Solving a pair by substitution
To split a bill of two items, you first find one amount in terms of the other, then put it in the second condition.
In the substitution method you write one variable in terms of the other from one equation, substitute it into the other equation, solve the single-variable equation, then find the second variable.
In the figure: x + y = 10 and x − y = 2: y = 10 − x, so x − (10 − x) = 2, x = 6, y = 4.
Solving a pair by substitution
In the app you can move this figure with your finger.
Open in the appRule 1: Follow the steps — Express, substitute, solve, then go back.
- Easy: y = 2x and x + y = 9: 3x = 9, x = 3, y = 6.
- Medium: x = 2y and x + 2y = 8: 4y = 8, y = 2, x = 4.
- Exam-type: 3x + y = 11 and y = x + 3: 4x + 3 = 11, x = 2, y = 5.
Rule 2: Check the answer — Put (x, y) into both original equations.
- Easy: (6, 4): 6 + 4 = 10 ✓ and 6 − 4 = 2 ✓.
- Medium: (2, 5): 6 + 5 = 11 ✓ and 5 = 2 + 3 ✓.
- Exam-type: A check in only one equation is not enough.
Always substitute into the OTHER equation, never the one you used to get the expression.
Substituting back into the same equation, which gives 0 = 0 and no answer.
"Express, put in the other, solve, return."
Practice questions
- 1Solve by substitution: x + 2y = 5 and 4x − y = −7. What is the value of y?
- 2Solve by substitution: x − y = −1 and 2x − 5y = −26. What is the value of y?
- 3Solve by substitution: 5x + y = 4 and −x + 5y = −6. What is the value of y?
- 4Solve by substitution: −x − 5y = 23 and −4x + 6y = −38. What is the value of x?
- 5Solve the pair −4x − y = −34 and x − 4y = 0 by substitution and find the value of x + y.
- 6Solve the pair x + 5y = 31 and −5x + 4y = −10 by substitution and find the value of xy.
- 7Solve the pair x − y = 14 and −3x − 3y = −6 by substitution and find the value of x + y.
- 8If x and y satisfy x − 7y = −32 and −5x − 2y = 12, find the value of 2x + 3y.
Show answers (try them yourself first)
- 13
- 28
- 3−1
- 42
- 510
- 630
- 72
- 84
Writing one variable in terms of the other
Pick the easiest equation and the easiest variable: the one with coefficient 1 is simplest to isolate.
To express a variable, move all other terms to the other side and divide by its coefficient. Prefer a variable whose coefficient is 1 or −1 to avoid fractions.
Formula: ax + by = c ⇒ y = (c − ax)/b
In the figure: x + 3y = 7 becomes x = 7 − 3y, a form ready to substitute.
Rule 1: Isolate a variable — Choose the variable with coefficient 1.
- Easy: x + 3y = 7 ⇒ x = 7 − 3y.
- Medium: 3x − y = 5 ⇒ y = 3x − 5.
- Exam-type: 2x + 3y = 12 ⇒ y = (12 − 2x)/3.
Rule 2: Use it in the other equation — Replace the variable by the whole expression in brackets.
- Easy: x = 7 − 3y in 2x − y = 7: 2(7 − 3y) − y = 7, 14 − 7y = 7, y = 1, x = 4.
- Medium: y = 3x − 5 in x + y = 7: x + 3x − 5 = 7, x = 3, y = 4.
- Exam-type: Use brackets so the minus sign is applied to every term.
Isolating a variable with coefficient 1 saves fractions.
Forgetting the brackets: 2 × 7 − 3y instead of 2(7 − 3y).
"Isolate the easy one; bracket the swap."
Practice questions
- 1From the equation 2x − 5y = −4, express x in terms of y.
- 2From the equation −6x − y = 13, express y in terms of x.
- 3From the equation x + 5y = −9, express x in terms of y.
- 4From the equation −7x − 7y = −13, express x in terms of y.
- 5From the equation 6x − y = −7, express y in terms of x.
- 6From the equation −x − y = 2, express y in terms of x.
- 7From the equation x + 6y = 13, express x in terms of y.
- 8From the equation −7x + y = −12, express y in terms of x.
Show answers (try them yourself first)
- 1(−4 + 5y)/2
- 2−13 − 6x
- 3−9 − 5y
- 4(13 − 7y)/7
- 57 + 6x
- 6−2 − x
- 713 − 6y
- 8−12 + 7x
Word problems by substitution
Many real situations hide two connected unknowns: numbers, ages, prices, lengths. Name them x and y, write two equations, substitute.
Read the problem, name the unknowns, write two equations, solve by substitution, and answer in words with units, checking against the story.
In the figure: Sum 25, difference 7: x + y = 25, x = y + 7, so y = 9, x = 16.
Rule 1: Numbers and digits — Digits: tens digit x, units digit y gives 10x + y.
- Easy: Two numbers: sum 25, difference 7 → 16 and 9.
- Medium: One number is 3 times the other and the sum is 20: 5 and 15.
- Exam-type: Digit sum 9, number 9 more than its reverse: x + y = 9, x − y = 1: number 54.
Rule 2: Money and shapes — Write a relation between the quantities, then a total.
- Easy: Pen = pencil + 2, pen + pencil = 8: pen 5, pencil 3.
- Medium: Rectangle: length = breadth + 3, perimeter 30: breadth 6 cm, length 9 cm.
