Solve (y/4) + 3 = 7

a) y = 40

b) y = 1

c) y = 25

d) y = 16

Question

Phoebe Blake

Mathematics

9 June, 14:43

Solve (y/4) + 3 = 7

a) y = 40

b) y = 1

c) y = 25

d) y = 16

+2

Answers (2)

Know the Answer?

Edith Castro · Answer 1

(Y/4) + 3 = 7

Subtract 3 from both sides:

Y/4 = 4

Multiply both sides by 4

Y = 16

Brandon Brennan · Answer 2

Option D

Step-by-step explanation:

=> (y/4) + 3 = 7

Subtracting 3 from both sides

=> y/4 = 7-3

=> y/4 = 4

Multiplying 4 to both sides

=> y = 4*4

=> y = 16

Solve (y/4) + 3 = 7a) y = 40b) y = 1c) y = 25d) y = 16

Solve (y/4) + 3 = 7

a) y = 40

b) y = 1

c) y = 25

d) y = 16