back into one of the original equations to find the other variable.
  5. Check: Verify your solution works in both original equations.

Example 1: Simple Substitution

Solve: y = 2x + 1 and 3x + y = 11

  1. Equation 1 is already solved for y: y = 2x + 1
  2. Substitute into Equation 2: 3x + (2x + 1) = 11
  3. Solve for x: 5x + 1 = 11 → 5x = 10 → x = 2
  4. Back-substitute into y = 2x + 1: y = 2(2) + 1 → y = 5
  5. Check: 5 = 2(2)+1 ✔ and 3(2)+5=11 ✔

Solution: (2, 5)

Example 2: Isolate First

Solve: 2x + y = 7 and x - y = 5

  1. Isolate y in the first equation: y = 7 - 2x
  2. Substitute into Equation 2: x - (7 - 2x) = 5
  3. Solve for x: x - 7 + 2x = 5 → 3x = 12 → x = 4
  4. Back-substitute: y = 7 - 2(4) → y = -1
  5. Check: 2(4)+(-1)=7 ✔ and 4-(-1)=5 ✔

Solution: (4, -1)

Common Mistakes to Avoid ⚠️

Distributing Negatives Incorrectly: When substituting an expression like (7 - 2x), remember the negative sign applies to both terms. x - (7 - 2x) = x - 7 + 2x.

Not Checking Your Answer: Always plug your final (x, y) back into both original equations. It's the best way to catch calculation errors.

Solving for the Wrong Variable: Choose to isolate the variable that's easiest (like a variable with a coefficient of 1).

Tips & Tricks

Look for the Easiest Path: If one equation is already solved for y or x, use that one for substitution immediately.

Stay Organized: Use parentheses when substituting to avoid sign errors. Write each step clearly.

Visualize: Remember, you are finding the intersection point of two lines. Your answer should be an (x, y) ordered pair.

How to Practice

  • Start with systems where one variable is already isolated.
  • Move on to systems where you need to do one simple step to isolate a variable.
  • Create your own systems and solve them, then check with a graphing calculator or app to see the intersection point.
  • Practice with a partner—one person solves, the other checks.

Practice problems

6 of the 101, worked through step by step — try them before opening the answer.

1 2x² - 3x - 5 = 0

Hint: For equations of the form ax² + bx + c = 0, you can use a specific formula that involves the coefficients a, b, and c along with square roots.

Show the answer

Answer: x = 2.5, x = -1

  1. Identify coefficients** The standard form of a quadratic is: a x² + b x + c = 0 Here: a = 2 b = -3 c = -5 --- **
  2. Use the quadratic formula** The quadratic formula is: x = [ -b ± sqrt(b² - 4ac) ] / (2a) Substitute a, b, c: x = [ -(-3) ± sqrt( (-3)² - 4 * 2 * (-5) ) ] / (2 * 2) --- **
  3. Simplify inside the square root** First, -(-3) = 3 Next, (-3)² = 9 Then, 4 * a * c = 4 * 2 * (-5) = -40 So b² - 4ac = 9 - (-40) = 9 + 40 = 49 --- **
  4. Take the square root** sqrt(49) = 7 So: x = [ 3 ± 7 ] / 4 --- **
  5. Solve for the two values** First solution: x = (3 + 7) / 4 = 10 / 4 = 5/2 = 2.5 Second solution: x = (3 - 7) / 4 = (-4) / 4 = -1 --- **Final Answer:** x = 2.5, x = -1

We are solving the quadratic equation: 2x² - 3x - 5 = 0 --- **

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

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

Show the answer

Answer: x = 3, y = 2

  1. Multiply the first equation by 2: 2(2x + 3y) = 2(12) → 4x + 6y = 24
  2. Multiply the second equation by 3: 3(3x - 2y) = 3(5) → 9x - 6y = 15
  3. Add the two equations: (4x + 6y) + (9x - 6y) = 24 + 15 → 13x = 39
  4. Solve for x: x = 39 ÷ 13 = 3
  5. Substitute x = 3 into the first equation: 2(3) + 3y = 12 → 6 + 3y = 12
  6. Solve for y: 3y = 6 → y = 2
  7. Verify with second equation: 3(3) - 2(2) = 9 - 4 = 5 ✓ The solution is x = 3, y = 2.

3 x² + y² = 41, y = x + 1. Solve system.

Hint: Substitute the linear expression for y into the circle equation, then solve the resulting quadratic for x. Remember to find both y-values by plugging each x back into the linear equation.

