Simultaneous Linear Equations in two variables | ICSE Class 9 Maths Notes

This note covers simultaneous linear equations in two variables, numerical coefficients, common solutions, substitution, elimination, cross multiplication, equations with decimal coefficients, and the formation and checking of equations for simple word problems.

What are simultaneous linear equations in two variables?

An equation states that two expressions have equal values. A variable is a letter representing a number whose value may be unknown. Here, x and y denote the two variables unless a particular problem gives them a more specific meaning.

A linear equation in two variables can be written as ax + by + c = 0. The letters a and b are numerical coefficients, meaning the numbers multiplying x and y. The letter c is the constant term, which contains no variable.

The coefficients a and b must not both be zero. Each variable appears as a number multiplied by that variable; the variables are neither squared nor multiplied together. A pair of simultaneous equations consists of two equations whose conditions must hold for the same values of the variables.

Definition: A solution of a pair of simultaneous linear equations is a pair of values that satisfies both equations. To satisfy an equation means to make its two sides equal when the values are substituted.

What does a common solution mean?

Consider x + y = 14 and x − y = 4. The pair x = 9, y = 5 makes both statements true: 9 + 5 = 14 and 9 − 5 = 4. It is therefore a common solution.

Writing this as the ordered pair (9, 5) puts the value of x first and the value of y second. The order records which number belongs to which variable. A final answer should identify both values, rather than leaving their roles uncertain.

Solving simultaneously is more restrictive than solving either equation separately. Each equation imposes a condition; the answer must meet both together. In a word problem, the two letters must retain their original meanings throughout the calculation and in the final statement.

How should equations be arranged before solving?

Arrange corresponding variable terms beneath one another and keep track of every sign. A term is a part of an expression separated by addition or subtraction. In 7x − 15y = 2, the coefficient of y is −15, including its negative sign.

Two useful arrangements are ax + by = a numerical constant, and the standard form ax + by + c = 0. Moving a term across the equality sign changes its sign because the same quantity is added to or subtracted from both sides.

Property: Equivalent equations preserve solutions

Equivalent equations have the same solutions. Multiplying every term on both sides of an equation by the same non-zero number gives an equivalent equation. Dividing every term by the same non-zero number also preserves its solutions.

This property explains why equations can be scaled before elimination. The multiplier applies to the entire equation, including the constant. Changing only the variable terms changes the condition being represented and can therefore produce a different answer.

These two equations will also illustrate cross multiplication. Their constant terms are negative in standard form, even though the original right-hand sides are positive. Copying the positive right-hand values directly into a standard-form formula would therefore give incorrect signs.

Note: A missing written coefficient does not mean zero. In the term x, the coefficient is 1. Read both equations in the same arrangement before listing their coefficients.

How does the substitution method find a common solution?

Substitution means replacing a variable by an equal expression or value. First express one variable in terms of the other. Then substitute that expression into the other equation, reducing the problem to an equation containing just one unknown.

1. Choose an equation from which one variable can be isolated conveniently.
2. Write that variable as an expression involving the other variable.
3. Substitute the whole expression into the other equation and solve it.
4. Substitute the value found into the rearranged equation to obtain the remaining variable.
5. Check the resulting pair in both original equations.

How are brackets used in substitution?

Worked example 1. Solve 7x − 15y = 2 and x + 2y = 3 by substitution.

Answer: From the second equation, x = 3 − 2y. Substitute into the first: 7(3 − 2y) − 15y = 2.

Expanding gives 21 − 14y − 15y = 2. Hence −29y = −19 and y = 19/29. Therefore x = 3 − 2(19/29) = 49/29.

Check: 7(49/29) − 15(19/29) = (343 − 285)/29 = 2. Also, 49/29 + 2(19/29) = 87/29 = 3. Thus both equations are satisfied.

The brackets show that 7 multiplies the entire expression 3 − 2y. Multiplying only the first term inside the brackets would not be a valid substitution. Keeping the brackets until expansion makes the operation easier to follow and check.

The answer contains fractions, which are exact values. A fraction's numerator is its top number and its denominator is its bottom number. Keep these values as fractions during checking rather than replacing them with rounded decimals.

