Problem Set 1 – Detailed Solutions | Linear Equations in Two Variables

Problem Set – 1

Question 1 • All 5 MCQs

Question 1 — Choose the correct alternative for each question.

(1) To draw graph of 4x + 5y = 19, find y when x = 1.

Solution

4(1) + 5y = 19
5y = 19 − 4
5y = 15
y = 3

Reason: Substitute the given value of x in the equation.

Answer: (B) 3

(2) For simultaneous equations, D_x = 49, D = 7. Find x.

Solution

By Cramer’s rule,
x = D_xD = 497 = 7

Reason: x = D_x/D when D ≠ 0.

Answer: (A) 7

(3) Find the value of 53−7−4.

Solution

5(−4) − 3(−7)
= −20 + 21
= 1

Reason: For a 2 × 2 determinant, ad − bc is used.

Answer: (D) 1

(4) For x + y = 3 and 3x − 2y − 4 = 0, find D.

Solution

Standard form: x + y = 3, 3x − 2y = 4.
D = 113−2
= 1(−2) − 1(3)
= −5

Answer: (C) −5

(5) ax + by = c and mx + ny = d, and an ≠ bm. Then these simultaneous equations have —

Solution

D = an − bm.
Given an ≠ bm, therefore D ≠ 0.

Reason: When D ≠ 0, a pair of simultaneous linear equations has one unique common solution.

Answer: (A) Only one common solution.

Problem Set – 1 • Question 2.1

Table completion

Question

Complete the following table to draw the graph of 2x − 6y = 3.

Solution

	First point	Second point
x	−5	32
y	−136	0
(x, y)	(−5, −136)	(32, 0)

Step 1: When x = −5
2(−5) − 6y = 3
−10 − 6y = 3
−6y = 3 + 10
−6y = 13
y = −136

Reason: Divide both sides by −6.

Step 2: When y = 0
2x − 6(0) = 3
2x = 3
x = 32

Reason: Substituting y = 0 gives the corresponding x-coordinate.

Answer: The ordered pairs are (−5, −13/6) and (3/2, 0).

Problem Set – 1 • Question 3.1

Graphical solution

(1) 2x + 3y = 12 ; x − y = 1

Solution

x − y = 1 → y = x − 1
2x + 3(x − 1) = 12
2x + 3x − 3 = 12
5x = 15 → x = 3
y = 3 − 1 → y = 2

Graphically, the two lines intersect at their common point.

Answer: (3, 2)

Problem Set – 1 • Question 3.2

Graphical solution

(2) x − 3y = 1 ; 3x − 2y + 4 = 0

Solution

x = 1 + 3y
3(1 + 3y) − 2y + 4 = 0
11y + 7 = 0
y = −711
x = 1 − 2111 = −1011

Reason: The common point where the two lines intersect gives the simultaneous solution.

Answer: (−10/11, −7/11)

Problem Set – 1 • Question 3.3

Graphical solution

(3) 5x − 6y + 30 = 0 ; 5x + 4y − 20 = 0

Solution

5x − 6y = −30
5x + 4y = 20
Subtract: 10y = 50
y = 5
5x + 20 = 20 → x = 0

Reason: The common point where the two lines intersect gives the simultaneous solution.

Answer: (0, 5)

Problem Set – 1 • Question 3.4

Graphical solution

(4) 3x − y − 2 = 0 ; 2x + y = 8

Solution

3x − y = 2
2x + y = 8
Add: 5x = 10
x = 2
4 + y = 8 → y = 4

Reason: The common point where the two lines intersect gives the simultaneous solution.

Answer: (2, 4)

Problem Set – 1 • Question 3.5

Graphical solution

(5) 3x + y = 10 ; x − y = 2

Solution

Add the equations:
4x = 12
x = 3
3 − y = 2 → y = 1

Reason: The common point where the two lines intersect gives the simultaneous solution.

Answer: (3, 1)

Problem Set – 1 • Question 4.1

Determinant

(1) 4327

Solution

= 4(7) − 3(2)
= 28 − 6
= 22

Reason: For a 2 × 2 determinant, use the rule ad − bc.

Answer: 22

Problem Set – 1 • Question 4.2

Determinant

(2) 5−2−31

Solution

= 5(1) − (−2)(−3)
= 5 − 6
= −1

Reason: For a 2 × 2 determinant, use the rule ad − bc.

Answer: −1

Problem Set – 1 • Question 4.3

Determinant

(3) 3−114

Solution

= 3(4) − (−1)(1)
= 12 + 1
= 13

Reason: For a 2 × 2 determinant, use the rule ad − bc.

