Solve: y = 2x + 1 and 3x + y = 11
Step 1 Substitute y into second equation
3x + (2x + 1) = 11
Step 2 Solve for x
5x + 1 = 11
5x = 10
x = 2
Step 3 Find y
y = 2(2) + 1 = 4 + 1 = 5
Step 4 Check
3(2) + 5 = 6 + 5 = 11 โ
This study notes covers Worked Example 2: Substitution within Simultaneous Equations for GCSE Mathematics. Revise Simultaneous Equations in Algebra for GCSE Mathematics with 15 exam-style questions and 12 flashcards. This is a high-frequency topic, so it is worth revising until the explanation feels precise and repeatable. It is section 6 of 8 in this topic. Use this study notes to connect the idea to the wider topic before moving on to questions and flashcards.
Practice questions for Simultaneous Equations
Which method is most efficient for solving the following simultaneous equations? 5x + 2y = 14 3x + 2y = 10
Ali solves the simultaneous equations 5x + y = 17 and x - y = 1 by elimination. Ben says he can also use substitution by rearranging x - y = 1 to get x = y + 1. Explain how elimination works for these two equations and state one advantage of Ben's substitution approach.