The equation x + 2y = 3 is a convenient starting point because the coefficient of x is 1. Either original equation could be rearranged, but this choice keeps the initial expression simple and avoids introducing a fraction before it is needed.

How does elimination remove one variable?

Elimination means removing one variable by adding or subtracting suitably arranged equations. Make the coefficients of one variable numerically equal, then choose the operation that cancels those terms. Solve the remaining equation and substitute back to find the other variable.

If the matching coefficients have the same sign, subtract the equations. If they have opposite signs, add them. The aim is cancellation, so inspect the signs before choosing the operation. Elimination is sometimes more convenient than substitution.

How are suitable multipliers chosen?

Worked example 2. Solve x + y = 5 and 2x − 3y = 4 by elimination.

Answer: Multiply the first equation by 3 to obtain 3x + 3y = 15. Add the second equation: 5x = 19, so x = 19/5.

Substitute into x + y = 5: y = 5 − 19/5 = 6/5. Check: 19/5 + 6/5 = 5, and 2(19/5) − 3(6/5) = 20/5 = 4.

The original coefficients of y were 1 and −3. After multiplication they become 3 and −3, so addition removes y. The right-hand side of the first equation also changes from 5 to 15 because the whole equation is multiplied by 3.

1. Write both equations with corresponding terms in the same order.
2. Choose a variable and multiply by suitable non-zero numbers to make its coefficients numerically equal.
3. Add or subtract the equations to remove that variable.
4. Solve for the remaining variable, then substitute into an original equation.
5. Check both values in the two equations before stating the answer.

When subtracting an equation, place its entire left-hand side in brackets if necessary. Subtraction affects every term in that expression. A sign error here can survive the later working, which makes checking against the original equations particularly useful.

How can equations with decimal coefficients be simplified?

Decimal coefficients are numerical coefficients written in decimal form. They do not require a new method. Multiplying the entire equation by an appropriate power of ten can convert its decimal coefficients into integers, which are whole numbers, their negatives, and zero.

The transformation must include both sides. Keep the original equations available for the final check. Solving a simplified pair is useful only when each simplification preserves the conditions of the pair that was given.

How does clearing decimals help elimination?

Worked example 3. Solve 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3.

Answer: Multiply both equations by 10. This gives 2x + 3y = 13 and 4x + 5y = 23.

Double the first new equation: 4x + 6y = 26. Subtract 4x + 5y = 23 to obtain y = 3. Then 2x + 9 = 13, giving x = 2.

Check in the original equations: 0.2(2) + 0.3(3) = 0.4 + 0.9 = 1.3. Also, 0.4(2) + 0.5(3) = 0.8 + 1.5 = 2.3.

After the decimals are removed, the coefficients of x are 2 and 4. Doubling the first equation makes these coefficients equal. Because both are positive, subtraction removes x and leaves a simple equation in y.

This example combines two operations with different purposes. Multiplication by 10 clears the decimals. The later multiplication by 2 prepares the equations for elimination. Keeping these stages separate shows why each new line follows from the preceding one.

Note: Clearing decimals changes the appearance of an equation, not the pair of values that satisfies it. In this example, substitution of x = 2 and y = 3 verifies the original decimal equations directly.

How does the cross-multiplication method work?

Cross multiplication uses differences of products of coefficients to obtain the variables. Begin with the standard forms a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

Here a₁, b₁ and c₁ are the coefficient of x, the coefficient of y and the constant in the first equation. Similarly, a₂, b₂ and c₂ refer to the second equation. The small numerals label the equations; they are not powers.

Define D = a₁b₂ − a₂b₁, the common denominator in the solution formulas. Provided D ≠ 0, where ≠ means “is not equal to”, the pair has a unique solution, meaning exactly one common solution.

Result: Cross-multiplication formulas

x = (b₁c₂ − b₂c₁)/D and y = (c₁a₂ − c₂a₁)/D, provided D ≠ 0. The order of subtraction matters in every numerator and in D. Do not reverse just one difference.

The familiar combined arrangement is x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁), wherever these displayed divisions are defined. The separate formulas above also handle a zero value of x or y when D is non-zero.

Why do these formulas follow from elimination?