Answer: 13

Problem Set – 1 • Question 5.1

Cramer’s method

(1) 6x − 3y = −10 ; 3x + 5y − 8 = 0

Solution

3x + 5y = 8.
D = 6−335 = 30 + 9 = 39.
D_x = −10−385 = −50 + 24 = −26.
D_y = 6−1038 = 48 + 30 = 78.
x = −2639 = −23, y = 7839 = 2.

Reason: By Cramer’s rule, x = D_x/D and y = D_y/D.

Answer: x = −2/3, y = 2

Problem Set – 1 • Question 5.2

Cramer’s method

(2) 4m − 2n = −4 ; 4m + 3n = 16

Solution

D = 4(3) − (−2)(4) = 20.
D_m = (−4)(3) − (−2)(16) = 20.
D_n = 4(16) − (−4)(4) = 80.
m = 20/20 = 1; n = 80/20 = 4.

Reason: Use Cramer’s rule: x = Dₓ/D and y = Dᵧ/D, when D ≠ 0.

Answer: m = 1, n = 4

Problem Set – 1 • Question 5.3

Cramer’s method

(3) 3x − 2y = 5/2 ; x/3 + 3y = −4/3

Solution

Multiply both equations by 2 and 3 respectively:
6x − 4y = 5
x + 9y = −4.
D = 6(9) − (−4)(1) = 58.
D_x = 5(9) − (−4)(−4) = 29.
D_y = 6(−4) − 5(1) = −29.
x = 29/58 = 1/2; y = −29/58 = −1/2.

Reason: Use Cramer’s rule: x = Dₓ/D and y = Dᵧ/D, when D ≠ 0.

Answer: x = 1/2, y = −1/2

Problem Set – 1 • Question 5.4

Cramer’s method

(4) 7x + 3y = 15 ; 12y − 5x = 39

Solution

Write: −5x + 12y = 39.
D = 7(12) − 3(−5) = 99.
D_x = 15(12) − 3(39) = 63.
D_y = 7(39) − 15(−5) = 348.
x = 63/99 = 7/11; y = 348/99 = 116/33.

Reason: Use Cramer’s rule: x = Dₓ/D and y = Dᵧ/D, when D ≠ 0.

Answer: x = 7/11, y = 116/33

Problem Set – 1 • Question 5.5

Cramer’s method

(5) x+y−82 = x+2y−143 = 3x−y4

Solution

Let the common value be k.
x + y = 2k + 8
x + 2y = 3k + 14
3x − y = 4k.
From first two: y = k + 6 and x = k + 2.
3(k+2) − (k+6) = 4k → 2k = 4k → k = 0.
Therefore x = 2 and y = 6.

Reason: Use Cramer’s rule: x = Dₓ/D and y = Dᵧ/D, when D ≠ 0.

Answer: x = 2, y = 6

Problem Set – 1 • Question 6.1

Simultaneous equations

(1) 2x + 23y = 16 ; 3x + 2y = 0

Solution

Let 1x = m, 1y = n.
12m + 4n = 1 ...(1)
3m + 2n = 0 ...(2)
2×(2): 6m + 4n = 0.
(1) − (2×2): 6m = 1 → m = 16.
Then n = −14.
Thus x = 6 and y = −4.

Reason: Substitution of suitable reciprocal variables reduces the equations to a pair of linear equations.

Answer: (6, −4)

Problem Set – 1 • Question 6.2

Simultaneous equations

(2) 72x+1 + 13y+2 = 27 ; 132x+1 + 7y+2 = 33

Solution

Let m = 1/(2x+1), n = 1/(y+2).
7m + 13n = 27 ...(1)
13m + 7n = 33 ...(2)
Add: m + n = 3.
Subtract: m − n = 1.
Hence m = 2, n = 1.
2x + 1 = 12 → x = −14.
y + 2 = 1 → y = −1.

Reason: Substitution of suitable reciprocal variables reduces the equations to a pair of linear equations.

Answer: (−1/4, −1)

Problem Set – 1 • Question 6.3

Simultaneous equations

(3) 148x + 231y = 527xy ; 231x + 148y = 610xy

Solution

Multiply by xy:
231x + 148y = 527 ...(1)
148x + 231y = 610 ...(2)
Add: 379x + 379y = 1137 → x + y = 3.
Subtract: 83x − 83y = −83 → x − y = −1.
Add the two: 2x = 2 → x = 1.
Then y = 2.

Reason: Substitution of suitable reciprocal variables reduces the equations to a pair of linear equations.

Answer: (1, 2)

Problem Set – 1 • Question 6.4

Simultaneous equations

(4) 7x−2yxy = 5 ; 8x+7yxy = 15

Solution

Rewrite: 7/x + 8/y = 15 and −2/x + 7/y = 5.
Let 1/x = m and 1/y = n.
−2m + 7n = 5 ...(1)
7m + 8n = 15 ...(2)
Solving gives m = 1 and n = 1.
Therefore x = 1 and y = 1.

