Math, asked by santoshk7749, 10 months ago

EXERCISE 6A
1. Solve the following equations by adding, subtracting, multiplying or dividing both
sides of the equations by the same number.
a. x + 5 = 3 b. a - 3 = 5 c. 7a = 14 d. 2+ 9 = 6
e. 8 = 16d
5a + a = 12
g. a - 3a = 8 h. 6p = -9 + 9 i. 4y = 2y - 5 j. 19a = -38
uring the solution d = 5.
f
.
the value of the​

Answers

Answered by jahnvirt
6

Answer:

a. x+ 5 = 3

x = 3 -5

x= -2

b.a - 3 = 5

a = 5+3

a = 8

c. 7a = 14

a = 14/7

a = 2

d. sorry ..

e.8 = 16d

d = 16/8

d = 2

f. 5a + a = 12

6a = 12

a = 12/6

a = 2

g. a - 3a = 8

-2a = 8

a = 8/-2

a = -4

h.6p = -9 +9

p = -6

I. 4y = 2y -5

4y - 2y = -5

2y = -5

y = -5/2

j. 19a = -38

a = -38/19

a = -2

hope it's help...

follow me..if it's help you

Answered by halamadrid
2

The correct answers are a) x = -2, b) a = 8, c) a = 2, d) 2+9 ≠ 6,

e) d = 0.5, f) a = 2, g) a = -4, h) p = 0, i) y =-5/2, and j) a = -2.

a. We have been given that  

x + 5 = 3

Taking 5 to the RHS, we have

x = 3-5

Using the operation of subtraction of integers, we have

x = -2.

b. We have been given that

a - 3 = 5

Taking -3 to the RHS, we have

a = 5+3

Using the operation of subtraction of integers, we have

a = 8.

c. We have been given that

7a = 14

Taking 7 to the RHS, we have

a = 14/7

Using the operation of the division of integers, we have

a = 2.

d. We have been given that

2+ 9 = 6

⇒ 11 = 6

This statement is false as it is an inequality.

∴ 2+9 ≠ 6.

e. We have been given that

8 = 16d

Taking 16 to the LHS, we have

8/16 = d

Using the operation of the division of real numbers, we have

d = 1/2 = 0.5

∴ d = 0.5

f. We have been given that

5a + a = 12

⇒ 6a = 12

Taking 6 to the RHS, we have

a = 12/6

Using the operation of the division of integers, we have

a = 2

g. We have been given that

a - 3a = 8

⇒ -2a = 8

Using the operation of the division of integers, we have

a = 8/(-2) = -4

a = -4

h. We have been given that

6p = -9 + 9

6p = 0

Using the operation of the division of integers, we have

p = 0/6 = 0

p = 0

i. We have been given that

4y = 2y - 5

Taking 2y to the RHS, we have

4y-2y = -5

⇒ 2y = -5

Using the operation of the division, we have

y =-5/2.

j. We have been given that

19a = -38

Taking 19 to the RHS and applying the operation of division, we have

a = -38/19 = -2

a = -2.

#SPJ2

Similar questions