Updated on: 2026-03-31 | Author: Aarti Kulkarni

Algebra (Math 1) Chapter 1 Linear Equations In Two Variables Solutions

Q1. `|(3, 5),(2, x)|` = 2 ∴ x = ______

4

Explanation:

`|(3, 5),(2, x)|` = 2

3x - 10 = 2
3x = 2 + 10
3x = 12

x = `12/3`

∴ x = 4

Q2. A person starts a job with a fixed salary and yearly increment. After 4 years his salary is ₹ 15000 and after 10 years it becomes ₹ 18000. Then find his monthly salary and increment

Let the fixed salary be ₹ x and the annual increment be ₹ y.

Then, salary after 4 years = x + 4y
salary after 10 years = x + 10y

According to the first condition,

x + 4y = 15000 ......(i)

According to the second condition:

x + 10y = 18000 ......(ii)

Subtracting equation (i) from (ii), we get

x + 10y = 18000
  • x + 4y = 15000

− − −____

6y = 3000
∴ y = `3000/6` = 500
Substituting y = 500 in equation (i), we get
x + 4(500) = 15000
∴ x + 2000 = 15000
∴ x = 13000
∴ The monthly salary is ₹ 13,000 and yearly increment is ₹ 500.

Q3. A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train was slower by 6 km/h it would have taken 6 hours more than the scheduled time. Find the length of the journey.

Let the speed of the train be x km/hr.

Let the time taken to travel certain distance be y hrs.

We know that, speed × time = distance
∴ Distance = xy km

According to the first condition, if the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time.

∴ (x + 6)(y – 4) = xy
∴ xy – 4x + 6y – 24 = xy
∴ – 4x + 6y – 24 = 0
∴ 2x – 3y = –12 ......(i)

According to the second condition, if the train was slower by 6 km/hr, it would have taken 6 hours more than the scheduled time.

∴ (x – 6)(y + 6) = xy
∴ xy + 6x – 6y – 36 = xy
∴ 6x – 6y – 36 = 0
∴ x – y = 6 ......(ii)

Multiplying both sides by 2, we get

2x – 2y = 12 ......(iii)

Subtracting equation (iii) from (i), we get

2x – 3y = –12

2x – 2y = 12

– + –

– y = – 24

∴ y = 24
Substituting y = 24 in equation (ii), we get
x – 24 = 6
∴ x = 30
∴ Length of the journey = xy

= 30 × 24

= 720 km

Q4. Ajay is younger than Vijay by 3 years. The sum of their ages is 25 years, what is the age of Ajay

Let the ages of Ajay and Vijay be x years and y years respectively.

According to the first condition,

Ajay is younger than Vijay by 3 years.

∴ y – x = 3
i.e., – x + y = 3 .....(i)

According to the second condition,

Sum of their ages is 25.

∴ x + y = 25 .....(ii)

Subtracting equation (i) from (ii), we get

x + y = 25
− x + y = 3

+ − −

2x = 22
∴ x = `22/2` = 11
∴ Ajay’s age is 11 years.

Alternate Method:

Let the age of Ajay be x years.

Ajay is younger than Vijay by 3 years.

∴ Vijay’s age = x + 3

According to the given condition,

Sum of their ages is 25.

∴ x + x + 3 = 25
∴ 2x + 3 = 25
∴ 2x = 22
∴ x = `22/2` = 11
∴ Ajay’s age is 11 years.

Q5. Complete the activity.

Length of base = y cm and length of side = x cm

Base of the triangle is 3cm more than twice its congruent sides.

∴ y = 2x + 3
∴ y – 2x = 3
i.e., – 2x + y = 3 ...(i)

Perimeter of the triangle is 35 cm.