Multiply the first standard-form equation by b₂ and the second by b₁, then subtract. The y terms cancel, giving (a₁b₂ − a₂b₁)x = b₁c₂ − b₂c₁. Dividing by non-zero D gives the formula for x.

To find y, multiply the second equation by a₁ and the first by a₂, then subtract the latter from the former. The x terms cancel, giving (a₁b₂ − a₂b₁)y = c₁a₂ − c₂a₁. Divide by D.

Worked example 4. Solve x + 3y = 6 and 2x − 3y = 12 by cross multiplication.

Answer: Standard forms are x + 3y − 6 = 0 and 2x − 3y − 12 = 0. Thus a₁ = 1, b₁ = 3, c₁ = −6, a₂ = 2, b₂ = −3 and c₂ = −12.

D = 1(−3) − 2(3) = −9. Hence x = [3(−12) − (−3)(−6)]/(−9) = −54/(−9) = 6.

Also y = [(−6)(2) − (−12)(1)]/(−9) = 0. Check: 6 + 3(0) = 6 and 2(6) − 3(0) = 12.

The value y = 0 is valid. However, the combined ratio arrangement would involve division by its zero numerator. Calculating x and y separately avoids that difficulty. If D itself is zero, use elimination or substitution to investigate the pair instead.

What happens when both variables disappear?

Elimination or substitution does not necessarily produce a single numerical answer for each variable. Sometimes all variable terms cancel. Then inspect the remaining numerical statement before deciding what the cancellation means.

A consistent pair has at least one common solution. An inconsistent pair has no common solution. A dependent pair consists of equivalent equations and has infinitely many common solutions. Such a pair is consistent, but it does not determine a unique answer.

Result: A true statement and a contradiction have different meanings

If valid elimination reduces a pair of linear equations to a true statement without variables, the equations have infinitely many common solutions. If it produces a contradiction, meaning a false statement, they have no common solution.

Worked example 5. Determine whether 2x + 3y = 9 and 4x + 6y = 18 determine unique values.

Answer: Double the first equation to obtain 4x + 6y = 18, exactly the second equation. Subtraction gives 0 = 0. The pair is dependent and consistent, with infinitely many solutions.

The solutions remain linked by x = (9 − 3y)/2. Cancelling the variables does not mean that every independently chosen pair of values is a solution.

Worked example 6. Find all possible solutions of 2x + 3y = 8 and 4x + 6y = 7.

Answer: Double the first equation to obtain 4x + 6y = 16. Subtract the second equation, giving 0 = 9. This is false, so the pair is inconsistent and has no solution.

In the first example, the second equation repeats the condition in the first. In the second example, the same expression is required to equal different numbers. No choice of x and y can fulfil those two requirements together.

These outcomes also explain why a zero value of D in cross multiplication needs further investigation. It does not by itself distinguish a dependent pair from an inconsistent one. The original equations supply the information needed to decide.

How are simple word problems translated into equations?

Forming equations means expressing the stated relationships using variables. Begin by defining the quantities represented by the letters. Then translate each condition separately. A correct calculation cannot repair an equation that represents the wrong relationship.

1. Identify the two unknown quantities and state their meanings clearly.
2. Write an equation for the first relationship, preserving its order and units.
3. Write a second equation for the other relationship.
4. Solve the pair by a suitable algebraic method.
5. Interpret the values and check both original conditions in words or numbers.

How are a count and a total cost linked?

Worked example 7. Akhila plays Hoopla, a ring-throwing game, half as many times as she rides the Giant Wheel. Each ride costs ₹3 and each game costs ₹4. She spends ₹20 altogether. Find both counts. The symbol ₹ denotes rupees.

Answer: Let x be the number of rides and y the number of games. Then y = x/2 and 3x + 4y = 20.

Substitute y = x/2 into the cost equation: 3x + 4(x/2) = 20. Thus 5x = 20, x = 4 and y = 2.

Akhila takes 4 rides and plays 2 games. Check: 2 is half of 4, and 3(4) + 4(2) = 12 + 8 = ₹20.

The first equation compares counts, while the second adds costs. Multiplying each count by its own price converts it into the appropriate contribution to the total. Adding the counts alone would not represent the amount spent.