- Exam-type: Ritu has 3 times Ravi's coins, total 60: Ravi 15, Ritu 45.
Check the solution against the words of the problem, not just the equations.
Writing "difference 7" as y − x = 7 when x is the larger number.
"Name, two equations, substitute, answer."
Practice questions
- 1The sum of the ages of a father and his son is 59 years, and the father is 29 years older than the son. Find the son's age.
- 2The sum of two numbers is 112 and their difference is 12. Find the smaller number.
- 3The sum of the ages of a father and his son is 36 years, and the father is 16 years older than the son. Find the father's age.
- 4The sum of two numbers is 131 and their difference is 21. Find the larger number.
- 5A father is 3 times as old as his son, and the sum of their ages is 32 years. Find the son's age.
- 6One number is 8 more than twice the other number, and their sum is 65. Find the larger number.
- 7A father is 5 times as old as his son, and the sum of their ages is 90 years. Find the father's age.
- 8The perimeter of a rectangle is 132 cm and its length is 16 cm more than its breadth. Find its length (in cm).
Show answers (try them yourself first)
- 115
- 250
- 326
- 476
- 58
- 646
- 775
- 841
Special cases: no solution or infinitely many
Sometimes the variables vanish while substituting. What is left tells whether the lines are the same or parallel.
If substitution leaves a true statement like 0 = 0, there are infinitely many solutions (coincident lines). If it leaves a false statement like 8 = 9, there is no solution (parallel lines).
In the figure: x + y = 4 and 2x + 2y = 8 give 8 = 8, so infinitely many solutions.
Rule 1: True leftover — The two equations describe the same line.
- Easy: x + y = 4, 2x + 2y = 8: y = 4 − x ⇒ 2x + 8 − 2x = 8, 0 = 0, infinite.
- Medium: x − 2y = 3, 3x − 6y = 9: x = 3 + 2y ⇒ 9 + 6y − 6y = 9, infinite.
- Exam-type: Any point of the line is a solution.
Rule 2: False leftover — The lines are parallel.
- Easy: x + y = 4, 2x + 2y = 9: 8 − 2x + 2x = 9 gives 8 = 9, no solution.
- Medium: 2x + y = 3, 4x + 2y = 7: 6 = 7, no solution.
- Exam-type: Match with the ratio test to confirm.
The ratio test and the substitution result must tell the same story.
Saying "no solution" because x disappeared; look at whether the leftover is true or false.
"True leftover: infinite. False leftover: none."
Practice questions
- 1Solve x + 4y = −10 and −5x − 5y = 15 by substitution. What does the result show?
- 2Solve x − 4y = 1 and 5x − 20y = 5 by substitution. What does the result show?
- 3Solve x + 5y = 2 and −x − 3y = −11 by substitution. What does the result show?
- 4Solve x + y = 7 and 3x + 3y = 24 by substitution. What does the result show?
- 5Solve x − 7y = −4 and 3x − 3y = 0 by substitution. What does the result show?
- 6Solve x + 6y = 12 and 3x + 18y = 38 by substitution. What does the result show?
- 7Solve x − 6y = 12 and −2x − 7y = 19 by substitution. What does the result show?
- 8Solve x + 4y = −3 and 2x + 8y = −3 by substitution. What does the result show?
Show answers (try them yourself first)
- 1Exactly one solution
- 2Infinitely many solutions
- 3Exactly one solution
- 4No solution
- 5Exactly one solution
- 6No solution
- 7Exactly one solution
- 8No solution
Using and checking a given solution
If you are told one answer, you can find a missing number in the equation, or check whether a pair really fits.
A pair (x, y) is a solution of an equation if putting the values makes both sides equal. A given solution can be put in an equation to find an unknown constant.
In the figure: If (2, 1) solves ax + 3y = 7, then 2a + 3 = 7, so a = 2.
Rule 1: Find the constant — Replace x and y by the given numbers.
- Easy: (1, −1) solves 3x + by = 1: 3 − b = 1, b = 2.
- Medium: (2, 3) solves ax + 2y = 12: 2a + 6 = 12, a = 3.
- Exam-type: (3, 2) solves x + y = c: c = 5.
Rule 2: Check a pair — Both equations must be true.
- Easy: (4, 3) in x + y = 7 and x − y = 1: both true ✓.
- Medium: (5, 1) in x + y = 6 and x − y = 3: second is 4 ≠ 3 ✗.
- Exam-type: One true equation is not enough to call it a solution of the pair.
A solution of the pair satisfies BOTH equations.
Checking in only one equation and calling it a solution of the pair.
"Put in both, both must hold."
Practice questions
- 1If x = 7, y = 2 is a solution of −3x + 7y = c, find c.
- 2If x = 1, y = −3 is a solution of −5x + 4y = c, find c.
- 3If x = −2, y = −1 is a solution of −6x − 3y = c, find c.
- 4If x = −6, y = −4 is a solution of 3x + 3y = c, find c.
- 5If x = −1, y = −2 is a solution of −4x + ky = −6, find k.
- 6If x = 4, y = −7 is a solution of −3x + ky = −19, find k.
- 7If x = −6, y = −6 is a solution of 3x + ky = −36, find k.
- 8x = 2, y = 3 is the solution of ax + by = 28 and ax − by = −8. Find a + b.
Show answers (try them yourself first)
- 1−7
- 2−17
- 315
- 4−30
- 55
- 61
- 73
- 811