10th Class Abhyasa Deepika Maths Solutions 4 Pair of Linear Equations in Two Variables Telangana SSC (English & Telugu Medium) – Complete Key & Exam Guide

Solution:
A linear equation must have degree 1.
Option B contains x², so it is not linear.

Answer: B

Solution:
Check each:
(3,1) → 3+1=4
(4,0) → 4+0=4
(1,4) → 1+4=5 ✔
(0,−5) → 0−5=−5

Answer: C

Solution:
log₂8 = 3
log₇49 = 2

a = 3 + 2 = 5

Answer: D

Solution:
For coincident lines:

a₁/a₂ = b₁/b₂ = c₁/c₂

3/6 = (−1)/(−k) = 8/16

1/2 = 1/k

k = 2

Answer: C

Solution:
Parallel condition:
a₁/a₂ = b₁/b₂

3/2 = 2k/5

15 = 4k

k = 15/4

Answer: C

Solution:
Dependent → equations are multiples.

Multiply given equation by −2:

10x − 14y + 4 = 0

Answer: D

Solution:
Parallel lines must satisfy:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Check C:

3/9 = −2/−6

1/3 = 1/3 ✔

Answer: C

Solution:

From x − y = −1
x = y − 1

Substitute in first equation:

3(y − 1) − 5y = −1
3y − 3 − 5y = −1
−2y − 3 = −1
−2y = 2
y = −1

x = y − 1
x = −1 − 1
x = −2

Answer: (−2, −1)

Solution:

From first equation:

x = −1 − 2y

Substitute in second equation:

2(−1 − 2y) − 3y = 12
−2 − 4y − 3y = 12
−2 − 7y = 12
−7y = 14
y = −2

Substitute in x = −1 − 2y

x = −1 − 2(−2)
x = −1 + 4
x = 3

Answer: (3, −2)

Solution:

Add both equations:

2x = 14
x = 7

Substitute:

7 + y = 9
y = 2

Answer: (7, 2)

Solution:

From first equation:

2x = 9 − 3y
x = (9 − 3y)/2

Substitute in second equation:

3(9 − 3y)/2 + 4y = 5

Multiply by 2:

27 − 9y + 8y = 10
27 − y = 10
y = 17

Substitute:

2x + 3(17) = 9
2x + 51 = 9
2x = −42
x = −21

Answer: (−21, 17)

Solution:

Multiply second equation by 5:

15x + 10y = 20

Multiply first equation by 2:

16x + 10y = 18

Subtract:

x = −2

Substitute:

3(−2) + 2y = 4
−6 + 2y = 4
2y = 10
y = 5

Answer: (−2, 5)

Solution:

Rewrite:

11x − 5y = −61
3x − 20y = 2

Multiply first equation by 4:

44x − 20y = −244

Subtract second equation:

41x = −246
x = −6

Substitute:

11(−6) − 5y = −61
−66 − 5y = −61
−5y = 5
y = −1

Answer: (−6, −1)

Solution:

Multiply second equation by 3:

−3x + 12y = 21

Add to first equation:

14y = 35
y = 5/2

Substitute:

3x + 2(5/2) = 14
3x + 5 = 14
3x = 9
x = 3

Answer: (3, 5/2)

Solution:

Multiply first equation by 6:

3x + 4y = −6

Multiply second equation by 3:

3x − y = 9

Subtract:

5y = −15
y = −3

Substitute:

x − (−3)/3 = 3
x + 1 = 3
x = 2

Answer: (2, −3)

Solution:

Rewrite equations:

x − y = −1
3x + 2y = 12

From first equation:

y = x + 1

Substitute in second equation:

3x + 2(x + 1) = 12
3x + 2x + 2 = 12
5x = 10
x = 2

y = 2 + 1 = 3

Hence, the lines intersect at (2, 3).

Solution:

Second equation is 2 × first equation.

Therefore, both equations represent the same line.

Hence, infinitely many solutions (coincident lines).

