You don't specify in your question but I am assuming that you want to solve for the variable "d". You have the following equation written:

`(d/5)=7`

You need to multiply both sides of the equation by 5 in order to solve for d.

`(d/5)*5=7*5`

The fives cancel each other out on the left hand side, leaving you with only d.

`d=35`

**The solution is d=35.**

**Question:-**

d/5 = 7

**Solution:-**

d/5 = 7

Multiply 5 to both sides, so that the denominator 5 can be cancelled:-

(5) . d/5 = 7. 5

5d/5 = 35

Cancel 5 from LHS, in numerator and denominator, then we get;

d = 35

Hence solved.

Therefore the value of d when inserted in the equation given, will prove that the answer is right.

Like;

d/5 = 7

35/5 = 7

7 = 7

LHS = RHS

Hence Proved that the value of d is correct.

d/5 = 7

In order to solve this we need to isolate the variable d, therefore

d/5 = 7 Multiplying both sides by 5 we get

d/5 * 5 = 7 * 5

**d = 35 Answer.**

Input the value of d in the above equation,

d/5 = 7

35/5 = 7

7 = 7

LHS = RHS

d/5=7

all you have to do it multiply both sides by 5 to get rid of the denominator of 5, the reason you do this is because the opposite of division is subtraction.

`5 xx (d/5)=7 xx 5 `

d =35

d / 5 = 7

To solve these equations , I always remember to do the opposite . For example if it is addition , I would do subtraction . Second you need to always get the variable by itself , which means for this question to get " d " by itself

Now , first you should do the opposite so you should multiply 5 on both sides . By multiplying 5 on both sides , you should get

d = 35 which is answers . As you can see " d " is alone by itself now on the left side of the equation .