Elimination Method for Solving Simultaneous Equations

A way of solving simultaneous equations, two equations with the same two unknowns, by adding or subtracting them so that one unknown cancels out. For 3x + y = 7 and x − y = 5, adding the equations removes y and leaves 4x = 12, so x = 3. American sites call the pair a system of equations; the method is the same.

Four steps. Line up the equations and, if needed, multiply one so a pair of terms match. Add or subtract to eliminate one unknown. Solve the equation that is left. Substitute back to find the other unknown, then check both in the original equations. For 3x + y = 7 and x − y = 5: adding gives 4x = 12, so x = 3, and then y = −2.

Use elimination and show every line: multiply an equation if the terms do not match, add or subtract, solve, substitute, check. Linear simultaneous equations appear on both Foundation and Higher papers, calculator and non-calculator; pairs with a quadratic are Higher only. Marks are given for the method, so write the working even when you can see the answer.

Add when the matching terms have opposite signs, so +y and −y cancel to zero. Subtract when they have the same sign, so +2x and +2x cancel. For 2x + 5y = −10 and 2x + y = 8, the x terms match, so subtract: 4y = −18, giving y = −4.5, and then 2x − 4.5 = 8 gives x = 6.25. Both routes give the same solution.

Transform one equation first: multiply every term on both sides by a number that makes a pair of coefficients match. In the video’s second example the y terms do not match, so the second equation is multiplied by 3; adding then eliminates y and gives x = 3, and substituting back gives y = 2. Multiply the whole equation, not just one side.

Substitute both values into the original equations, not the transformed ones, and confirm each one balances: for x = 3 and y = −2, 3 × 3 + (−2) = 7 is right. The two common slips are adding before the coefficients match, so nothing cancels, and skipping this check. Multiply first, then eliminate, then check.