The words “half as many” must attach to the correct quantity. Here the number of games is half the number of rides. Defining x and y before forming the equations helps prevent that relationship from being reversed.

Because the unknowns count rides and games, the final answer is expressed as counts. The money appears in the cost check. Keeping those roles distinct makes the interpretation clear and shows that both pieces of information have been used.

How should age and income problems be modelled?

In an age problem, define the variables as present ages unless another time is explicitly chosen. Moving to an earlier or later date changes both ages by the same number of years. A comparison at that date must use both changed ages.

How does the time reference affect both ages?

Worked example 8. Seven years ago, Aftab was seven times as old as his daughter. Three years from now, he will be three times as old as she will be. Find their present ages.

Answer: Let s be Aftab's present age and t his daughter's present age, both in years. The equations are s − 7 = 7(t − 7) and s + 3 = 3(t + 3).

These simplify to s − 7t + 42 = 0 and s − 3t = 6. From the second, s = 3t + 6. Substitute into the first: 3t + 6 − 7t + 42 = 0.

Therefore 4t = 48, t = 12 and s = 42. Seven years ago their ages were 35 and 5, with 35 = 7 × 5. Three years from now they will be 45 and 15, with 45 = 3 × 15.

The brackets in 7(t − 7) multiply the daughter's entire earlier age by seven. The present age cannot be used in that comparison. The final answer reports present ages because that is what s and t were defined to represent.

How are ratios connected to savings?

A ratio compares quantities by division; the colon separates the compared quantities. In an income problem, expenditure is the amount spent, and savings are income minus expenditure. Ratios alone give relative sizes, so keep the stated savings condition as well.

Worked example 9. Two persons' monthly incomes are in the ratio 9 : 7 and their monthly expenditures in the ratio 4 : 3. Each saves ₹2000 per month. Find their monthly incomes.

Answer: Let their incomes be ₹9x and ₹7x, and their expenditures ₹4y and ₹3y. Here x and y are the respective common scaling amounts. The equations are 9x − 4y = 2000 and 7x − 3y = 2000.

Multiply the first equation by 3 and the second by 4: 27x − 12y = 6000 and 28x − 12y = 8000. Subtraction gives x = 2000. Substitution then gives y = 4000.

The monthly incomes are ₹18,000 and ₹14,000. Their expenditures are ₹16,000 and ₹12,000, leaving ₹2000 each. The income ratio is 9 : 7 and the expenditure ratio is 4 : 3.

The variables x and y are scaling amounts, not the requested incomes themselves. Ending the answer at x = 2000 would therefore leave the problem unfinished. Recover each income from its defining expression before stating the result.

How are digit problems solved and answers checked?

A digit is a symbol used to write a number. In a two-digit number, let x denote the tens digit and y the units digit. The number is then 10x + y, because the tens digit contributes ten times its own value.

Reversing the digits exchanges their places, giving 10y + x. Distinguish the number itself from the sum of its digits, x + y. These expressions describe different quantities, so read the relationship carefully before forming an equation.

What if the larger digit is not specified?

Worked example 10. A two-digit number and the number obtained by reversing its digits have a sum of 66. The digits differ by 2. Find all possible numbers.

Answer: With x as the tens digit and y as the units digit, (10x + y) + (10y + x) = 66. Therefore 11(x + y) = 66 and x + y = 6.

If x − y = 2, adding this to x + y = 6 gives 2x = 8, so x = 4 and y = 2. The number is 42.

If y − x = 2, solving with x + y = 6 gives y = 4 and x = 2. The number is 24. Check: 42 + 24 = 66, and 4 − 2 = 2. Both numbers qualify.

The phrase “digits differ by 2” does not specify which digit is larger. Both cases are needed. Writing only x − y = 2 would find one valid number but omit another allowed by the wording.

What should the final check establish?

First verify the pair in the original equations. Then return to the problem's quantities: ages must refer to the requested time, income variables may need conversion into incomes, and digit variables must be assembled into numbers.

Use substitution when a variable is easily isolated, elimination when coefficients can conveniently cancel, and cross multiplication when standard-form coefficients are clear and D is non-zero. If a method is specified, show that method's working.