∴ x + x + y = 35
∴ 2x + y = 35 ...(ii)

Adding equations (i) and (ii), we get

– 2x + y = 3
+ 2x + y = 35
2y = 38
∴ y = `38/2` = 19 cm
∴ Length of base = 19 cm

Q6. Complete the activity to find the value of x. 3x + 2y = 11 …(i) and 2x + 3y = 4 …(ii) Solution: Multiply equation (i) by _______ and equation (ii) by _______. `square` × (3x + 2y = 11) ∴ 9x + 6y = 33 …(iii) `square` × (2x + 3y = 4) ∴ 4x + 6y = 8 …(iv) Subtract (iv) from (iii), `square` x = 25 ∴ x = `square`

3x + 2y = 11 …(i) and 2x + 3y = 4 …(ii)

Multiply equation (i) by 3 and equation (ii) by 2.

3 × (3x + 2y = 11) ∴ 9x + 6y = 33 …(iii)
2 × (2x + 3y = 4) ∴ 4x + 6y = 8 …(iv)

Subtract (iv) from (iii),

5 x = 25
∴ x = `25/5`
∴ x = 5

Q7. Complete the following table to draw the graph of 3x − 2y = 18 x 0 4 2 −1 y − 9 ______ ______ ______ (x, y) (0, −9) (______, _______) (______, _______) ______

Substituting x = 4 in 3x – 2y = 18,
We get 3(4) – 2y = 18
∴ 12 – 2y = 18
∴ 2y = 12 – 18 = – 6
∴ y = `(-6)/2` = – 3
Substituting x = 2 in 3x – 2y = 18,
We get 3(2) – 2y = 18
∴ 6 – 2y = 18
∴ 2y = 6 – 18 = – 12
∴ y = `(-12)/2` = – 6
Substituting x = – 1 in 3x – 2y = 18,
We get 3(–1) – 2y = 18
∴ – 3 – 2y = 18
∴ 2y = – 3 – 18

= – 21

∴ y = `(-21)/2`
x 0 4 2 −1
y − 9 − 3 − 6 `(-21)/2`
(x, y) (0, −9) (4, − 3)  (2, − 6) `(-1, (-21)/2)`

Q8. Complete the table to draw the graph of 2x – 3y = 3, x − 6 `square` y `square` 1 (x, y) `square` `square`

Diagram: Refer textbook

Q9. Decide whether (0, 2) is the solution of the equation 5x + 3y = 6

L.H.S. = 5x + 3y

= 5(0) + 3(2)

= 0 + 6

= 6

= R.H.S

∴ (0, 2) is the solution of the given equation.

Q10. Decide whether x = 2 and y = − 1 is the solution of the equation 2x + y = 3 or not?

L.H.S. = 2x + y

= 2(2) – 1

= 4 – 1

= 3

= R.H.S

∴ x = 2 and y = – 1 is the solution of the equation 2x + y = 3.

Q11. Difference between two numbers is 3. The sum of three times the bigger number and two times the smaller number is 19. Then find the numbers

Let the bigger number be x and the smaller number be y.

According to the first condition, difference between two numbers is 3.

∴ x – y = 3 .....(i)

According to the second condition, the sum of three times the bigger number and two times the smaller number is 19.

∴ 3x + 2y = 19 .....(ii)

Multiplying equation (i) by 2, we get

2x – 2y = 6 .....(iii)

Adding equations (ii) and (iii), we get

3x + 2y = 19
+ 2x – 2y = 6
5x = 25
∴ x = `25/5`
∴ x = 5
Substituting x = 5 in equation (i), we get
x – y = 3
∴ 5 – y = 3
∴ 5 – 3 = y
∴ y = 2
∴ The required numbers are 5 and 2.

