Secondary Mathematics: Simultaneous Equations, Elimination and Modelling

Secondary Mathematics · Worked Repair Guide 17

Two equations are useful only when they are allowed to speak about the same quantities. A value that fits one condition but contradicts the other is not a simultaneous solution. The task is to find the complete pair that makes both original statements true, then decide whether that pair also makes sense in the question’s context.

This guide is a worked problem-solving companion. It moves from checking ordered pairs to choosing an efficient method, handling fractions and signs, building models and recognising inconsistent information. The examples are deliberately varied: the same algebra appears as purchases, mixtures, rectangles, coordinate intersections and a final planning problem. The aim is to make the method survive a change of wording.

For the shorter explanation of why the topic becomes difficult, read Why Are Simultaneous Equations Difficult to Understand? This page supplies the extended worked practice rather than replacing that diagnostic article. All situations and numerical data below are invented for teaching; they are not official examination questions or actual prices.

Use the guide: start with the entry check, study the substitution and elimination methods, continue to modelling, then complete the independent practice before reading its answers.

1. Check the meaning before checking the method

Consider x + y = 11 and 2x − y = 4. Test the pairs (5, 6), (4, 7) and (6, 5). Every pair satisfies the first equation. Only (5, 6) satisfies the second: 2(5) − 6 = 4. The first coordinate must consistently represent x and the second y.

This is an excellent opening check because it requires no elimination. A learner who cannot explain why (4, 7) fails is not yet reading the two equations as simultaneous conditions. More procedural practice would hide that gap rather than repair it. A learner who explains the conditions but makes a negative-sign error needs a different intervention.

Now solve x + y = 9 and x − y = 3 without looking ahead. The answer is x = 6, y = 3. Finally, decide whether x + y = 9 and 2x + 2y = 20 can both hold. They cannot: doubling the first equation would give 18, not 20. These checks separate understanding a pair, finding a pair and recognising that no pair exists.

2. One linear equation usually leaves a choice

The equation 2x + y = 12 can be rewritten y = 12 − 2x. Choosing x determines a corresponding y. For example, x = 0 gives y = 12, x = 3 gives y = 6, and x = 8 gives y = −4. Without additional restrictions, there are infinitely many real pairs.

A graph represents that collection as a line. A second linear equation represents another collection. Their shared points satisfy both. Two distinct non-parallel lines meet once; distinct parallel lines never meet; coincident lines repeat the same condition. This standard interpretation is also explained in OpenStax’s treatment of two-variable linear systems.

Be precise about the domain. If x and y count boxes, negative and fractional values are not permitted. A line may contain infinitely many real points but only a limited collection of feasible whole-number pairs. The symbolic system and the real situation must both be checked; neither replaces the other.

3. Substitution replaces a quantity with an equal expression

Solve y = 3x − 4 together with 2x + y = 16. The first equation gives an expression equal to y. Replace y in the second equation by that whole expression:

2x + (3x − 4) = 16
5x − 4 = 16
5x = 20
x = 4.

Return to y = 3x − 4 to obtain y = 8. Check both originals: 8 = 3(4) − 4 and 2(4) + 8 = 16. The pair is (4, 8), not simply the number 4. Solving for the first variable is an intermediate result.

The brackets record that the complete expression replaces y. Although they are not essential after a plus sign, writing them prepares the learner for cases with multiplication or subtraction. Substitution is particularly economical when a variable already stands alone, or when isolating it does not create an awkward fraction.

4. A negative multiplier must reach the whole substitution

Now solve y = 5 − 2x and 3x − 4y = 13. Substitution produces 3x − 4(5 − 2x) = 13. Expanding gives 3x − 20 + 8x = 13, hence 11x = 33 and x = 3. Then y = −1.

Check the original second equation carefully: 3(3) − 4(−1) = 9 + 4 = 13. The two negatives in the product are doing a different job from the subtraction between terms. An answer y = 1 would fail the first equation as well.

The faulty line 3x − 20 − 2x = 13 reveals a distribution error, not a failure to understand simultaneous equations. Repair that specific line by writing the two products, (−4)(5) and (−4)(−2x), separately. Then try a changed example before returning to the whole system.

5. Elimination creates a useful equation without discarding the system

Solve 4x + 3y = 31 and 2x − 3y = 5. Adding the equations cancels the opposite y-terms, giving 6x = 36 and x = 6. Substitute into the second original equation: 12 − 3y = 5, so y = 7/3.

