Access Selina Solutions Concise Mathematics Class 6 Chapter 22 Simple (Linear) Equations
Exercise 22(A)
1. Solve:
(i) x + 2 = 6
(ii) x + 6 = 2
(iii) y + 8 = 5
(iv) x + 4 = – 3
(v) y + 2 = – 8
Solution:
(i) x + 2 = 6
x = 6 – 2
We get,
x = 4
Hence, the value of x for x + 2 is 4
(ii) x + 6 = 2
x = 2 – 6
We get,
x = = – 4
Hence, the value of x for x + 6 = 2 is – 4
(iii) y + 8 = 5
y = 5 – 8
We get,
y = – 3
Hence, the value of y for y + 8 = 5 is – 3
(iv) x + 4 = – 3
x = – 3 – 4
We get,
x = – 7
Hence, the value of x for x + 4 = – 3 is – 7
(v) y + 2 = – 8
y = – 8 – 2
We get,
y = – 10
Hence, the value of y for y + 2 = – 8 is – 10
2. Solve:
(i) x – 3 = 2
(ii) m – 2 = – 5
(iii) b – 5 = 7
(iv) a – 2.5 = – 4
(v) y – 3 (1 / 2) = 6
Solution:
(i) x – 3 = 2
x = 2 + 3
We get,
x = 5
Therefore, the value of x for x – 3 = 2 is 5
(ii) m – 2 = – 5
m = – 5 + 2
We get,
m = – 3
Therefore, the value of m for m – 2 = – 5 is – 3
(iii) b – 5 = 7
b = 7 + 5
We get,
b = 12
Therefore, the value of b for b – 5 = 7 is 12
(iv) a – 2. 5 = – 4
a = – 4 + 2. 5
We get,
a = – 1. 5
Therefore, the value of a for (a – 2. 5) = – 4 is – 1. 5
(v) y – 3 (1 / 2) = 6
This can be written as,
y – (7 / 2) = 6
y = 6 + (7 / 2)
y = (12 + 7) / 2
y = 19 / 2
y =
Therefore, the value of y for y – 3 (1 / 2) = 6 is
3. Solve:
(i) 3x = 12
(ii) 2y = 9
(iii) 5z = 8.5
(iv) 2.5m = 7.5
(v) 3.2p = 16
Solution:
(i) 3x = 12
x = 12 / 3
We get,
x = 4
Hence, the value of x for 3x = 12 is 4
(ii) 2y = 9
y = 9 / 2
We get,
y = 4.5
Hence, the value of y for 2y = 9 is 4.5
(iii) 5z = 8.5
z = 8.5 / 5
We get.
z = 1.7
Hence, the value of z for 5z = 8.5 is 1.7
(iv) 2.5m = 7.5
m = 7.5 / 2.5
We get,
m = 3
Hence, the value of m for 2.5m = 7.5 is 3
(v) 3.2p = 16
p = 16 / 3.2
p = (16 × 10) / 32
p = 160 / 32
p = 5
Hence, the value of p for 3.2p = 16 is 5
4. Solve:
(i) x / 2 = 5
(ii) y / 3 = – 2
(iii) a / 5 = – 15
(iv) z / 4 = 3 (1 / 4)
(v) m / 6 = 2 (1 / 2)
Solution:
(i) x / 2 = 5
x = 5 × 2
We get,
x = 10
Hence, the value of x for x / 2 = 5 is 10
(ii) y / 3 = – 2
y = – 2 × 3
We get,
y = – 6
Hence, the value of y for y / 3 = – 2 is – 6
(iii) a / 5 = – 15
a = – 15 × 5
We get,
a = – 75
Hence, the value of a for a / 5 = – 15 is – 75
(iv) z / 4 = 3 (1 / 4)
This can be written as,
z / 4 = 13 / 4
z = 13 / 4 × 4
We get,
z = 13
Hence, the value of z for z / 4 = 3 (1 / 4) is 13
(v) m / 6 = 2 (1 / 2)
This can be written as,
m / 6 = 5 / 2
m = 5 / 2 × 6
m = 5 × 3
We get,
m = 15
Hence, the value of m for m / 6 = 2 (1 / 2) is 15
5. Solve:
(i) – 2x = 8
(ii) – 3.5y = 14
(iii) – 5z = 4
(iv) – 5 = a + 3
(v) 2 = p + 5
Solution:
(i) – 2x = 8
x = – 8 / 2
We get,
x = – 4
Therefore, the value of x for – 2x = 8 is – 4
(ii) – 3.5y = 14
y = – 14 / 3. 5
We get,
y = – 4
Therefore, the value of y for – 3.5y = 14 is – 4
(iii) – 5z = 4
z = – 4 / 5
We get,
z = – 0.8
Therefore, the value of z for – 5z = 4 is – 0.8
(iv) – 5 = a + 3
– 5 – 3 = a
On calculating, we get
a = – 8
Therefore, the value of a for – 5 = a + 3 is – 8
(v) 2 = p + 5
2 – 5 = p
We get,
p = – 3
Therefore, the value of p for 2 = p + 5 is – 3