Show the answer

Answer: x = 4, y = 5 and x = -5, y = -4

  1. Substitute y = x + 1 into x² + y² = 41.
  2. x² + (x + 1)² = 41
  3. Expand: x² + (x² + 2x + 1) = 41
  4. Combine like terms: 2x² + 2x + 1 = 41
  5. Subtract 41 from both sides: 2x² + 2x - 40 = 0
  6. Divide by 2: x² + x - 20 = 0
  7. Factor: (x + 5)(x - 4) = 0
  8. Solve: x = -5 or x = 4
  9. For x = -5, y = -5 + 1 = -4. For x = 4, y = 4 + 1 = 5.
  10. Verify: (-5)² + (-4)² = 25 + 16 = 41 ✓ and 4² + 5² = 16 + 25 = 41 ✓ The solutions are x = 4, y = 5 and x = -5, y = -4.

4 x² + y² = 61, y = x + 1. Solve system.

Hint: Substitute the linear expression for y into the quadratic equation. This will give you a quadratic in x. Solve it by factoring or using the quadratic formula, then find the corresponding y values.

Show the answer

Answer: x = -6, y = -5 and x = 5, y = 6

  1. Substitute y = x + 1 into x² + y² = 61.
  2. x² + (x + 1)² = 61
  3. Expand (x + 1)² = x² + 2x + 1
  4. x² + x² + 2x + 1 = 61 → 2x² + 2x + 1 = 61
  5. Subtract 61 from both sides: 2x² + 2x - 60 = 0
  6. Divide by 2: x² + x - 30 = 0
  7. Factor: (x + 6)(x - 5) = 0
  8. So x = -6 or x = 5
  9. For x = -6, y = -6 + 1 = -5
  10. For x = 5, y = 5 + 1 = 6
  11. Verify: For (-6, -5): (-6)² + (-5)² = 36 + 25 = 61 ✓; For (5, 6): 5² + 6² = 25 + 36 = 61 ✓ The solutions are x = -6, y = -5 and x = 5, y = 6.

5 x² + y² = 85, y = 2x - 5. Solve system.

Hint: Substitute the linear expression for y into the circle equation, then solve the resulting quadratic for x. Use each x-value to find the corresponding y.

Show the answer

Answer: x = 6, y = 7 and x = -4, y = -13

  1. Substitute y = 2x - 5 into x² + y² = 85: x² + (2x - 5)² = 85.
  2. Expand (2x - 5)² = 4x² - 20x + 25.
  3. Combine: x² + 4x² - 20x + 25 = 85 → 5x² - 20x + 25 = 85.
  4. Subtract 85: 5x² - 20x - 60 = 0.
  5. Divide by 5: x² - 4x - 12 = 0.
  6. Factor: (x - 6)(x + 4) = 0 → x = 6 or x = -4.
  7. For x = 6, y = 2(6) - 5 = 12 - 5 = 7.
  8. For x = -4, y = 2(-4) - 5 = -8 - 5 = -13.
  9. Verify: (6)² + (7)² = 36 + 49 = 85 ✓; (-4)² + (-13)² = 16 + 169 = 85 ✓. The solutions are (6, 7) and (-4, -13).

6 x² + y² = 52, y = 2x + 2. Solve system.

Hint: Use substitution: replace y in the first equation with the expression from the second equation, then solve the resulting quadratic for x.

Show the answer

Answer: x = 2, y = 6 and x = -22/5, y = -34/5

  1. Substitute y = 2x + 2 into x² + y² = 52.
  2. x² + (2x + 2)² = 52 → x² + (4x² + 8x + 4) = 52 → 5x² + 8x + 4 = 52.
  3. Subtract 52 from both sides: 5x² + 8x - 48 = 0.
  4. Solve the quadratic using the quadratic formula: x = [-8 ± sqrt(64 - 4(5)(-48))] / (2*5) = [-8 ± sqrt(64 + 960)] / 10 = [-8 ± sqrt(1024)] / 10 = [-8 ± 32] / 10.
  5. Two solutions: x = (-8 + 32)/10 = 24/10 = 12/5 = 2.4, and x = (-8 - 32)/10 = -40/10 = -4.
  6. For x = 12/5, y = 2(12/5) + 2 = 24/5 + 10/5 = 34/5 = 6.8.
  7. For x = -4, y = 2(-4) + 2 = -8 + 2 = -6.
  8. Verify: For (12/5, 34/5): (144/25) + (1156/25) = 1300/25 = 52 ✓. For (-4, -6): 16 + 36 = 52 ✓. The solutions are x = 12/5, y = 34/5 and x = -4, y = -6.
Practise this topic — 10 free problems, no signup →