Solution:

Rewrite:

3x + y = 5
2x − y = 5

Add equations:

5x = 10
x = 2

Substitute:

3(2) + y = 5
6 + y = 5
y = −1

Intersection point: (2, −1)

Solution:

Multiply first equation by 2:

4x + 6y = 18

Subtract from second equation:

(−10x + 6y) − (4x + 6y) = −22 − 18

−14x = −40
x = 20/7

Substitute:

2(20/7) + 3y = 9

40/7 + 3y = 9
3y = 63/7 − 40/7
3y = 23/7
y = 23/21

Intersection point: (20/7, 23/21)

Solution:

Let a = 1/x, b = 1/y

Then:

2a + 3b = 2
a − b/2 = 1/3

Multiply second equation by 2:

2a − b = 2/3

Now subtract:

(2a + 3b) − (2a − b) = 2 − 2/3

4b = 4/3
b = 1/3

Substitute:

2a + 1 = 2
2a = 1
a = 1/2

Thus,

x = 2
y = 3

Solution:

Let a = 1/(x + y), b = 1/(x − y)

Then:

20a + 3b = 7
8b − 15a = 5

Multiply first equation by 5:

100a + 15b = 35

Multiply second equation by 1:

−15a + 8b = 5

Solve simultaneously:

a = 1/5
b = 1

So,

x + y = 5
x − y = 1

Add:

2x = 6
x = 3

y = 2

Answer: (3, 2)

Solution:

Let a = 1/√x
b = 1/√y

Then:

2a + 3b = 2
4a + 9b = 5

Multiply first equation by 2:

4a + 6b = 4

Subtract from second:

3b = 1
b = 1/3

Substitute:

2a + 1 = 2
2a = 1
a = 1/2

So,

1/√x = 1/2 → √x = 2 → x = 4
1/√y = 1/3 → √y = 3 → y = 9

Answer: (4, 9)

Solution:

Given equations:

2x + 3y = 9
4x + 6y = 18

Divide second equation by 2:

2x + 3y = 9

Both equations are identical.

So,

a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the lines are coincident.

Therefore, the system has infinitely many solutions.

Yes, the student is correct. The system is consistent and dependent.

Solution:

Write in standard form:

3x + 2y − 8 = 0
6x − 4y − 9 = 0

Compute ratios:

a₁/a₂ = 3/6 = 1/2
b₁/b₂ = 2/(−4) = −1/2

Since a₁/a₂ ≠ b₁/b₂

The lines are intersecting.

Hence, they have a unique solution.

Solution:

For no solution:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Here,

a₁ = k, b₁ = 3, c₁ = −1
a₂ = 12, b₂ = k, c₂ = −2

Condition:

k/12 = 3/k

Cross multiply:

k² = 36

k = 6 or k = −6

Now check second condition:

k/12 ≠ (−1)/(−2)

(−1)/(−2) = 1/2

If k = 6:

6/12 = 1/2 (equal) → infinitely many solutions ❌

If k = −6:

−6/12 = −1/2 ≠ 1/2 ✔

Hence,

k = −6

Solution:

Perimeter = AB + BC + CD + DE + EA

= 5 + (x − y) + (x + y) + 7 + 5

Simplify:

5 + x − y + x + y + 7 + 5

= 2x + 17

Given perimeter = 27

2x + 17 = 27

2x = 10
x = 5

Now,

BC = x − y
CD = x + y

Since BE ∥ CD and BE = 7,

CD = 7

So,

x + y = 7
5 + y = 7
y = 2

Hence,

x = 5
y = 2

Solution:

Let son’s age = x

Father’s age = 3x

x + 3x = 48

4x = 48

x = 12

Father’s age = 3 × 12 = 36

Hence,

Son = 12 years
Father = 36 years

Solution:

Let tens digit = x
Units digit = y

Number = 10x + y
Reversed number = 10y + x

Given:

7(10x + y) = 4(10y + x)

70x + 7y = 40y + 4x

66x = 33y

2x = y

Also given difference of digits = 3

y − x = 3

Substitute y = 2x:

2x − x = 3

x = 3

Then y = 6

Number = 36

Check:

7 × 36 = 252
4 × 63 = 252 ✔

Hence, the number is 36.