Exercise 4.2 — Algebraic Methods of Solving
Two powerful algebraic methods — Substitution and Elimination — to solve pairs of linear equations without drawing graphs.
Method 1 — Substitution Method
In the Substitution Method, we express one variable in terms of the other from one equation, substitute that expression into the second equation, and solve the resulting single-variable equation. The method works in five clear steps.
Substitution Method — Worked Example
Solve: 2x + y − 4 = 0 …(1) and x + 2y − 5 = 0 …(2)
Step 1 — Express one variable in terms of the other
From (1): y = 4 − 2x
Step 2 — Substitute into the second equation
x + 2(4 − 2x) − 5 = 0
Step 3 — Simplify and find x
x + 8 − 4x − 5 = 0 ⟹ −3x + 3 = 0 ⟹ x = 1
Step 4 — Substitute x back to find y
2(1) + y − 4 = 0 ⟹ y − 2 = 0 ⟹ y = 2
Step 5 — Verify in both original equations
2(1) + 2 − 4 = 0 ✓ 1 + 2(2) − 5 = 0 ✓
∴ x = 1, y = 2
Method 2 — Elimination Method
In the Elimination Method, we multiply each equation by suitable numbers so that the coefficient of one variable becomes equal in both equations. We then add or subtract the equations to eliminate that variable and solve for the remaining one.
Elimination Method — Worked Example
Solve: 3x + 2y = 11 …(1) and 2x + 3y = 4 …(2)
Step 1 — Write in ax + by = c form (already done)
3x + 2y = 11 and 2x + 3y = 4
Step 2 — Make coefficients of x equal
(1) × 2 ⟹ 6x + 4y = 22
(2) × 3 ⟹ 6x + 9y = 12
Step 3 — Subtract to eliminate x (same sign → subtract)
6x + 4y = 22
6x + 9y = 12
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−5y = 10 ⟹ y = −2
Step 4 & 5 — Substitute y = −2 into (1)
3x + 2(−2) = 11 ⟹ 3x − 4 = 11 ⟹ 3x = 15 ⟹ x = 5
∴ x = 5, y = −2
Exercise 4.2 — Word Problems
Question 1
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹2000 per month, find their monthly incomes.
Setting Up Equations
Let incomes = 9x and 7x (in ratio 9:7)
Let expenditures = 4y and 3y (in ratio 4:3)
Income − Expenditure = Savings = ₹2000
9x − 4y = 2000 …(1)
7x − 3y = 2000 …(2)
Method 1 — Substitution
Solution by Substitution
From (1): 4y = 9x − 2000 ⟹ y = (9x − 2000) / 4
Substitute in (2): 7x − 3(9x − 2000)/4 = 2000
⟹ 28x − 3(9x − 2000) = 8000
⟹ 28x − 27x + 6000 = 8000
⟹ x = 2000
y = (9 × 2000 − 2000) / 4 = 16000 / 4 = 4000
∴ Incomes = 9(2000) = ₹18,000 and 7(2000) = ₹14,000
Method 2 — Elimination
Solution by Elimination
(1) × 7 ⟹ 63x − 28y = 14000
(2) × 9 ⟹ 63x − 27y = 18000
Subtract: −y = −4000 ⟹ y = 4000
Substitute in (2): 7x − 3(4000) = 2000 ⟹ 7x = 14000 ⟹ x = 2000
∴ Monthly incomes = ₹18,000 and ₹14,000
Question 2
The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number. How many such numbers are there?
Solution
Let units digit = x, tens digit = y
Original number = 10y + x, Reversed = 10x + y
Sum condition: (10y + x) + (10x + y) = 66
⟹ 11x + 11y = 66 ⟹ x + y = 6 …(1)
Digit difference: x − y = 2 …(2)
From (1): x = 6 − y. Substitute in (2): 6 − y − y = 2 ⟹ −2y = −4 ⟹ y = 2
x = 6 − 2 = 4
∴ Numbers are 42 and 24. There are two such numbers.
Question 3
The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Recall: Two angles are supplementary if they add up to 180°.
Solution
Let larger angle = x, smaller angle = y
x + y = 180° …(1) (supplementary)
x − y = 18° …(2) (differ by 18°)
Adding (1) + (2): 2x = 198° ⟹ x = 99°
From (1): y = 180° − 99° = 81°
∴ The angles are 99° and 81°.
Question 4
Taxi charges in Hyderabad: fixed charge + per-km charge. First 10 km costs ₹166; 15 km costs ₹256. (i) Find the fixed charge and charge per km. (ii) Find the cost for 25 km.
Note: The fixed charge covers the first 3 km. Beyond 3 km, per-km charge applies. So for 10 km, the variable distance = 10 − 3 = 7 km. For 15 km, variable distance = 15 − 3 = 12 km.