Q12. Find the value of `|(5, 2),(0, -1)|`

`|(5, 2),(0, -1)|` = 5 × (–1) – 2 × 0

= – 5 – 0

= – 5

Q13. Find the value of D x for the equation 4x + 3y = 19 and 4x − 3y = −11

− 24

Explanation:

Here, a1 = 4, b1 = 3, c1 = 19
a2 = 4, b2 = -3, c2 = -11
∴ Dx = `|(c_1, b_1),(c_2, b_2)| = |(19, 3),(-11, -3)|`

= -57 + 33

= -24

Q14. For an A.P., t 17 = 54 and t 9 = 30 find the first term(a) and common difference(d)

t17 = 54 and t9 = 30 ......[Given]
tn = a + (n – 1)d
t17 = 54
∴ a + (17 – 1)d = 54
∴ a + 16d = 54 ......(i)
t9 = 30
∴ a + (9 – 1)d = 30
∴ a + 8d = 30 ......(ii)

Subtracting equation (ii) from (i), we get

a + 16d = 54
a + 8d = 30

− − −

8d = 24
∴ d = `24/8` = 3
Substituting d = 3 in equation (ii), we get
a + 8(3) = 30
∴ a + 24 = 30
∴ a = 6
∴ first term (a) = 6 and common difference (d) = 3

Q15. For equations 5x + 3y + 11 = 0 and 2x + 4y = −10 find D.

14

Explanation:

Here, a1 = 5, b1 = 3
a2 = 2, b2 = 4
∴ D = `|(a_1, b_1),(a_2, b_2)| = |(5, 3),(2, 4)|`

= 20 − 6

= 14

Q16. For the equation 3x − 2𝑦 = 17, find the value of x when y = −1 and find the value of y when x = 3

Substituting y = – 1 in 3x – 2y = 17, we get
3x – 2(–1) = 17
∴ 3x + 2 = 17
∴ 3x = 17 – 2
∴ 3x = 15
∴ x = `15/3` = 5
Substituting x = 3 in 3x – 2y = 17, we get
3(3) – 2y = 17
∴ 9 – 2y = 17
∴ 2y = 9 – 17
∴ 2y = – 8
∴ y = `(-8)/2` = – 4
∴ The value of x is 5 when y = −1 and the value of y is – 4 when x = 3.

Q17. For the equation 4x + 5y = 20 find y when x = 0

Substituting x = 0 in 4x + 5y = 20, we get
4(0) + 5y = 20
∴ 5y = 20
∴ y = `20/5`

= 4

Q18. For the equation a + 2b = 7, find a when b = 4

Substituting b = 4 in a + 2b = 7, we get
a + 2(4) = 7
∴ a + 8 = 7
∴ a = 7 – 8

= – 1

Q19. For the equations with variables x and y, if Dx = 26, Dy = −39 and D = 13 then x = ?

2

Explanation:

Dx = 26, Dy = −39 and D = 13

Using Cramer's rule,

x = `"Dx"/"x" = 26/13 = 2`
x = 2

Q20. For the equations y + 2x = 19 and 2x – 3y = − 3, find the value of D.

The given equations are

y + 2x = 19
i.e., 2x + y = 19

2x – 3y = – 3

Comparing the given equations with a1x + b1y = c1 and a2x + b2y = c2,

we get

a1 = 2, b1 = 1, c1 = 19
a2 = 2, b2 = – 3, c2 = – 3
∴ D = `|("a"_1, "b"_1),("a"_2, "b"_2)|`

= `|(2, 1),(2, -3)|`

= 2 × (– 3) – 1 × 2

= – 6 – 2

= – 8

Q21. Form the simultaneous linear equations using the determinants D = `|(4, -3),(2, 5)|`, D x = `|(5, -3),(9, 5)|`, D y = `|(4, 5),(2, 9)|`

If a1x + b1y = c1 and a2x + b2y = c2 are linear equations in two variables, then

D = `|("a"_1, "b"_1),("a"_2, "b"_2)|`, Dx = `|("c"_1, "b"_1),("c"_2, "b"_2)|`, Dy = `|("a"_1, "c"_1),("a"_2, "c"_2)|` ......(i)

Given, D = `|(4, -3),(2, 5)|`, Dx = `|(5, -3),(9, 5)|`, Dy = `|(4, 5),(2, 9)|`

Comparing these determinants with equation (i), we get

a1 = 4, b1 = – 3, c1 = 5
a2 = 2, b2 = 5, c2 = 9
∴ The required equations are 4x – 3y = 5 and 2x + 5y = 9.