Reason: Substitution of suitable reciprocal variables reduces the equations to a pair of linear equations.

Answer: (1, 1)

Problem Set – 1 • Question 6.5

Simultaneous equations

(5) 12(3x+4y) + 15(2x−3y) = 14 ; 53x+4y − 22x−3y = −32

Solution

Let m = 1/(3x+4y), n = 1/(2x−3y).
10m + 4n = 5 ...(1)
10m − 4n = −3 ...(2)
Add: 20m = 2 → m = 110.
Then n = 1.
So 3x+4y=10 and 2x−3y=1.
3×first: 9x+12y=30.
4×second: 8x−12y=4.
Add: 17x=34 → x=2; then y=1.

Reason: Substitution of suitable reciprocal variables reduces the equations to a pair of linear equations.

Answer: (2, 1)

Problem Set – 1 • Question 7.1

Word problem

(1) Two-digit number

Question

A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.

Solution

Let unit digit = x and tens digit = y.
Original number = 10y + x.
Interchanged number = 10x + y.
10y+x+10x+y=143 → x+y=13 ...(I)
x=y+3 → x−y=3 ...(II)
Add: 2x=16 → x=8.
y=13−8=5.

Reason: The equations are formed directly from the conditions given in the word problem.

Original number = 10×5+8 = 58.

Problem Set – 1 • Question 7.2

Word problem

(2) Tea and sugar

Question

Kantabai bought 112 kg tea and 5 kg sugar. With ₹50 return fare, total expense was ₹700. Next month she ordered 2 kg tea and 7 kg sugar for ₹880. Find the rates.

Solution

Let tea = ₹x/kg, sugar = ₹y/kg.
32x + 5y + 50 = 700 → 3x+10y=1300 ...(I)
2x+7y=880 ...(II)
2×(I): 6x+20y=2600.
3×(II): 6x+21y=2640.
Subtract: y=40.
2x+280=880 → x=300.

Reason: The equations are formed directly from the conditions given in the word problem.

Tea = ₹300/kg; Sugar = ₹40/kg.

Problem Set – 1 • Question 7.3

Word problem

(3) Number of notes

Question

Anushka had ₹100 and ₹50 notes. Total amount was ₹2500. On interchanging their numbers, she would receive ₹500 less. Find the number of each note.

Solution

Let ₹100 notes = x and ₹50 notes = y.
100x+50y=2500 → 2x+y=50 ...(I)
After interchange: 50x+100y=2000 → x+2y=40 ...(II)
2×(II): 2x+4y=80.
Subtract (I): 3y=30 → y=10.
2x+10=50 → x=20.

Reason: The equations are formed directly from the conditions given in the word problem.

₹100 notes = 20; ₹50 notes = 10.

Problem Set – 1 • Question 7.4

Word problem

(4) Present ages

Question

Sum of present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times Savita’s age. Find their present ages.

Solution

Let Manish = x years, Savita = y years.
x+y=31 ...(I)
x−3=4(y−3) → x−4y=−9 ...(II)
Subtract: 5y=40 → y=8.
x=31−8=23.

Reason: The equations are formed directly from the conditions given in the word problem.

Manish = 23 years; Savita = 8 years.

Problem Set – 1 • Question 7.5

Word problem

(5) Skilled and unskilled workers

Question

The ratio of salary of skilled and unskilled workers is 5:3. Total salary for one day is ₹720. Find daily wages.

Solution

Let skilled = ₹x and unskilled = ₹y.
3x−5y=0 ...(I)
x+y=720 ...(II)
5×(II): 5x+5y=3600.
Add (I): 8x=3600 → x=450.
y=720−450=270.

Reason: The equations are formed directly from the conditions given in the word problem.

Skilled = ₹450/day; Unskilled = ₹270/day.

Problem Set – 1 • Question 7.6

Word problem

(6) Hamid and Joseph

Question

A and B are 30 km apart. Hamid travels A→B and Joseph B→A. They meet after 20 minutes. If Joseph travels in the opposite direction, Hamid catches him after 3 hours. Find their speeds.

Solution

Let Hamid = x km/h and Joseph = y km/h.
20 min = 13 h.
x3 + y3 = 30 → x+y=90 ...(I)
In 3 h: 3x−3y=30 → x−y=10 ...(II)
Add: 2x=100 → x=50.
y=90−50=40.

Reason: The equations are formed directly from the conditions given in the word problem.

Hamid = 50 km/h; Joseph = 40 km/h.