The fractional answer is not evidence of a mistake. The checks are 24 + 7 = 31 and 12 − 7 = 5. Unless the question requires whole-number quantities, an exact fraction is an entirely valid coordinate.

There is an important logical detail: the summed equation alone does not determine y. Elimination is used together with an original equation. Replacing the second equation by the sum while retaining the first preserves the full information, because subtracting the first recovers the original second equation. Keeping only 6x = 36 and discarding both originals would lose a condition.

6. Scale entire equations to create matching coefficients

Consider 3x + 2y = 23 and 5x − 3y = 13. Neither variable cancels immediately. Multiply the first equation by 3 and the second by 2:

9x + 6y = 69
10x − 6y = 26.

Add to obtain 19x = 95, so x = 5. Substitution gives 15 + 2y = 23, so y = 4. Check the other original: 25 − 12 = 13. All three terms of each equation, including the constant, were multiplied.

The choice of 6 as the matching y-coefficient comes from the least common multiple of 2 and 3. Multiplying by larger factors would also work, but creates unnecessary arithmetic. Equally, eliminating x is valid: multiply by 5 and 3 and subtract. Method choice should reduce the likelihood of an error, not merely imitate a preferred classroom layout.

7. Subtraction acts on every term of the second equation

For 5x − 2y = 21 and 3x − 2y = 11, subtract the second equation from the first. Write the grouped left side as (5x − 2y) − (3x − 2y). It becomes 2x, because −2y − (−2y) = 0. The right side is 10, so x = 5 and y = 2.

A frequent faulty line is 2x − 4y = 10. That line subtracts 3x but adds the second negative y-term instead of subtracting it. The cure is not a new elimination rule; it is disciplined subtraction of an entire expression.

For a quick planning check, inspect the signs before combining. Equal coefficients with equal signs cancel by subtraction. Equal magnitudes with opposite signs cancel by addition. State which situation you have. This takes less space than recovering from a line in which no variable actually disappeared.

8. Clear numerical denominators one equation at a time

Solve x/2 + y/3 = 4 and x/3 − y/2 = −5/6. Multiply each entire equation by 6. The system becomes 3x + 2y = 24 and 2x − 3y = −5. Multiplying the first by 3 and the second by 2 then gives 9x + 6y = 72 and 4x − 6y = −10.

Add: 13x = 62, hence x = 62/13. In 3x + 2y = 24, we get 2y = 126/13, so y = 63/13. Check the first fraction equation: 31/13 + 21/13 = 4. Check the second: 62/39 − 63/26 = (124 − 189)/78 = −5/6.

Nothing requires both equations to use the same initial clearing multiplier. Choose a convenient non-zero common multiple separately for each equation. Numerical denominators cause no variable restrictions. A denominator such as x − 1 would require a separate domain check and may lead beyond a linear system after clearing.

9. Decimal coefficients can be rescaled exactly

Consider 0.4x + 0.3y = 2.9 and 0.2x − 0.1y = 0.7. Multiplying both equations by 10 gives 4x + 3y = 29 and 2x − y = 7. The second gives y = 2x − 7. Substitute: 4x + 3(2x − 7) = 29, so x = 5 and y = 3.

The original checks are 2 + 0.9 = 2.9 and 1 − 0.3 = 0.7. Multiplication by 10 has changed the representation, not the solutions. For a pure algebra exercise, the terminating decimals are treated as the stated exact numbers.

If the numbers are rounded measurements instead, an exact intersection of their displayed equations does not remove the measurement uncertainty. That distinction becomes important in the later section on almost-parallel lines. An equation can be solved exactly while the model used to build it remains approximate.

10. No solution and infinitely many solutions are successful conclusions

The pair 2x + 3y = 12 and 4x + 6y = 25 is inconsistent. Doubling the first gives 4x + 6y = 24, which contradicts the second. Elimination produces 0 = 1. There is no pair to find.

Changing 25 to 24 gives a different situation. The second equation is now a duplicate of the first. Elimination produces 0 = 0. That does not mean x = 0 and y = 0; those values would fail 2x + 3y = 12. It means the second equation supplies no additional restriction.

One way to describe all real solutions is x = 6 − 3t/2 and y = t for any real t. Each choice of t produces a pair on the common line. In a non-negative-integer context, enumerate only the permitted pairs. A repeated equation does not create an extra independent piece of evidence merely because it is printed on another line.

