Learning Objectives
Choose elimination or substitution to solve a pair of equations.
Interpret one, no or infinitely many common solutions.
Solve a factorable linear–quadratic system and retain both coordinate pairs.
Check parameter cases and contextual constraints.
Prerequisites: Section 1B Lessons 01–03, including factorisation and linear parameter cases. A point (x, y) records its horizontal and vertical coordinates. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
Simultaneous equations describe conditions that must hold at the same time. With two variables, the answer is usually an ordered pair, not a single number. This lesson explains how two equations work together to determine, restrict or rule out that pair.
We begin with elimination and substitution for linear equations, then connect the algebra to intersections of graphs. Next we allow one equation to be quadratic, where two pairs may be possible. Finally, we examine parameters and word problems, where the number or meaning of solutions needs a further check.
1. Solving Two Conditions at Once
One equation in two variables generally leaves several possible pairs. A second equation narrows those possibilities. The aim is to combine the information while retaining the requirement that every final pair satisfies both original conditions.
1.1 A common solution: one pair satisfying both equations
A simultaneous solution is a pair of values that makes both equations true at once. Solving either equation alone usually leaves many possibilities. Substitution replaces one variable using a relation from the other equation; elimination combines equations to remove a variable.
1.2 Elimination and substitution: choosing a useful route
If the coefficients of one variable already match or are opposites, elimination often keeps the arithmetic short. If one equation already gives y in terms of x, substitution is usually convenient. Neither method changes the need to check both original equations.
Elimination adds or subtracts suitable multiples of the equations so one variable cancels. Substitution instead replaces a variable by an equal expression obtained from the other equation. Both methods reduce the problem to one variable before recovering the other.
Worked Example: Eliminate a Variable
Question: Solve 2x + 3y = 13 and 3x − 2y = 0.
Method: Multiply the equations so the y coefficients are opposites, add to find x, and then recover y.
Multiply the first equation by 2 and the second by 3: 4x + 6y = 26 and 9x − 6y = 0. Add them to obtain 13x = 26, so x = 2.
Substitute into 3x − 2y = 0: 6 − 2y = 0, hence y = 3. Check: 4 + 9 = 13 and 6 − 6 = 0. The solution is (2, 3).
Worked Example: Substitution
Question: Solve y = 2x + 1 and x + y = 10.
Method: Use the equation already giving y to replace it in the second equation.
Substitute 2x + 1 for y in the second equation: x + 2x + 1 = 10. Thus x = 3 and y = 7.
Both original equations hold: 7 = 2(3) + 1 and 3 + 7 = 10.
2. Graph Intersections and the Number of Solutions
The algebra has a geometric interpretation: the graphs show all pairs satisfying each separate equation, and their shared points satisfy both. This viewpoint explains the one-solution, no-solution and infinitely-many-solution cases.
2.1 Intersecting, parallel and coincident lines
A graph of an equation consists of the points satisfying it. Two genuine straight-line equations can represent intersecting lines, distinct parallel lines or the same line. Their common points correspond respectively to one solution, no solutions or infinitely many solutions.
Coincident lines are the same geometric line described by equivalent equations. They have every point on that line in common, whereas distinct parallel lines have no common point.
2.2 Constant results after elimination: contradiction or repetition
In elimination, a false constant statement such as 0 = 1 means no common pair exists. An identity such as 0 = 0 means that elimination has removed a repeated condition; keep the remaining equation to describe the solutions. It does not mean every point in the plane works.
2.3 Reading the schematic diagrams
The sketches below are qualitative. They show the possible relationships, not numerical solutions to a particular system. Exact coordinates should come from the algebra.
Figure 1. Common points of two lines give one, no or infinitely many solutions. These qualitative sketches have no numerical scale; the equations determine exact coordinates.
Worked Example: Repeated or Inconsistent
Question: Compare x + 2y = 3 with 2x + 4y = 6, then replace the second right side by 7.
Method: Compare the entire equations, including their constants, to distinguish repeated information from inconsistent information.
In the first pair, the second equation is twice the first. All pairs with x = 3 − 2y work, so there are infinitely many solutions along one line.
With right side 7 instead, subtracting twice the first equation gives 0 = 1. No pair can satisfy both equations. The corresponding lines are distinct and parallel.
3. Linear–Quadratic Systems: Keeping Every Coordinate Pair
A straight line and a quadratic curve can meet more than once. The substitution method still works, but the equation left in one variable may now have two roots. Each root must be converted into its own complete coordinate pair.
3.1 Factorisation and the zero-product rule
Substitution can leave a quadratic equation. When that quadratic factorises, use the zero-product rule: a product of real numbers is zero exactly when at least one factor is zero. Solve each factor equation and then find the matching value of the other variable.
3.2 Recovering y separately for each x
Do not mix coordinates from different solutions. If the x values are a and b, the values of y must be found separately from the same original relation. General quadratic techniques and repeated-root conditions are developed in Section 1C.
Think of the substitution as two branches of the calculation. For each root, return to the same relation for y, form the ordered pair, and test it in both equations. This prevents an x from one branch being paired with a y from another.
