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§ Algebra

Systems of Equations

§ Algebra

Systems of Equations

CCSS.8.EECCSS.HSA.REI3 min read

A system of equations consists of two or more equations that share the same variables and must be solved together to find values that satisfy all equations simultaneously. The solution represents the point where the equations intersect when graphed. In Year 9 of the UK National Curriculum, pupils learn to solve simultaneous linear equations using substitution and elimination methods.

§ 01

Why it matters

Systems of equations model countless real-world scenarios where multiple constraints exist simultaneously. A school canteen manager might use them to determine how many portions of chips (£1.50) and pies (£2.20) to prepare when the budget is £150 and they need 80 items total. Engineers use systems to calculate forces in bridge designs where multiple structural requirements must be met. In economics, they model supply and demand equilibrium points. The elimination and substitution techniques learned here form the foundation for solving more complex systems in A-level Further Maths and university courses. Linear programming, used in business optimisation, extends these principles to systems with dozens or hundreds of variables.

§ 02

How to solve systems of equations

Systems of Equations

  • Write both equations.
  • Use substitution or elimination to solve for one variable.
  • Substitute back to find the other.
  • Verify in both equations.

Example: x + y = 5, x − y = 1 → x = 3, y = 2.

§ 03

Worked examples

Beginner§ 01

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

Answer: x = 1, y = 1

  1. Label the equations → (1) x + y = 2 (2) x = 1 — Number each equation so we can refer to them.
  2. Solve equation (1) for y → y = 2 − 1x — Isolate y in the simpler equation to use substitution.
  3. Substitute into equation (2) → Substitute y into (2) and solve for x — Replace y in equation (2) with the expression from equation (1), then solve for x.
  4. Find x → x = 1 — Solving gives x = 1.
  5. Substitute x back to find y → In (1): 1·1 + 1·y = 2 → 1 + 1·y = 2 → 1·y = 1 → y = 1 — Plug x = 1 into equation (1) and solve for y.
  6. Write the solution → x = 1, y = 1 — The intersection point of the two lines.
  7. Verify in both equations → (1) 1·1 + 1·1 = 2 = 2 ✓ (2) 1·1 + 0·1 = 1 = 1 ✓ — Substitute the solution into both original equations to confirm.
Easy§ 02

Two siblings have a combined age of 2. 3 times the older sibling's age minus the younger's age is 2. How old is each?

Answer: older = 1, younger = 1

  1. Define variables → Let x = older sibling's age, y = younger sibling's age x + y = 2 3x − y = 2 — Translate ages into a system of equations.
  2. Label the equations → (1) x + y = 2 (2) 3x − 1y = 2 — Number each equation so we can refer to them.
  3. Solve equation (1) for y → y = 2 − 1x — Isolate y in the simpler equation to use substitution.
  4. Substitute into equation (2) → Substitute y into (2) and solve for x — Replace y in equation (2) with the expression from equation (1), then solve for x.
  5. Find x → x = 1 — Solving gives x = 1.
  6. Substitute x back to find y → In (1): 1·1 + 1·y = 2 → 1 + 1·y = 2 → 1·y = 1 → y = 1 — Plug x = 1 into equation (1) and solve for y.
  7. Write the solution → x = 1, y = 1 — The intersection point of the two lines.
  8. Verify in both equations → (1) 1·1 + 1·1 = 2 = 2 ✓ (2) 3·1 + -1·1 = 2 = 2 ✓ — Substitute the solution into both original equations to confirm.
Medium§ 03

Solve the system: 4x + 2y = -12 3x − 2y = -2

Answer: x = -2, y = -2

  1. Label the equations → (1) 4x + 2y = -12 (2) 3x − 2y = -2 — Number each equation so we can refer to them.
  2. Choose a variable to eliminate → Multiply equations to align coefficients of y — We'll eliminate y by making its coefficients equal in magnitude, then subtract.
  3. Subtract to eliminate y → Solve the resulting equation for x — After eliminating y, we get a simple equation in x only.
  4. Find x → x = -2 — Solving gives x = -2.
  5. Substitute x back to find y → In (1): 4·-2 + 2·y = -12 → -8 + 2·y = -12 → 2·y = -4 → y = -2 — Plug x = -2 into equation (1) and solve for y.
  6. Write the solution → x = -2, y = -2 — The intersection point of the two lines.
  7. Verify in both equations → (1) 4·-2 + 2·-2 = -12 = -12 ✓ (2) 3·-2 + -2·-2 = -2 = -2 ✓ — Substitute the solution into both original equations to confirm.
§ 04

Common mistakes

  • When solving x + y = 7 and 2x - y = 5, writing the solution as x = 4, y = 2 without verification, which would give 4 + 2 = 6 ≠ 7 in the first equation.
  • In elimination, adding equations x + 2y = 8 and 3x - 2y = 4 to get 4x = 12, then stating y = 12 instead of x = 3.
  • When substituting y = 5 - x into 2x + 3y = 17, expanding to get 2x + 15 - x = 17, then solving as x = 2 instead of x + 15 = 17, giving x = 2.
§ 05

Frequently asked questions

What is the difference between substitution and elimination methods?
Substitution involves solving one equation for a variable, then replacing that variable in the other equation. Elimination involves adding or subtracting equations to cancel out one variable. Substitution works well when one equation is already solved for a variable, whilst elimination is efficient when coefficients align easily.
How do you know if a system of equations has no solution?
A system has no solution when the equations represent parallel lines that never intersect. This occurs when simplifying leads to a contradiction like 0 = 5. Graphically, the lines have the same slope but different y-intercepts, such as y = 2x + 1 and y = 2x + 3.
Can a system of equations have more than one solution?
Yes, when both equations represent the same line, there are infinitely many solutions. This happens when one equation is a multiple of the other, such as x + y = 4 and 2x + 2y = 8. Every point on the line satisfies both equations simultaneously.
Why do we verify solutions in both original equations?
Verification catches arithmetic errors that can occur during substitution or elimination. Even if the algebraic steps appear correct, a computational mistake might produce values that satisfy the derived equation but not the originals. Checking both equations ensures the solution is mathematically sound.
When should I use elimination instead of substitution?
Use elimination when coefficients of one variable are already opposites (like 3x and -3x) or can easily be made equal through multiplication. Choose substitution when one equation is already solved for a variable or can be rearranged simply, such as x = 5 or y = 2x + 1.
§ 06

See also

MultipleVariableEquation
§ 06

Where to next?

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