The methods seek the same common solution. A disagreement between answers is a reason to recheck signs, arithmetic and the original equations. Checking the full pair is stronger than confirming a single intermediate calculation made during the solution.

Glossary

• Variable — A letter representing a number whose value may be unknown in an equation.
• Coefficient — The numerical factor multiplying a variable, including its positive or negative sign.
• Constant term — A term containing no variable, whose numerical value stays fixed in the equation.
• Linear equation in two variables — An equation expressible as ax + by + c = 0, with a and b not both zero.
• Simultaneous equations — Equations whose conditions must hold together for the same values of their variables.
• Common solution — A pair of variable values that makes both equations true at the same time.
• Substitution — Replacing a variable by an equal expression or value to simplify another equation.
• Elimination — Removing one variable by adding or subtracting equations after suitable multiplication where necessary.
• Cross multiplication — A method using differences of coefficient products to calculate a pair's unique solution.
• Equivalent equations — Equations that have the same solutions despite being written in different forms.
• Unique solution — Exactly one pair of values that satisfies both equations being considered simultaneously.
• Consistent pair — A pair of equations with at least one common solution for its variables.
• Inconsistent pair — A pair of equations for which no values satisfy both conditions simultaneously.
• Dependent pair — A pair of equivalent linear equations having infinitely many distinct common solutions.

Common errors and misconceptions

• Misconception: A pair is solved once one equation is satisfied. Correct: The same values must satisfy both original equations.
• Misconception: Multiplying an equation means multiplying only its variable terms. Correct: Multiply every term on both sides, including the constant.
• Misconception: Equal signed coefficients should be added to eliminate a variable. Correct: Subtract when the signs match; add when numerically equal coefficients have opposite signs.
• Misconception: In cross multiplication, the original right-hand constant can be used unchanged. Correct: First write both equations with zero on the right and identify the signed constants.
• Misconception: Cancellation of both variables proves that both variables are zero. Correct: Inspect the remaining statement. A true statement indicates dependence; a false statement indicates inconsistency.
• Misconception: An age comparison from seven years ago uses one present age. Correct: Subtract seven years from both present ages before applying that comparison.
• Misconception: “The digits differ by 2” means the tens digit is larger. Correct: Consider both possible orders unless the wording specifies the larger digit.
• Misconception: The values of ratio scaling variables are the requested incomes. Correct: Use those values in the expressions defining the incomes, then state the amounts.

Exam-style questions with model answers

Q1. Verify that x = 9 and y = 5 solve x + y = 14 and x − y = 4 simultaneously. [2 marks]

1. Substitution into the first equation gives 9 + 5 = 14, so the first condition is satisfied.
2. Substitution into the second gives 9 − 5 = 4. Both equations hold for the same pair, so it is a simultaneous solution.

Q2. Solve 7x − 15y = 2 and x + 2y = 3 by substitution. [3 marks]

1. Use the second equation to express x in terms of y: x = 3 − 2y. Substitute this expression into the first equation to obtain 7(3 − 2y) − 15y = 2.
2. Expand and collect terms: 21 − 14y − 15y = 2, giving −29y = −19. Therefore y = 19/29.
3. Substitute back into x = 3 − 2y. Then x = 3 − 38/29 = 49/29. Thus the solution is x = 49/29, y = 19/29.

Q3. Solve 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 by elimination, and check your answer. [4 marks]

1. Multiply each complete equation by 10 to clear the decimals. The resulting equations are 2x + 3y = 13 and 4x + 5y = 23.
2. Double the first resulting equation to obtain 4x + 6y = 26. Subtract the second equation, leaving y = 3.
3. Substitute y = 3 into 2x + 3y = 13. This gives 2x + 9 = 13, hence x = 2.
4. Check the originals: 0.2(2) + 0.3(3) = 1.3 and 0.4(2) + 0.5(3) = 2.3. Both conditions hold.