Q22. From the railway station I took a rickshaw to go home. It is decided that I have to pay ₹ X for the first kilometre and for each kilometre ₹ Y for the next. For 10 kilometres the fare is ₹ 40 and for 16 kilometres fare is ₹ 58. Find the fare for the first kilometre

I pay ₹ X for the first kilometre and for each kilometre ₹ Y for the next.

For 10 kilometres, the fare will be ₹ X + 9Y.

For 16 kilometres, the fare will be ₹ X + 15Y.

For 10 kilometres, the fare is ₹ 40.

∴ X + 9Y = 40 ......(i)

For 16 kilometres, the fare is ₹ 58.

X + 15Y = 58 ......(ii)

Subtracting equation (i) from (ii), we get

X + 15Y = 58
X + 9Y = 40

− − −

6Y = 18
∴ Y = `18/6` = 3
Substituting Y = 3 in equation (i), we get
X + 9(3) = 40
∴ X + 27 = 40
∴ X = 40 – 27 = 13
∴ The fare for the first kilometre is ₹ 13.

Q23. I held a number 75 in my mind. Write any condition showing the relation between their digits. Write the condition showing the relation between the number and the number obtained by interchanging the digits

Answer

Condition showing the relation between their digits.

The digit in the ten’s place is 2 more than the digit in the unit’s place.

Condition showing relation between the number and the number obtained by interchanging the digits.

The sum of the number and the number obtained by interchanging the digits is 132.

Q24. If (2, 0) is the solution of 2x + 3y = k then finds the value of k by completing the activity Solution: (2, 0) is solution of the equation 2x + 3y = k Putting x = `square` and y = `square` ∴ `2(square) + 3 xx 0` = k ∴ 4 + 0 = k ∴ k = `square`

(2, 0) is solution of the equation 2x + 3y = k
Putting x = 2 and y = 0
2(2) + 3 × 0 = k
∴ 4 + 0 = k
∴ k = 4

Q25. If (2, −5) is the solution of the equation 2x − ky = 14, then find k = ?

(2, – 5) is the solution of 2x – ky = 14.
Substituting x = 2, y = – 5 in 2x – ky = 14, we get
2(2) – k(– 5) = 14
∴ 4 + 5k = 14
∴ 5k = 14 – 4 = 10
∴ k = `10/5` = 2

Q26. If 49x – 57y = 172 and 57x – 49y = 252 then x + y = ?

10

Explanation:

49x - 57y = 172
57x - 49y = 252
  • + -
  • 8x - 8y = - 80
∴ x + y = `(-80)/(-8) = 10`

Q27. If 52x + 65y = 183 and 65x + 52y = 168, then find x + y = ?

52x + 65y = 183 .......(i)
65x + 52y = 168 ......(ii)

Adding equations (i) and (ii), we get

52x + 65y = 183
+ 65x + 52y = 168
117x + 117y = 351
∴ 117x + 117y = 351
∴ 117(x + y) = 351
∴ x + y = `351/117` ......[Dividing both sides by 117]
∴ x + y = 3

Q28. If `"a"/4 + "b"/3` = 4, write the equation in standard form

Answer

`"a"/4 + "b"/3` = 4

∴ `(3"a" + 4"b")/12` = 4
∴ 3a + 4b = 12 × 4
∴ 3a + 4b = 48

Q29. If Dx = 24 and x = – 3, then find the value of D

x = `"D"_x/"D"`

∴ D = `"D"_x/x`

= `24/(-3)`

= – 8

Q30. If x + 2y = 5 and 2x + y = 7, then find the value of x + y

x + 2y = 5
+ 2x + y = 7
3x + 3y = 12
∴ x + y = `12/3` = 4

Q31. In the equation 2x – y = 2 if x = 3, then find y = ?

