Choosing multipliers to eliminate
To remove one kind of item from two bills, make the amounts of that item equal, then subtract the bills.
To eliminate a variable, multiply the equations by suitable numbers so that its coefficients are equal in size. Then subtract if the signs are the same, add if the signs are opposite.
In the figure: 2x + 3y = 8 and 3x + 2y = 7: multiply by 3 and by 2 to get 6x in both, then subtract: 5y = 10, y = 2, x = 1.
Rule 1: Pick the multipliers — Choose the variable that needs the smaller multipliers.
- Easy: x + y = 5 and x − y = 1: no multiplier, just add: 2x = 6.
- Medium: 3x + 2y = 12 and x + y = 5: multiply the second by 2 and subtract: x = 2, y = 3.
- Exam-type: 5x − 3y = 7 and 2x + y = 5: multiply the second by 3 and add: 11x = 22, x = 2, y = 1.
Rule 2: Add or subtract — Same signs: subtract. Opposite signs: add.
- Easy: 3x + 2y and x − 2y: opposite signs, add.
- Medium: 4x + 3y and 2x + 3y: same signs, subtract.
- Exam-type: Subtracting a negative term changes its sign.
Multiply BOTH sides of the whole equation, not just one term.
Multiplying only the variable terms and forgetting the constant on the right.
"Equal the coefficients, then add or subtract."
Practice questions
- 1If the coefficients of a variable are equal in magnitude and have opposite signs in two equations, we ___ the equations to eliminate it.
- 2To eliminate x directly from 6x − 7y = 18 and −6x − 6y = −5, what should we do?
- 3To eliminate y directly from −x + 2y = 24 and 7x − 2y = 0, what should we do?
- 4To eliminate y directly from −5x + 3y = 6 and 4x − 3y = −18, what should we do?
- 5To eliminate y directly from −4x + 3y = 12 and 5x − 3y = 29, what should we do?
- 6To eliminate y directly from x + 2y = 13 and −x − 2y = 20, what should we do?
- 7To make the coefficients of x numerically equal in 2x + 2y = 2 and −6x − y = −11, by which smallest positive numbers should the first and second equations be multiplied (in that order)?
- 8To make the coefficients of y numerically equal in −5x + 8y = 10 and −5x + 2y = −11, by which smallest positive numbers should the first and second equations be multiplied (in that order)?
Show answers (try them yourself first)
- 1add
- 2Add the two equations
- 3Add the two equations
- 4Add the two equations
- 5Add the two equations
- 6Add the two equations
- 73 and 1
- 81 and 4
The equation after adding or subtracting
After you add or subtract, one variable disappears and a short equation is left: this is the step that makes the problem easy.
Add (or subtract) the left sides and the right sides separately. The result is an equation in one variable that you can solve directly.
In the figure: Adding x + 2y = 7 and 3x − 2y = 5 removes y: 4x = 12.
Rule 1: Write the new equation — Add or subtract term by term.
- Easy: (x + y = 9) + (x − y = 3): 2x = 12.
- Medium: (5x + 2y = 16) − (3x + 2y = 10): 2x = 6.
- Exam-type: (2x + 3y = 12) − (2x − y = 4): 4y = 8.
Rule 2: Solve it — Divide both sides by the remaining coefficient.
- Easy: 2x = 12 ⇒ x = 6.
- Medium: 2x = 6 ⇒ x = 3.
- Exam-type: 4y = 8 ⇒ y = 2.
After subtracting, take care with every sign, especially minus signs on the second equation.
Subtracting only the first term of the second equation.
"Left with left, right with right."
