A triple of threes

as a=b+4, 3(b+4)-3b+3, by putting the value of a in 3a-3b+3 = 3b+12-3b+3 =15

Let's start the answer:

If it says that a=b+4 then:

$3a-3b+3= 3(b+4)-3b+3=$ $=3b+12-3b+3= 12+3 =15$ The result is $\boxed{15}$

If a=b+4
then a-b=4

3a-3b+3
3(a-b+1)............................a-b=4
3(4+1)
15

• a - b = 4
• 3 ( a - b ) = 12
• 3a - 3b = 12
• so, 3a - 3b + 3 = 12 + 3 = 15

b + 4= a b - a= - 4 a - b= 4 3 (a - b ) = 3 (4) 3a - 3b = 12 12 + 3 = 15

if a=b+4 then eq : 3a-3b+3......... (1)

put the value of "a" in eq : 3(b+4)-3b+3 ......... (2) 3b+12-3b+3 +3b-3b+12+3 0 + 15 = 15 ans : 15

when I put the value of a in eq b was cancelled and value came was 15

Given a=b+4
i.e, a-b=4
So, 3a-3b+3=3(a-b)+3
= $3*4+3$
=15

if a=b+4 put in 3a-3b+3 3(b+4)-3b+3 3b+12-3b+3 15ans

3a - 3b +3 = 3(b+4) -3b +3 = 3b + 12 - 3b +3 = 3b - 3b +15 =15

3a-3b+3=3b+12-3b+3=15

a=b+4; a-b=4; 3a-3b+3=3(a-b)+3 =(3*4)+3 =15

= 3(b+4) - 3b + 3 = 3b + 12 - 3b + 3 = cancel 3b - 3b since it is equal to zero then add 12 + 3 = 15

If a = b + 4 , then substitute a from the equation 3a - 3b + 3 .
3(b + 4) - 3b + 3
3b + 12 - 3b + 3
12 + 3
$\boxed{15}$

a=b+4 substitute: 3a-3b+3 it will become: 3(b+4)-3b+3 multiply 3(b+4) then solve, it will become something like this: 3b+12-3b+3 then solve common term (3b-3b) will be eliminated then, 12+3=15 that's your answer

Substitute a=b+4 in 3a-3b+3. Then 3(b+4)-3b+3,Then you will get answer

If a = b + 4

So 3a - 3b + 3 = 3(b+4) - 3b +3 = 3b +12 -3b + 3 = 15

substitute the value of a to 3a - 3b + 3 to get the value of b,

3a - 3b + 3

=3(b+4) - 3b + 3 = 3b + 12 - 3b + 3

= 3b + 3b = 3 - 12

= 6b = -9

= b = -9/6 or -3/2

now substitute the value of b to a = b + 4 to get the value of a,

a = b + 4

a = (-3/2) + 4

a = 5/2

now that we have the value of a and b we substitute it to 3a - 3b +3,

3a - 3b + 3

= 3(5/2) - 3(-9/2) + 3

= 15/2 + 9/2 + 3

= 15

the value of 3a - 3b +3 is 15

3a-3b+3= 3b+12-3b+3= 3b-3b+12+3= 12+3= 15

=3(b+4)-3b+3 =3b+12-3b+3 ~cancel the two 3b =12+3 =15

3a-3b+3 since a=b+4 3(b+4)-3b+3 ie, 3b+12-3b+3 = 12+3 =15

3(b+4)-3b+3 3b+12-3b+3 (+3b CANCEL -3b) 15

firstly,put the value of a=b + 4 then,multiply 3 by (b + 4) then,the equation becomes 3b + 12 - 3b + 3 then,solve 3b-3b=0 then,12 + 3 remains and the answer will be 15

a=b=4 =>a-b=4 =>3(a-b)=3*4 =>3a-3b=12 so, 3a-3b+3=15

3(b+4)-3b+3? 3b+12-3b+3? 12+3=15

a=b+4 so a-b=4 in the givn equation take 3 common n de equation is 3(a-b+1) so substitue a-b=4 in dat equation 3(4+1) 3(5)=15

3a-3b-3=0.......since, a=b+4

so, 3(b+4)-3b+3 3b+12-3b+3 3b-3b+12+3 0+15 15 so, the answer is 15.........

We want to find the value of 3a - 3b + 3 .We are given that a = b + 4 so when we give the value of a in the euation to find,we get 3(b +4) - 3b + 3 = 3b + 12 - 3b + 3 = 12 +3 = 15 . Therefore, 15 is the answer.

a-b=4

3 (a-b)+3=3 4+3=15

1. 3(b+4) - 3b + 3
2. 3b + 12 - 3b + 3
3. (3b - 3b) + 12 + 3
4. 0 + 12 + 3 =15

a=b+4 3(b+4)-3b+3 3b+12-3b+3 15

a=b+4 (given) then 3a=3(b+4)

3a-3b+3=(3b+12)-3b+3

3b+12-3b+3=15

3a - 3b + 3 = 3(b + 4) - 3b + 3 = 3b + 12 - 3b + 3 = 15

a=b+4,

a-b=4

so, 3a - 3b + 3

= 3(a-b) + 3

= 3 x 4 + 3

= 12 + 3

= 15

by substituting the value of a, you would get 3(b+4)-3b+3. then it will be 3b+12-3b+3. then by Addition Property, the answer will be 15.

3a-3b+3 = 3(a-b) +3 but from the question a=b+4 and a-b=4 so, 3(a-b)+3= 3(4)+3= 12+ 3 = 15 where 15 is the correct answer

if a = b+4, also 3a-3b+4 as eq(1) so putting value of a in eq(1): 3(b+4)-3b+3 then 3b-3b+12+3 =15