Substituting x = 3 in 2x – y = 2,
we get 2(3) – y = 2
∴ 6 – y = 2
∴ y = 6 – 2
∴ y = 4

Q32. Show the condition using variables x and y: Two numbers differ by 3

x – y = 3, where x > y

OR

y – x = 3, where y > x

Q33. Solve 0.4x + 0.3y = 1.7; 0.7 x − 0.2y = 0.8

0.4x + 0.3y = 1.7
∴ 4x + 3y = 17 ......(i)[Multiplying both sides by 10]
0.7x – 0.2y = 0.8
∴ 7x – 2y = 8 ......(ii)[Multiplying both sides by 10]

Multiplying equation (i) by 2, we get

8x + 6y = 34 ......(iii)

Multiplying equation (ii) by 3, we get

21x – 6y = 24 ......(iv)

Adding equations (iii) and (iv), we get

8x + 6y = 34
+21x – 6y = 24
29x = 58
∴ x = `58/29` = 2
Substituting x = 2 in equation (i), we get
4(2) + 3y = 17
∴ 8 + 3y = 17
∴ 3y = 9
∴ y = `9/3` = 3
∴ x = 2 and y = 3 is the solution of the equation 0.4x + 0.3y = 1.7 and 0.7x – 0.2y = 0.8.

Q34. Solve: 4m – 2n = – 4, 4m + 3n = 16

The given equations are

4m − 2n = – 4 ...(i)

4m + 3n = 16 ...(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with a1m + b1n = c1 and a2m + b2n = c2, we get
a1 = 4, b1 = − 2, c1 = − 4 and
a2 = 4, b2 = 3, c2 = 16
∴ D = `|("a"_1, "b"_1),("a"_2, "b"_2)|`

= `|(4, -2),(4, 3)|`

= (4 × 3) − (– 2 × 4)

= 12 – (– 8)

= 12 + 8

= 20 ≠ 0

Dm = `|("c"_1, "b"_1),("c"_2, "b"_2)|`

= `|(-4, -2),(16, 3)|`

= (–4 × 3) – (– 2 × 16)

= –12 – (– 32)

= –12 + 32

= 20

Dn = `|("a"_1, "c"_1),("a"_2, "c"_2)|`

= `|(4, -4),(4, 16)|`

= (4 × 16) – (– 4 × 4)

= 64 – (– 16)

= 64 + 16

= 80

∴ By Cramer’s rule, we get

m = `("D"_"m")/"D"` and n = `("D"_"n")/"D"`

∴ m = `20/20` and n = `80/20`
∴ m = 1 and n = 4
∴ (m, n) = (1, 4) is the solution of the given equations.

Q35. Solve: 99x + 101y = 499, 101x + 99y = 501

The given equations are

99x + 101y = 499 .....(i)
101x + 99y = 501 .....(ii)

Adding equations (i) and (ii), we get

99x + 101y = 499
+ 101x + 99y = 501
200x + 200y = 1000
∴ x + y = `1000/200` ......[Dividing both sides by 200]
∴ x + y = 5 ......(iii)

Subtracting equation (ii) from (i), we get

99x + 101y = 499
101x + 99y = 501

− − −

−2x + 2y =−2

∴ x – y = `(-2)/(-2)` ......[Dividing both sides by – 2]
∴ x − y = 1 .......(iv)

Adding equations (iii) and (iv), we get

x + y = 5
+ x − y = 1
2x = 6
∴ x = `6/2` = 3
Substituting x = 3 in equation (iii), we get
x + y = 5
∴ 3 + y = 5
∴ y = 5 − 3 = 2
∴ (x, y) = (3, 2) is the solution of the given equations.