Practice questions
- 1Subtracting the second equation from the first 7x − 5y = −13 and 7x − 2y = −1 (to eliminate x) we get
- 2Adding the two equations x + 6y = 8 and 5x − 6y = 11 (to eliminate y) we get
- 3Adding the two equations −6x + 2y = 7 and 4x − 2y = 13 (to eliminate y) we get
- 4Adding the two equations 4x + 6y = −3 and −5x − 6y = 25 (to eliminate y) we get
- 5Subtracting the second equation from the first 7x + 4y = −3 and x + 4y = 22 (to eliminate y) we get
- 6Adding the two equations 2x − 6y = 10 and −2x − 5y = −11 (to eliminate x) we get
- 7Subtracting the second equation from the first 6x + 2y = 21 and 6x + 4y = 28 (to eliminate x) we get
- 8Subtracting the second equation from the first 6x + 2y = −11 and 6x − 7y = 20 (to eliminate x) we get
Show answers (try them yourself first)
- 1−3y = −12
- 26x = 19
- 3−2x = 20
- 4−x = 22
- 56x = −25
- 6−11y = −1
- 7−2y = −7
- 89y = −31
Solving a pair by elimination
First remove one unknown, then find the other, then go back and check. Elimination avoids fractions and brackets.
Steps of the elimination method: (1) make coefficients equal, (2) add or subtract, (3) solve for one variable, (4) put it back to find the other, (5) check in both equations.
In the figure: 3x + 2y = 11 and x − 2y = 1: add to get 4x = 12, x = 3, y = 1.
Solving a pair by elimination
In the app you can move this figure with your finger.
Open in the appRule 1: Solve completely — Find both values, then check.
- Easy: 3x + 2y = 11, x − 2y = 1: x = 3, y = 1 (check: 3 − 2 = 1 ✓).
- Medium: 2x + y = 7, 3x − y = 8: add 5x = 15, x = 3, y = 1.
- Exam-type: 4x + 3y = 17, 2x + y = 7: multiply the second by 2, subtract: y = 3, x = 2.
Rule 2: Choose the quicker variable — Eliminate the variable that needs the simplest multipliers.
- Easy: If one pair has opposite signs and equal coefficients, add directly.
- Medium: If one coefficient divides the other, multiply one equation only.
- Exam-type: Otherwise multiply both by each other's coefficients.
Elimination is the best method when the coefficients are not 1.
Finding x only and writing it as the final answer; y is also needed.
"Eliminate, solve, return, check."
Practice questions
- 1Solve 5x − 4y = −56 and 5x + 4y = 16. What is the value of y?
- 2Solve by the elimination method: −3x − 3y = −30 and −3x + y = −10. What is the value of x?
- 3Solve 3x + 2y = 23 and 3x − 2y = 19. What is the value of y?
- 4Solve by the elimination method: 3x − 4y = 22 and 4x − y = 12. What is the value of x?
- 5Solve by the elimination method: −4x + 3y = −32 and −x + 3y = −17. What is the value of x?
- 6Solve by the elimination method: x + 3y = −8 and 2x + 5y = −13. What is the value of x?
- 7Solve 2x − 5y = −13 and x − 7y = −11 by elimination and find the value of xy.
- 8Solve 4x + 2y = 46 and −6x + 6y = 12 by elimination and find the value of 2x + 3y.
Show answers (try them yourself first)
- 19
- 25
- 31
- 42
- 55
- 61
- 7−4
- 841
Word problems: cost, fixed charges, digits
Shop bills, taxi fares and two-digit numbers all give two equations. Elimination finds both unknowns fast.
Name the unknowns, write two equations (cost: total = items × price; fixed charge: f + n × rate; digits: 10x + y), eliminate, then give the answer in words.
Formula: total = fixed + number × rate
In the figure: Taxi: fixed f, rate k per km. 10 km costs 105 and 15 km costs 155: f + 10k = 105, f + 15k = 155, so k = 10, f = 5.
Rule 1: Cost problems — Each statement is one equation.
- Easy: 2 pencils + 3 erasers = 9 and 4 pencils + 5 erasers = 17: pencil 3, eraser 1.
- Medium: 3 kg apples + 1 kg grapes = 480, 2 kg apples + 1 kg grapes = 340: apples 140, grapes 60.
- Exam-type: Subtract the equations to remove the equal item.
Rule 2: Fixed charge and digits — Fixed + rate × quantity; and 10x + y for digits.
- Easy: Taxi: f + 10k = 105 and f + 15k = 155: k = 10, f = 5.
- Medium: Digits sum 9, reversed number 27 greater: y − x = 3, x + y = 9: number 36.
