4. Solve each of the following equations using trial and error method
5. (i) a + 9 = 14 (ii) p – 2 = 1 (iii) 6x = 24 (iv) 8 – x = 5 (viii) m/3 = 1
Answers
Step-by-step explanation:
Ncert solutions
Grade 7
Mathematics
Science
Chapters in NCERT Solutions - Mathematics, Class 7
Exercises in Simple Equations
Question 3
Q3) Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4
Solution
Transcript
Solution 3:
(i) putting p = -3 in equation
5(-3) +2 = 17
-15+2 = 17
-13\ne17−13
=17
putting p = -2 in equation
5\left(-2\right)+2=175(−2)+2=17
-10+2=17−10+2=17
-8\ne17−8
=17
putting p =-1
5\times\left(-1\right)+2\ =\ 17\5×(−1)+2 = 17
-5+2=17−5+2=17
-3\ne17−3
=17
putting 0 in the equation
5\times0\ +2\ =\ 17\5×0 +2 = 17
2\ne172
=17
putting p =1 n the equation
5\times1+2=175×1+2=17
5+2=175+2=17
7\ne177
=17
puttin p = 2 in the equation
5\times2\ +2\ =\ 17\5×2 +2 = 17
10+2\ =\ 17\10+2 = 17
12\ne1712
=17
putting p = 3 in the equation
5\times3+2=175×3+2=17
15+2=1715+2=17
17=17\17=17
As LHS=RHS
p=3 is the solution
(ii) 3m-14 = 4
putting m = -2
3(-2) -14 = 4
-6-14=4
-20\ \ne\ 4−20
= 4
putting m = -1
3(-1) -14 = 4
-3-14 = 4
-17\ne4−17
=4
putting m = 0
3(0) -14=4
0-14 = 4
-14\ne4−14
=4
putting m =1
3(1) -14 = 4
3-14 = 4
-11\ne4−11
=4
putting m = 2
3(2) - 14 = 4
6-14=4
-8\ne4−8
=4
putting m = 3
3(3) - 14 = 4
9-14 = 4
-5\ne4−5
=4
putting m = 4
3(4)- 14 = 4
12-14 = 4
-2\ne4−2
=4
putting m = 5
3(5)-14=4
15-14=4
1\ne41
=4
putting m = 6
3(6)-14 = 4
18-14
4=4
LHS = RHS
therefore m = 6 is a solution.
Step-by-step explanation:
I)a+9=14
a=14-9
a=5
ii)p-2=1
p=3
iii)6x=24
x=24÷6=4
iv)8-x=5
-x= -8+5=-3
m/3=1
m=1×3=3