Q36. Solve by Cramer’s rule, 3x – 4y = 10, 4x + 3y = 5

The given equations are

3x – 4y = 10 ......(i)
4x + 3y = 5 .....(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with a1x + b1y = c1 and a2x + b2y = c2, we get
a1 = 3, b1 = – 4, c1 = 10 and
a2 = 4, b2 = 3, c2 = 5
∴ D = `|("a"_1, "b"_1),("a"_2, "b"_2)|`

= `|(3, -4),(4, 3)|`

= (3 × 3) – (– 4 × 4)

= 9 − (−16)

= 9 + 16

= 25 ≠ 0

Dx = `|("c"_1, "b"_1),("c"_2, "b"_2)|`

= `|(10, -4),(5, 3)|`

= (10 × 3) – (– 4 × 5)

= 30 − (−20)

= 30 + 20

= 50

Dy = `|("a"_1, "c"_1),("a"_2, "c"_2)|`

= `|(3, 10),(4, 5)|`

= (3 × 5) – (10 × 4)

= 15 – 40

= – 25

∴ By Cramer’s rule, we get

x = `("D"_x)/"D"` and y = `("D"_y)/"D"`

∴ x = `50/25` and y = `(-25)/25`
∴ x = 2 and y = –1
∴ (x, y) = (2, –1) is the solution of the given equations.

Q37. Solve the following equations by graphical method, x − y = 1, 2x + y = 8

Diagram: Refer textbook

Q38. Solve the following to find the value of following determinant. `|(3, -2),(4, 5)| = 3 xx square - square xx 4 = square + 8 = square`

`|(3, -2),(4, 5)|` = 3 × \[\boxed{5}\] − \[\boxed{-2}\] × 4 = \[\boxed{15}\] + 8 = \[\boxed{23}\]

Q39. Solve the given simultaneous equations graphically x + y = 5 and y = 5

Diagram: Refer textbook

Q40. State with reason whether the equation 3x 2 − 7y = 13 is a linear equation with two variables?

Answer

No

Reason: The degree of variable x is 2.

Q41. State with reason whether the point (3, −2) will lie on the graph of the equation 5m – 3n = − 21

Answer

L.H.S. = 5m – 3n

= 5(3) – 3(– 2)

= 15 + 6

= 21

≠ R.H.S

∴ The point (3, – 2) does not lie on the graph of 5m – 3n = – 21.

Q42. The cost of the book is 5 rupees more than twice the cost of a pen. Show this using linear equation by taking cost of book(x) and cost of a pen(y)

x = 2y + 5
∴ x – 2y = 5

Q43. The difference between an angle and its complement is 10° find measure of the larger angle.

Let the angle be x and its complement be y, where x > y.

Sum of complementary angles is 90°.

∴ x + y = 90° ......(i)

According to the given condition,

The difference between an angle and its complement is 10°

∴ x – y = 10° ......(ii)

Adding equations (i) and (ii), we get

x + y = 90°
+ x – y = 10°
2x = 100°
∴ x = `(100^circ)/2` = 50°
∴ The larger angle is 50°.

Q44. The graph of the equations 2x − y − 4 = 0 and x + y + 1 = 0 intersect each other in point P(a, b), then find the coordinates of P?

Diagram: Refer textbook

Q45. The length of the rectangle is 5 more than twice its breadth. The perimeter of a rectangle is 52 cm, then find the length of the rectangle

Let the length and breadth of the rectangle be x cm and y cm respectively.

According to the first condition,

The length of the rectangle is 5 more than twice its breadth.

∴ x = 2y + 5
∴ x – 2y = 5 ......(i)

According to the second condition,

Perimeter of rectangle is 52 cm.

∴ 2(x + y) = 52
∴ 2x + 2y = 52 .....(ii)

Adding equations (i) and (ii), we get

x – 2y = 5
+ 2x + 2y = 52
3x = 57
∴ x = `57/3` = 19
∴ The length of the rectangle is 19 cm.