11. Distinguish a parameter from an unknown to be found

As an optional extension, consider x + y = 5 and 2x + ky = 12, where k is a fixed but unspecified parameter. Subtract twice the first equation from the second: (k − 2)y = 2. If k ≠ 2, then y = 2/(k − 2), and x = 5 − 2/(k − 2).

If k = 2, the transformed equation reads 0 = 2, so no solution exists. Dividing by k − 2 without mentioning this case would hide precisely the value at which the system changes behaviour.

This is not a requirement to use advanced matrix methods. It is an extension of ordinary division discipline: before dividing, ask whether the divisor can be zero. The algebraic fractions and formulae guide provides the prerequisite domain reasoning.

12. Build each equation from a named quantity

An imaginary workshop has standard packs costing $4 and extended packs costing $7. Thirty-five packs cost $188 altogether. Let s and e be the numbers of standard and extended packs. The count relationship is s + e = 35. The cost relationship is 4s + 7e = 188.

Subtract four times the count equation from the cost equation: 3e = 48, so e = 16 and s = 19. Check both conditions: 19 + 16 = 35 and 4(19) + 7(16) = 76 + 112 = 188. Both answers are non-negative integers, as pack counts require.

The first equation counts packs; the second counts dollars. Do not add the two rows of quantities without preserving those meanings. The coefficients in the cost equation are prices per pack, so multiplying them by pack counts gives costs. A unit label beside the model can prevent the common confusion between a price and a quantity.

13. Unit-price problems reverse the role of the variables

Suppose three notebooks and two folders cost $16, while two notebooks and five folders cost $18. Let n and f represent the prices of one notebook and one folder in dollars. The equations are 3n + 2f = 16 and 2n + 5f = 18.

Multiply the first by 5 and the second by 2: 15n + 10f = 80 and 4n + 10f = 36. Subtract to get 11n = 44, so n = 4. Then f = 2. The quantities being bought are known; the unit prices are unknown.

Compare this with the pack problem. The algebraic appearance is similar, but the variables have exchanged jobs. Define them before choosing a method. A final sentence should say “one notebook costs $4”, not “four notebooks were bought”. Correct numerical values can still answer the wrong question when variable meanings drift during the working.

14. Mixtures require a total and a component total

In a purely mathematical mixture model, combine x litres of a 10% solution with y litres of a 25% solution to obtain 12 litres at 15%. Assume volumes add and the percentages refer to compatible volume fractions. This is an algebra example, not a practical mixing instruction.

The total-volume equation is x + y = 12. The component-volume equation is 0.10x + 0.25y = 0.15(12) = 1.8. Multiplying by 100 gives 10x + 25y = 180. Subtract 10x + 10y = 120 to obtain 15y = 60. Thus y = 4 and x = 8.

Check the component amount: 0.8 + 1.0 = 1.8 litres out of 12, or 15%. Averaging 10% and 25% would give 17.5%, but that assumes equal volumes. The weighted relationship is what makes the mixture work. The target also lies between the original percentages, a useful feasibility check before solving.

15. Geometry supplies constraints, not an exemption from algebra

A rectangle has perimeter 46 cm and length 5 cm more than its width. Let l and w be its dimensions in centimetres. The equations are 2l + 2w = 46 and l − w = 5. Divide the first by 2 to obtain l + w = 23. Adding gives 2l = 28, hence l = 14 and w = 9.

Check the difference and perimeter separately. The positive dimensions satisfy both. A sketch can organise the labels, but the drawn proportions are not part of the evidence.

Now replace the perimeter condition with an area condition. The equation lw = 126 contains a product of variables, so the resulting system is not linear. Substituting l = w + 5 gives a quadratic. That is a legitimate next topic, addressed in Quadratic Equations, Factorisation and Roots, but it should not be forced into a linear-elimination recipe.

16. An impossible context must not be repaired by rounding

Change the workshop total from $188 to $189 while retaining 35 packs at $4 or $7. The same subtraction gives 3e = 49, so e = 49/3. There is a unique real solution to the equations, but it is not a feasible allocation of whole packs.

Rounding e to 16 does not solve the problem; it produces a total of $188. Rounding to 17 produces $191. The exact stated data do not fit the whole-number model.

In an exercise, state the inconsistency. In a real modelling investigation, one would check whether the recorded total, fixed prices or assumptions were wrong. Do not silently change a supplied number merely to manufacture a tidy answer. This is mathematical judgement rather than a refusal to calculate.

17. Nearly parallel lines explain sensitivity

As an optional modelling extension, compare y = 2x + 1 with y = 2.01x. Their intersection satisfies 0.01x = 1, so x = 100 and y = 201. Change the first intercept from 1 to 1.1. The new intersection is x = 110, y = 221.1.

