Algebra

How to Solve Simultaneous Equations: Substitution vs Elimination

By Math Solving Space · · 2 min read

On this page
  1. The example we will use
  2. Method 1: Substitution
  3. Method 2: Elimination
  4. Always check
  5. What it means on a graph
  6. Three equations, three unknowns
  7. Common mistakes
  8. Practice questions
  9. Frequently asked questions

Simultaneous equations are two (or more) equations that must be true at the same time. To solve a pair, either substitute one equation into the other, or eliminate a variable by adding or subtracting the equations. On a graph, the solution is where the two lines cross.

The example we will use

{2x+3y=13x−y=−1\begin{cases} 2x + 3y = 13 \\ x - y = -1 \end{cases}

Method 1: Substitution

  1. Rearrange one equation to get one variable on its own. From the second: x=y−1x = y - 1.
  2. Substitute into the other equation:
2(y−1)+3y=13⇒5y−2=13⇒y=32(y - 1) + 3y = 13 \Rightarrow 5y - 2 = 13 \Rightarrow y = 3
  1. Put y=3y = 3 back into x=y−1x = y - 1: x=2x = 2.

Substitution is best when one variable already has a coefficient of 1 or −1.

Method 2: Elimination

  1. Make the coefficients of one variable match. Multiply the second equation by 3:
3x−3y=−33x - 3y = -3
  1. Add it to the first equation so yy cancels:
(2x+3y)+(3x−3y)=13+(−3)⇒5x=10⇒x=2(2x + 3y) + (3x - 3y) = 13 + (-3) \Rightarrow 5x = 10 \Rightarrow x = 2
  1. Substitute back: 2−y=−12 - y = -1, so y=3y = 3.

Always check

Put x=2x = 2, y=3y = 3 into both original equations:

  • 2(2)+3(3)=4+9=132(2) + 3(3) = 4 + 9 = 13 ✓
  • 2−3=−12 - 3 = -1 ✓

What it means on a graph

Each linear equation is a straight line. The solution (2,3)(2, 3) is the point where the lines cross. This explains the special cases:

Lines Number of solutions
Cross once One solution
Parallel No solution
The same line Infinitely many

Example with no solution: x+2y=4x + 2y = 4 and 2x+4y=32x + 4y = 3. Doubling the first gives 2x+4y=82x + 4y = 8, which contradicts 2x+4y=32x + 4y = 3. The lines are parallel.

Three equations, three unknowns

The same idea extends to three equations in xx, yy and zz: eliminate one variable to get two equations in two unknowns, then solve those. For

{x+y+z=62x−y+z=3x+2y−z=2\begin{cases} x + y + z = 6 \\ 2x - y + z = 3 \\ x + 2y - z = 2 \end{cases}

the solution is x=1x = 1, y=2y = 2, z=3z = 3. With three or more equations, writing the coefficients as a matrix and row-reducing (Gauss–Jordan elimination) keeps things organised.

Common mistakes

  • Multiplying only one side of an equation when scaling it.
  • Losing a minus sign when subtracting equations.
  • Stopping after finding one variable.

Practice questions

  1. x+y=10x + y = 10 and x−y=4x - y = 4
  2. 3x+2y=163x + 2y = 16 and x+2y=8x + 2y = 8
  3. y=2x+1y = 2x + 1 and 3x+y=113x + y = 11

Answers: 1) x=7,y=3x = 7, y = 3 2) x=4,y=2x = 4, y = 2 3) x=2,y=5x = 2, y = 5

Check your answers, and see the lines cross, with the simultaneous equations solver. For single equations use the linear equation solver, and for the matrix method try the matrix calculator.

Frequently asked questions

Which method is quicker?

Substitution is quicker when a variable already stands alone or has coefficient 1. Elimination is usually quicker otherwise.

How do I know there is no solution?

If eliminating leads to a false statement like 0 = 5, the equations are inconsistent and there is no solution.

Can simultaneous equations be non-linear?

Yes, for example a line and a circle. Substitution is usually the best method then.

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