Solution — Part (i)
Let fixed charge = ₹x, charge per km = ₹y
x + 7y = 166 …(1)
x + 12y = 256 …(2)
Subtract (1) from (2): 5y = 90 ⟹ y = 18
From (1): x = 166 − 7(18) = 166 − 126 = 40
∴ Fixed charge = ₹40, Charge per km = ₹18
Solution — Part (ii)
For 25 km: variable distance = 25 − 3 = 22 km
Total charge = x + 22y = 40 + 22(18) = 40 + 396
∴ Charge for 25 km = ₹436
Question 5
A fraction equals 4/5 if 1 is added to both numerator and denominator. If 5 is subtracted from both numerator and denominator, the fraction equals 1/2. Find the fraction.
Solution
Let numerator = x, denominator = y (fraction = x/y)
Condition 1: (x+1)/(y+1) = 4/5
⟹ 5(x+1) = 4(y+1) ⟹ 5x + 5 = 4y + 4
⟹ 5x − 4y = −1 …(1)
Condition 2: (x−5)/(y−5) = 1/2
⟹ 2(x−5) = 1(y−5) ⟹ 2x − 10 = y − 5
⟹ 2x − y = 5 …(2)
(1) × 2 ⟹ 10x − 8y = −2
(2) × 5 ⟹ 10x − 5y = 25
Subtract: −3y = −27 ⟹ y = 9
From (2): 2x − 9 = 5 ⟹ 2x = 14 ⟹ x = 7
∴ The required fraction is 7/9.
Question 6
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. Find the speeds of both cars.
Key Formula: Distance = Speed × Time. When moving in the same direction, relative distance = difference of distances covered. When moving towards each other, relative distance = sum of distances covered.
Solution
Let speed of car from A = x km/hr, car from B = y km/hr
Case 1 (same direction, 5 hrs): 5x − 5y = 100 ⟹ x − y = 20 …(1)
Case 2 (opposite direction, 1 hr): x + y = 100 …(2)
Adding (1) + (2): 2x = 120 ⟹ x = 60
From (2): y = 100 − 60 = 40
∴ Speed of car from A = 60 km/hr, Speed of car from B = 40 km/hr.
Question 7
Two angles are supplementary. The larger angle is 3° less than twice the measure of the smaller angle. Find the measure of each angle.
Solution
Let larger angle = x, smaller angle = y
x + y = 180° …(1) (supplementary)
x = 2y − 3° …(2) (larger is 3° less than twice smaller)
Substitute (2) in (1): (2y − 3°) + y = 180°
⟹ 3y = 183° ⟹ y = 61°
x = 2(61°) − 3° = 122° − 3° = 119°
∴ The angles are 119° and 61°.
Question 8
A dictionary has a total of 1382 pages. It is broken into two parts. The second part has 64 pages more than the first part. How many pages are in each part?
Solution
Let pages in Part 1 = x, pages in Part 2 = y
x + y = 1382 …(1)
y = x + 64 …(2)
Substitute (2) in (1): x + (x + 64) = 1382
⟹ 2x = 1318 ⟹ x = 659
y = 659 + 64 = 723
∴ Part 1 has 659 pages and Part 2 has 723 pages.
Question 9
A chemist has two solutions of hydrochloric acid — one is 50% and the other is 80%. How much of each should be used to obtain 100 ml of a 68% solution?
Solution
Let quantity of 50% solution = x ml, 80% solution = y ml
x + y = 100 …(1) (total volume)
50% of x + 80% of y = 68% of 100
⟹ x/2 + 4y/5 = 68
⟹ 5x + 8y = 680 …(2) (multiply by 10)
(1) × 5 ⟹ 5x + 5y = 500
Subtract from (2): 3y = 180 ⟹ y = 60
From (1): x = 100 − 60 = 40
∴ 50% solution = 40 ml and 80% solution = 60 ml.
Question 10
You have ₹12,000 saved and want to invest it in two schemes yielding 10% and 15% interest. How much should be invested in each scheme to get an overall 12% interest?
Solution
Let amount at 10% = ₹x, amount at 15% = ₹y
x + y = 12000 …(1)
10% of x + 15% of y = 12% of 12000
⟹ x/10 + 3y/20 = 1440
⟹ 2x + 3y = 28800 …(2) (multiply by 20)
(1) × 2 ⟹ 2x + 2y = 24000
Subtract from (2): y = 4800
From (1): x = 12000 − 4800 = 7200
∴ Invest ₹7,200 at 10% and ₹4,800 at 15%.
Substitution vs Elimination — Quick Comparison
| Feature | Substitution Method | Elimination Method |
|---|---|---|
| Approach | Express one variable in terms of the other | Make one variable's coefficient equal, then add/subtract |
| Best when | One equation can easily give y = f(x) | Coefficients are easy to equalize by multiplication |
| Key step | Substitute expression into the other equation | Eliminate one variable by addition/subtraction |
| Result | Both methods give the same answer | Both methods give the same answer |