A small change in the constant has moved the intersection by ten horizontal units. The algebra explains why: solving requires division by the small gradient difference 0.01. When input coefficients are approximate, many calculator digits in the solution may conceal sensitivity to those inputs.

This example does not claim that every system is unreliable. It shows why checking a model involves more than checking arithmetic. Exact school coefficients lead to exact algebraic answers; measured coefficients need an interpretation that retains the accuracy of the data.

18. Capstone: solve the model and test a proposed change

An invented activity uses small kits containing two cards and three counters, and large kits containing five cards and four counters. All 76 cards and 74 counters are used. How many kits of each type are assembled?

Let s be small kits and l large kits. Cards give 2s + 5l = 76. Counters give 3s + 4l = 74. Multiply the first equation by 3 and the second by 2: 6s + 15l = 228 and 6s + 8l = 148. Subtract to obtain 7l = 80. This would give a fractional kit count, so the stated exact inventory is not feasible.

That conclusion is useful, but now consider a deliberately different inventory: 76 cards and 81 counters. The equations become 2s + 5l = 76 and 3s + 4l = 81. The scaled second equation is 6s + 8l = 162, so 7l = 66, again infeasible. Trying nearby numbers without a model is not a dependable repair.

Instead build a feasible comparison transparently. For 8 small kits and 12 large kits, the required resources are 2(8) + 5(12) = 76 cards and 3(8) + 4(12) = 72 counters. Solving with those stated totals gives 7l = 84, l = 12, s = 8. The initial 74-counter inventory would leave two counters unused if these kits were assembled, contradicting only the original “all used” condition. Distinguish solving the exact problem from proposing a changed constraint.

19. Independent practice: twenty tasks

For each system, state both coordinates and check both equations. For contextual tasks, include the meaning of each variable. Questions about inconsistency are intentional; a valid solution does not always end in a pair.

1. Solve x + y = 13 and x − y = 5.
2. Solve y = 2x + 3 and 3x + y = 18.
3. Solve y = 8 − 3x and 2x − y = 7.
4. Solve 2x + 3y = 19 and 4x − 3y = 11.
5. Solve 3x + 2y = 22 and 5x − 2y = 26.
6. Solve 2x + 5y = 29 and 3x − 2y = 15.
7. Solve 5x − 3y = 4 and 2x − 3y = −5.
8. Solve x/2 + y/3 = 5 and x − y = 5.
9. Solve 0.3x + 0.2y = 2.1 and 0.1x − 0.2y = −0.5.
10. Classify 2x + y = 8 and 4x + 2y = 17.
11. Describe all real solutions of x + y = 6 and 3x + 3y = 18.
12. Check whether (2, 5) solves x + y = 7 and 3x − y = 2.
13. Four pencils and three pens cost $17. Two pencils and five pens cost $19. Find each unit price.
14. A rectangle has perimeter 52 cm and length 6 cm more than width. Find both dimensions.
15. There are 24 items priced at $3 or $5, totalling $92. Find the number of each.
16. Under additive-volume assumptions, combine 10% and 30% solutions to make 20 litres at 25%. Find both volumes.
17. A learner multiplies x + 2y = 7 by 3 and writes 3x + 2y = 21. Repair the line.
18. For x + y = 4 and 2x + ky = 11, find the exceptional k giving no solution.
19. Find the intersection of y = −x + 9 and y = 2x − 3.
20. Ten tickets cost either $4 or $7 and total $53. Is there a feasible whole-ticket allocation?

20. Worked answers and independent checks

1. x = 9, y = 4. Add the equations to obtain 2x = 18, then recover y = 13 − 9. The original sum is 13 and difference is 5. The checks test the complete pair, not just x.

2. x = 3, y = 9. Substitute to get 3x + 2x + 3 = 18. Thus 5x = 15. Check 9 = 2(3) + 3 and 3(3) + 9 = 18.

3. x = 3, y = −1. Write 2x − (8 − 3x) = 7, giving 5x − 8 = 7. The second check is 6 − (−1) = 7. Retaining the substitution brackets protects the sign.

4. x = 5, y = 3. Addition gives 6x = 30. Then 10 + 3y = 19. Checks: 10 + 9 = 19 and 20 − 9 = 11.

5. x = 6, y = 2. Adding gives 8x = 48; substitution gives 18 + 2y = 22. The second original becomes 30 − 4 = 26.