Q46. The semi perimeter of a rectangular shape garden is 36 m. The length of the garden is 4 m more than its breadth. Find the length and the breadth of the garden

Let the length and breadth of the garden be x m and y m respectively.

According to the first condition,

The semi perimeter of a rectangular shape garden is 36 m.

∴ x + y = 36 ......(i)

According to the second condition,

The length of the garden is 4 m more than its breadth.

∴ x = y + 4
∴ x – y = 4 ......(ii)

Adding equations (i) and (ii), we get

x + y = 36
+ x – y = 4
2x = 40
∴ x = `40/2` = 20
Substituting x = 20 in equation (i), we get
20 + y = 36
∴ y = 36 – 20 = 16
∴ The length and breadth of the garden are 20 m and 16 m respectively.

Q47. The solution of the equation 2x – y = 2 is ______

(2, 2)

Q48. The solution of the equation ax + by + 5 = 0 and bx − ay − 12 = 0 is (2, – 3). Find the values of a and b

Since (2, – 3) is the solution of ax + by + 5 = 0 and bx – ay – 12 = 0, the point (x, y) = (2, – 3) satisfies the given equations.
ax + by + 5 = 0
∴ ax + by = – 5 ......(i)
bx – ay – 12 = 0
∴ bx – ay = 12 ......(ii)
Putting x = 2 and y = – 3 in equations (i) and (ii), we get

2a – 3b = – 5 ......(iii)

2b + 3a = 12
i.e., 3a + 2b = 12 ......(iv)

Multiplying equation (iii) by 2, we get

4a – 6b = – 10 .......(v)

Multiplying equation (iv) by 3, we get

9a + 6b = 36 .......(vi)

Adding equations (v) and (vi), we get

4a – 6b = –10

+ 9a + 6b = 36
13a = 26
∴ a = `26/13` = 2
Substituting a = 2 in equation (iv), we get
3(2) + 2b = 12
∴ 6 + 2b = 12
∴ 2b = 6
∴ b = `6/2` = 3
∴ a = 2 and b = 3

Q49. The solution of the equation x − y = 10 and x + y = 70 is ______

(40, 30)

Q50. The sum of the two-digit number and the number obtained by interchanging the digits is 132. The digit in the ten’s place is 2 more than the digit in the unit’s place. Complete the activity to find the original number. Activity: Let the digit in the unit’s place be y and the digit in the ten’s place be x. ∴ The number = 10x + y ∴ The number obtained by interchanging the digits = `square` ∴ The sum of the number and the number obtained by interchanging the digits = 132 ∴ 10x + y + 10y + x = `square` ∴ x + y = `square` .....(i) By second condition, Digit in the ten’s place = digit in the unit’s place + 2 ∴ x – y = 2 ......(ii) Solving equations (i) and (ii) ∴ x = `square`, y = `square` Ans: The original number = `square`

Let the digit in the unit’s place be y and the digit in the ten’s place be x.

∴ The number = 10x + y
∴ The number obtained by interchanging the digits = 10y + x
The sum of the number and the number obtained by interchanging the digits = 132
∴ 10x + y + 10y + x = 132
∴ 11x + 11y = 132
∴ x + y = `132/11` ......[Dividng both side by 11]
∴ x + y = 12 .....(i)

By second condition,

Digit in the ten’s place = digit in the unit’s place + 2
∴ x = y + 2
∴ x – y = 2 ......(ii)

Adding equations (i) and (ii), we get

x + y = 12
+ x – y = 2
2x = 14
∴ x = `14/2` = 7
Substituting x = 7 in equation (i), we get
7 + y = 12
∴ y = 12 – 7 = 5

Solving equations (i) and (ii),

∴ x = 7, y = 5
The original number = 10x + y

= 10(7) + 5

= 70 + 5

= 75

Q51. To draw the graph of 4x + 5y = 19, if x = 1 is taken then what will be the value of y?

