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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