Worked Example: Two Intersections
Question: Solve y = x² and y = x + 2.
Method: Equate the two expressions for y, factor the resulting quadratic, and recover y for each root.
Equal values of y give x² = x + 2, so x² − x − 2 = (x − 2)(x + 1) = 0. Therefore x = 2 or x = −1.
Using y = x + 2 gives the pairs (2, 4) and (−1, 1). Check in y = x²: 4 = 2² and 1 = (−1)². Both pairs are solutions.
4. Parameter Cases and Solutions in Context
The earlier methods apply to a family of equations as well as one fixed system. They also apply to models of quantities such as ticket counts. In both settings, the elimination result needs to be interpreted using the original assumptions.
4.1 Zero parameter coefficients and contextual restrictions
If a coefficient contains a parameter, elimination may produce a factor that is sometimes zero. Separate that case before dividing. For a word problem, also check whether the solution is nonnegative, integral, and consistent with the defined units.
4.2 Two different checks after the algebra
For a parameter, ask whether a step divided by a quantity that could vanish and return to that special case. For a word problem, ask whether the pair represents permitted quantities, such as nonnegative whole-number ticket counts. These checks serve different purposes even though both happen after elimination.
Worked Example: A Parameter Changes the Number of Solutions
Question: Classify x + y = 3 and 2x + ky = 6 for real k.
Method: Eliminate x, then split the case k = 2 before dividing by the remaining coefficient.
Subtract twice the first equation from the second: (k − 2)y = 0. For k ≠ 2, y = 0 and x = 3, giving one solution.
For k = 2, the second equation repeats the first, so every pair on x + y = 3 works. In both cases (3, 0) works, but it is not the only solution when k = 2.
Worked Example: Two Ticket Prices
Question: There are 10 tickets costing £8 or £5 each, and the total is £62. Find each count.
Method: Define each ticket count, write one equation for the total count and one for the total price, and eliminate a variable.
Let a count £8 tickets and b count £5 tickets. Then a + b = 10 and 8a + 5b = 62. Subtract five times the first equation: 3a = 12, so a = 4 and b = 6.
Both counts are nonnegative integers, and 4 × £8 + 6 × £5 = £62.
Common Mistakes
A common solution must satisfy both equations, not merely the equation left after an erroneous elimination.
Multiply every term of an equation when matching coefficients.
If elimination gives 0 = 0, keep the remaining relation; do not claim that every pair works.
For a quadratic substitution, retain every allowed root and pair each with its own y value.
5. Practice
Solve the first two systems by a method you can justify. Then classify repeated or inconsistent equations, retain both pairs in a quadratic system, and check a parameter case and a counting model. Substitute every reported pair into both original equations.
1. Solve 3x + 2y = 12 and x − y = 1.
2. Solve y = 3x − 2 and 2x + y = 8.
3. Classify x − y = 2 and 2x − 2y = 5.
4. Describe every solution of 2x + y = 4 and 6x + 3y = 12.
5. Solve y = x² and y = 3x − 2.
6. For which k does x + y = 4 and 2x + ky = 9 have no solution? Give the solution for other k.
7. A bag contains 12 coins, each worth 20p or 50p, totalling £3.60. Find the counts.
Worked Solutions
1. Use substitution and keep exact fractions
From x − y = 1, x = y + 1. Then 3(y + 1) + 2y = 12 gives y = 9/5 and x = 14/5. Check: 42/5 + 18/5 = 12 and 14/5 − 9/5 = 1.
2. Substitute the given expression for y
Substitute to get 2x + 3x − 2 = 8, so x = 2 and y = 4. Both original equations hold: 4 = 6 − 2 and 4 + 4 = 8.
3. Recognise inconsistent equations
Subtract twice the first equation from the second to get 0 = 1. There are no solutions.
4. Describe the remaining line of solutions
The second equation is three times the first. All pairs (t, 4 − 2t), for real t, work. Substitution gives 2t + 4 − 2t = 4 and 6t + 3(4 − 2t) = 12.
5. Keep both roots and their paired outputs
x² − 3x + 2 = (x − 1)(x − 2) = 0. Thus the pairs are (1, 1) and (2, 4). Both satisfy y = x² and y = 3x − 2.
6. Separate the exceptional parameter value
Elimination gives (k − 2)y = 1. At k = 2 this is impossible, so there is no solution. Otherwise y = 1/(k − 2) and x = 4 − 1/(k − 2). The first equation is automatic; the second gives 8 + (k − 2)/(k − 2) = 9.
7. Model counts and values in consistent units
Use pence. Let a count 20p coins and b count 50p coins: a + b = 12 and 20a + 50b = 360. Subtract 20 times the first to get 30b = 120. Thus b = 4 and a = 8. Check: 160p + 200p = 360p.
6. Summary
- Choose elimination or substitution to simplify the arithmetic.
- Interpret the result as common points, retaining a relation when there are infinitely many.
- Keep both roots and their correct coordinate partners in a factorable quadratic system.
- Check parameter cases and the physical meaning of a solution.