6. x = 7, y = 3. Multiply the first by 2 and the second by 5. Adding 4x + 10y = 58 and 15x − 10y = 75 gives 19x = 133. Both original equations then hold.

7. x = 3, y = 11/3. Subtract the second equation from the first: 3x = 9. Recover y from 15 − 3y = 4. The second equation gives 6 − 11 = −5, so the fractional value is valid.

8. x = 8, y = 3. Clear the first denominator to get 3x + 2y = 30. Use y = x − 5, giving 5x − 10 = 30. Check the original fractions: 4 + 1 = 5.

9. x = 4, y = 4.5. Adding the original equations gives 0.4x = 1.6. Then 1.2 + 0.2y = 2.1. The second original is 0.4 − 0.9 = −0.5.

10. No solution. Twice the first left side equals the second left side, but twice 8 is 16, not 17. Elimination gives a false constant statement. No rounding or further division is appropriate.

11. Infinitely many real solutions. The second equation repeats the first. Write x = t, y = 6 − t for any real t. A single example demonstrates membership, but the parameter description captures the whole solution set.

12. No. The pair satisfies 2 + 5 = 7 but gives 3(2) − 5 = 1, not 2. A pair satisfying only one original condition is not a simultaneous solution.

13. A pencil costs $2 and a pen $3. Let their prices be p and q. Double 2p + 5q = 19 and subtract 4p + 3q = 17 to obtain 7q = 21. Then p = 2. Check both purchase totals.

14. Length 16 cm, width 10 cm. Half the perimeter gives l + w = 26 and the difference is l − w = 6. Add to find l. Both dimensions are positive and 2(16 + 10) = 52.

15. Fourteen $3 items and ten $5 items. If all 24 were $3, the total would be $72. Each $5 item adds $2, so the $20 excess requires ten. This baseline method is equivalent to eliminating one variable.

16. Five litres at 10%, fifteen at 30%. Set x + y = 20 and 0.1x + 0.3y = 5. Subtract 0.1 times the total equation to get 0.2y = 3. The component check is 0.5 + 4.5 = 5 litres.

17. 3x + 6y = 21. Multiplication applies to both terms on the left and to the right side. Scaling only the selected coefficient breaks equivalence before elimination has begun.

18. k = 2. Subtract twice the first equation to get (k − 2)y = 3. At k = 2 this is impossible. For every other real k, division by k − 2 yields one pair.

19. (4, 5). Set −x + 9 = 2x − 3, giving x = 4. Both expressions produce y = 5. The intersection is a point, so report both coordinates.

20. No feasible whole-ticket allocation. All ten at $4 would cost $40. Each $7 ticket adds $3, but the $13 excess is not divisible by 3. Equivalently, the expensive-ticket count would be 13/3.

21. Turn a correction into a fresh test

Record the first failed decision, not just the last wrong answer. Useful labels include “did not preserve the negative substitution”, “scaled only one term”, “forgot the second variable”, “used the wrong unit for the unknown”, and “rounded an infeasible count”. Each label suggests a different next exercise.

A proposed practice cycle is one explained repair, two nearby variants and one mixed problem without a method label. Later, return to a similar structure with different numbers. Record whether the learner chose the method and completed both checks without a prompt. This is an instructional suggestion, not a validated prediction about how quickly any particular learner will improve.

For a stronger challenge, ask the learner to invent a system with a chosen solution, such as (2, −3), then construct a second version with no solution. Building examples reveals whether the meaning of a system is understood independently of the familiar procedure.

22. Continue through the BTT learning routes

Return to the BTT Mathematics Hub or the BTT Mathematical Lab for a persistent difficulty. Use Equations, Balance and Checking when a single equation is unstable.

Within Batch 05, continue to Quadratic Equations, Factorisation and Roots when products or squares change the model; Vectors, Magnitude, Direction and Geometric Reasoning when two routes locate the same point; or Quartiles, Cumulative Frequency, Box Plots and Standard Deviation when the given numbers are observations to interpret rather than exact constraints.

Sources and scope

Definitions and the standard classification of linear systems can be checked against OpenStax, College Algebra 2e, Section 7.1. The worked examples, counterexamples and practice problems on this page are original, with their mathematical checks displayed. No claim of official examination alignment, peer-reviewed instructional effectiveness or guaranteed improvement is made.

Use the MOE secondary curriculum doorway and the learner’s school topic list to select relevant sections. The parameter and sensitivity discussions are optional extensions, not a claim that every subject level studies them.