# Solve the following equations.

(a) 2y + 5/2 = 37/2 (b) 5t + 28 = 10

(c) a/5 + 3 = 2 (d) q/4 +7=5

(e) 5/2 x = −5 (f) 5/2 x = 25/4

(g) 7m + 19/2 = 13 (h) 6z + 10 = -2

(i) 3l/2 = 2/3 (j) 2b/3 − 5 = 3

**Solution:**

We use the concept of simple linear equations to solve the problem.

(a) 2y + 5/2 = 37/2

Transposing 5/2 to R.H.S we get,

2y = 37/2 − 5/2

2y = 32/2

y = 16

(b) 5t + 28 =10

Transposing 28 to R.H.S we get,

5t = 10 − 28

5t = −18

t = −18/5

(c) a/5 + 3 = 2

Transposing 3 to R.H.S we get,

a/5 = 2 − 3

a/5 = − 1

a = - 5

(d) q/4 + 7 = 5

Transposing 7 to R.H.S we get,

q/4 = 5 − 7

q = −8

(e) 5/2 x = −5

5x = − 5 × 2

5x = −10

x = −2

(f) 5/2 x = 25/4

5x = 25/4 × 2

x = 25/(2 × 5)

x = 5/2

(g) 7m + 19/2 = 13

Transposing 19/2 to the R.H.S.

7m = 13 − 19/2

7m = (26 −19)/2

7m = 7/2

m = 1/2

(h) 6z + 10 = −2

Transposing 10 to the R.H.S.

6z = −2 −10

z = −12/6

z = −2

(i) 3/2 l = 2/3

l = 2/3 × 2/3

l = 4/9

(j) 2b/3 − 5 = 3

2b/3 = 3 + 5

2b/3 = 8

b = 8 × 3/2

b = 12

**☛ Check: **NCERT Solutions for Class 7 Maths Chapter 4

**Video Solution:**

## Solve the following equations: (a) 2y + 5/2 = 37/2 (b) 5t + 28 =10 (c) a/5 + 3 = 2 (d) q/4 +7=5 (e) 5/2 x = −5 (f) 5/2 x = 25/4 (g) 7m + 19/2 =13 (h) 6z + 10 = -2 (i) 3l/2 = 2/3 (j) 2b/3 − 5=3

NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.3 Question 1

**Summary:**

We have solved the following equations: (a) 2y + 5/2 = 37/2; y = 16 (b) 5t + 28 =10; t = -18/5 (c) a/5 +3 = 2; a = - 5 (d) q/4 +7 = 5; q = - 8 (e) 5/2 x = −5; x = - 2 (f) 5/2 x = 25/4; x = 5/2 (g) 7m + 19/2 =13; m = 1/2 (h) 6z + 10 = -2; z = - 2 (i) 3l/2 = 2/3; l = 4/9 (j) 2b/3 − 5 = 3; b = 12

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