- Exam-type: Check: 63 − 36 = 27 ✓.
Write the final answer in a sentence with the unit (₹, km, kg).
Writing the digit number as xy (a product) instead of 10x + y.
"Cost: total; fixed: f plus; digits: ten x plus y."
Practice questions
- 1Two numbers add up to 959 and differ by 189. Find the larger number by elimination.
- 22 pencils and 3 erasers cost ₹145, while 5 pencils and 1 erasers cost ₹174. Find the cost of one eraser.
- 31 adult tickets and 3 child tickets cost ₹25, while 4 adult tickets and 1 child tickets cost ₹34. Find the price of one child ticket.
- 4Two numbers add up to 812 and differ by 150. Find the larger number by elimination.
- 5Two numbers add up to 288 and differ by 48. Find the larger number by elimination.
- 6A hostel charges a fixed monthly amount plus a charge for food per day. A student who eats for 4 days pays ₹273, and another who eats for 9 days pays ₹473. Find the food charge per day (in ₹).
- 7A taxi charges a fixed fare plus a rate per km. For a 5 km ride the fare is ₹252, and for a 10 km ride it is ₹442. Find the rate per km (in ₹).
- 8A hostel charges a fixed monthly amount plus a charge for food per day. A student who eats for 7 days pays ₹500, and another who eats for 17 days pays ₹730. Find the fixed charge (in ₹).
Show answers (try them yourself first)
- 1574
- 229
- 36
- 4481
- 5168
- 640
- 738
- 8339
Harder applications and special cases
Some pairs hide special answers or have 1/x and 1/y. A change of variable turns them into ordinary pairs.
If elimination gives 0 = 0 there are infinitely many solutions; if it gives a false statement like 0 = −1, there is no solution. For equations in 1/x and 1/y, put a = 1/x, b = 1/y.
In the figure: 2/x + 3/y = 13 and 5/x − 4/y = −2 become 2a + 3b = 13, 5a − 4b = −2, giving a = 2, b = 3, so x = 1/2, y = 1/3.
Rule 1: Special cases — Look at what is left after eliminating.
- Easy: 2x + 3y = 6 and 4x + 6y = 12: 0 = 0, infinitely many.
- Medium: 2x + y = 3 and 4x + 2y = 7: 0 = −1, no solution.
- Exam-type: Consistent with the ratio test.
Rule 2: Reciprocal equations — Substitute a = 1/x and b = 1/y, solve, then invert.
- Easy: 2a + 3b = 13, 5a − 4b = −2: a = 2, b = 3.
- Medium: Then x = 1/2 and y = 1/3.
- Exam-type: Check in the original: 2/(1/2) + 3/(1/3) = 4 + 9 = 13 ✓.
Remember x and y are the reciprocals of a and b, not a and b themselves.
Stopping at a and b without converting back to x and y.
"Zero equals zero: infinite; zero equals non-zero: none."
Practice questions
- 1In the elimination method, if both variables disappear and we get 0 = 0, the pair has ___.
- 2If 6x + 7y = −55 and 7x + 6y = −49, find the value of x + y.
- 3If 2x + 7y = −21 and 7x + 2y = −51, find the value of x + y.
- 4The incomes of two persons are in the ratio 11 : 6 and their expenditures are in the ratio 7 : 2. If each saves ₹12000, find the income (in ₹) of the first person.
- 5The incomes of two persons are in the ratio 10 : 8 and their expenditures are in the ratio 7 : 2. If each saves ₹3600, find the income (in ₹) of the first person.
- 6If 6x + 5y = −29 and 5x + 6y = −26, find the value of x + y.
- 7The incomes of two persons are in the ratio 8 : 7 and their expenditures are in the ratio 7 : 4. If each saves ₹13600, find the income (in ₹) of the first person.
- 8The incomes of two persons are in the ratio 11 : 10 and their expenditures are in the ratio 3 : 2. If each saves ₹2400, find the income (in ₹) of the first person.
Show answers (try them yourself first)
- 1infinitely many solutions
- 2−8
- 3−8
- 433000
- 55000
- 6−5
- 719200
- 83300