10 4. Solve the following equations: (a) 10p = 100 (b) 10p + 10 = 100 (c) P =5 4 (d) -P =5 3 3P = 6 6 (f) 3s = -9 (e) 4 (i) 2q = 6 (h) 3s = 0 (g) 3s + 12 = 0 (k) 2q +6=0 (1) 29 – 6=0 (1) 2q + 6 = 12
Answers
Answer:
Solution:
We use the basic concepts of simple linear equations to solve the questions given.
(a) 10p = 100
Dividing both the sides by 10 we get,
10p/10 = 100/10
p = 10
(b) 10p + 10 =100
Subtracting 10 from both sides we get,
10p + 10 − 10 = 100 −10
10p = 90
Dividing both the sides by 10 we get,
10p/10 = 90/10
p = 9
(c) p/4 = 5
Multiplying both the sides by 4 we get,
(p/4) × 4 = 5 × 4
p = 20
(d) − p/3 = 5
Multiplying both the sides by -3 ,
(−p/3) × (-3) = 5 × (-3)
p = -15
(e) 3p/4 = 6
Multiplying both the sides by 4, we get
(3p/4) × 4 = 6 × 4
3p = 24
Dividing both the sides by 3 we get,
3p/3 = 24/3
p = 8
(f) 3s = -9
Dividing both the sides by 3,
3s/3 = -9/3
s = -3
(g) 3s + 12 = 0
Subtracting 12 from both the sides of the equation we get,
3s + 12 - 12 = 0 - 12
3s = -12
Dividing both the sides by 3 we get,
3s/3 = -12/3
s = - 4
(h) 3s = 0
Dividing both the sides by 3 we get,
3s/3 = 0/3
s = 0
(i) 2q = 6
Dividing both the sides by 2 we get,
2q/2 = 6/2
q = 3
(j) 2q - 6 = 0
Adding 6 to both sides of the equation we get,
2q - 6 + 6 = 0 + 6
2q = 6
Dividing both the sides by 2 we get,
2q/2 = 6/2
q = 3
(k) 2q + 6 = 0
Subtracting 6 from both the sides of the equation we get,
2q + 6 − 6 = 0 - 6
2q = - 6
Dividing both the sides by 2 we get,
2q/2 = - 6/2
q = -3
(l) 2q + 6 =12
Subtracting 6 from both the sides of the equation we get,
2q + 6 − 6 = 12 - 6
2q = 6
Dividing both the sides by 2 we get,
2q/2 = 6/2
q = 3
Answer:
answer is 3
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