Q4. Use cross multiplication to solve x + 3y = 6 and 2x − 3y = 12. Show the standard forms and verify both values. [5 marks]

1. Write the standard forms as x + 3y − 6 = 0 and 2x − 3y − 12 = 0. The coefficient triples are (1, 3, −6) and (2, −3, −12).
2. The common denominator, denoted by D, is the difference of the coefficient products: D = 1(−3) − 2(3) = −9. It is non-zero, so the formulas apply.
3. For x, the numerator is 3(−12) − (−3)(−6) = −54. Therefore x = −54/(−9) = 6.
4. For y, the numerator is (−6)(2) − (−12)(1) = 0. Therefore y = 0/(−9) = 0.
5. Substitution gives 6 + 3(0) = 6 and 2(6) − 3(0) = 12. The solution satisfies both original equations.

Q5. Use elimination to show whether 2x + 3y = 8 and 4x + 6y = 7 have a common solution. [3 marks]

1. Multiply the first equation by 2, including its right-hand side. This gives 4x + 6y = 16, while the second equation remains 4x + 6y = 7.
2. Subtract the second equation from the new first equation. Both variable terms cancel, leaving the numerical statement 0 = 9.
3. This statement is false, so no pair of values can satisfy both original equations. The pair is inconsistent and has no common solution.

Q6. Seven years ago, Aftab was seven times as old as his daughter. Three years from now, he will be three times as old as she will be. Find and check their present ages. [5 marks]

1. Let s be Aftab's present age and t his daughter's present age, in years. The earlier comparison gives s − 7 = 7(t − 7), or s − 7t + 42 = 0.
2. The future comparison gives s + 3 = 3(t + 3). Simplifying gives s − 3t = 6, so s = 3t + 6.
3. Substitute this into the earlier equation: 3t + 6 − 7t + 42 = 0. Hence 4t = 48 and t = 12.
4. Then s = 3(12) + 6 = 42. Aftab is presently 42 years old and his daughter is 12 years old.
5. Seven years ago, 35 = 7 × 5 verifies the comparison. Three years from now, 45 = 3 × 15 verifies the other comparison.

Q7. A two-digit number and the number obtained by reversing its digits add to 66. Its digits differ by 2. Find all possible numbers. [5 marks]

1. Let x and y denote the tens and units digits. The number is 10x + y and the reversed number is 10y + x.
2. The sum condition gives (10x + y) + (10y + x) = 66. Simplifying and dividing by 11 gives x + y = 6.
3. If x − y = 2, add this to the sum equation to obtain 2x = 8. Thus x = 4, y = 2, and the number is 42.
4. If y − x = 2, solving with the sum equation gives y = 4 and x = 2. This produces the number 24.
5. Both possibilities must be retained because the larger digit was not specified. Their sum is 66 and the digits differ by 2, so both satisfy the conditions.

Key takeaways

• A simultaneous solution gives values of both variables that satisfy the two original equations together.
• Substitution replaces one variable with an equal expression, producing an equation in the other variable.
• Elimination uses suitable multiplication followed by addition or subtraction to cancel one variable correctly.
• Cross multiplication requires standard-form coefficients with their signs and a non-zero common denominator for a unique solution.
• A true numerical statement after elimination indicates dependence; a false numerical statement indicates that no common solution exists.
• Define every unknown before translating a word problem, and preserve the stated relationships between quantities.
• Check both original conditions and express the final answer as the quantities actually requested in the problem.
• When a digit difference does not specify which digit is larger, consider both possible orders.

Test yourself

What must a pair of values do to solve simultaneous equations?

It must satisfy both original equations when the same values are substituted into each.

Why is x + 2y = 3 convenient for substitution?

The coefficient of x is 1, so it rearranges directly to x = 3 − 2y.

When numerically equal coefficients have opposite signs, which operation eliminates the variable?

Add the equations so that the terms with opposite signs cancel each other.

What is the constant term when x + 3y = 6 is written with zero on the right?

The standard form is x + 3y − 6 = 0, so the constant term is −6.

What does 0 = 9 after valid elimination tell you?

It is a false statement, so the pair is inconsistent and has no common solution.

Why does a zero common denominator in cross multiplication need further investigation?

Division by zero is undefined. Use elimination or substitution to distinguish infinitely many common solutions from no common solution.

If a present age is t years, how is the age seven years ago expressed?

It is t − 7 years. Both people's ages must refer to the same earlier time when compared.

If x is the tens digit and y the units digit, what is the reversed number?

The reversed number is 10y + x, because y becomes the tens digit and x becomes the units digit.