3

Q52. To find the values of x and y for the equations x − 2y = 5 and 2x + 3y = 10 complete the activity. D = `|(1, -2),(2, 3)|` = 3 + 4 = 7 D x = `|(5, -2),(10, 3)| = square` D y = `|(1, 5),(2, 10)| = square` By Cramer’s rule x = `("D"_x)/"D" = square,` y = `("D"_y)/"D" = square`

D = `|(1, -2),(2, 3)|` = 3 + 4 = 7
Dx = `|(5, -2),(10, 3)|` = 15 + 20 = 35
Dy = `|(1, 5),(2, 10)|` = 10 – 10 = 0

By Cramer’s rule,

x = `("D"_x)/"D" = 35/7` = 5, y = `("D"_y)/"D" = 0/7` = 0

Q53. Using the determinants given below form two linear equations and solve them. D = `|(5, 7),(2, -3)|`, Dy = `|(5, 4),(2, -10)|`

If a1x + b1y = c1 and a2x + b2y = c2 are linear equations in two variables, then

D = `|("a"_1, "b"_1),("a"_2, "b"_2)|`

Dx = `|("c"_1, "b"_1),("c"_2, "b"_2)|`

Dy = `|("a"_1, "c"_1),("a"_2, "c"_2)|` ......(i)

Given, D = `|(5, 7),(2, -3)|`

Dy = `|(5, 4),(2, -10)|`

Comparing these determinants with equation (i), we get

a1 = 5, b1 = 7, c1 = 4
a2 = 2, b2 = –3, c2 = –10
∴ The equations are
5x + 7y = 4 ......(ii)

2x– 3y = –10 ......(iii)

Multiplying equation (ii) by 3 and equation (iii) by 7, we get

15x + 21y = 12 ......(iv)

14x – 21y = –70 ......(v)

Adding equations (iv) and (v), we get

15x + 21y = 12

+ 14x – 21y = – 70

29x = – 58

∴ x = `-58/29` = – 2

Substituting x = –2 in equation (ii), we get

5(–2) + 7y = 4
∴ –10 + 7y = 4
∴ 7y = 14
∴ y = `14/7` = 2
∴ x = –2 and y = 2 is the solution of the equations 5x + 7y = 4 and 2x – 3y = –10.

Q54. Using variables a and b write any two equations whose solution is (0, 2).

Answer

a + b = 2 and 2a + b = 2 are the equations whose solution is (0, 2).

Reason:

a + b = 2 ...(i)
2a + b = 2 ...(ii)

Subtracting equation (i) from (ii), we get

2a + b = 2
a + b = 2

− − −

a = 0
Substituting a = 0 in equation (i), we get
0 + b = 2
∴ b = 2

Q55. Which of the following is linear equation in two variables?

x + 2y = 5 – 3y

Q56. Which of the following is not the solution of 3x + 6y = 12?

(3, 1)

Explanation:

3x + 6y = 3(3) + 6(1)

= 9 + 6

= 15 ≠ 12

Q57. Write any two linear equations in two variables in which the value of one variable is 12 and the other 10

Answer

x + y = 22 ......(i)
x – y = 2 ......(ii)

Adding equations (i) and (ii), we get

x + y = 22
+ x – y = 2
2x = 24
∴ x = `24/2` = 12
Substituting x = 12 in equation (i), we get
12 + y = 22
∴ y = 22 – 12 = 10
∴ x + y = 22 and x – y = 2 are the required linear equations in two variables.

Q58. Write any two solutions of the equation a – b = – 3

Answer

When b = 1, a = – 3 + 1 = – 2
When b = 4, a = – 3 + 4 = 1
∴ (– 2, 1) and (1, 4) are the solutions of a – b = – 3.

Q59. Write any two solutions of the equation x + y = 7

Answer

When x = 1, y = 7 – 1 = 6
When x = 3, y = 7 – 3 = 4
∴ (1, 6) and (3, 4) are the solutions of x + y = 7.

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