To "b" or Not To "b"

$ab + bc = 44 + 112 = 156$ . Note that you can factorise $b$ out. Therefore, the above equation becomes: $b (a + c) = 156$ . Note that since $a+c = 13$ , $b$ is simply $\frac{156}{13} = \boxed{12}$ .

a + c = 13,
44/b + 112/b = 13,
b = 156/13,
b =12 answer

We have ab=44 so a=44/b and bc = 112 so c = 112/b So a+c =13 = (44/b)+(112/b) = 156/b 13 = 156/b Therefore b = 156/13 = 12 So b=12

$ab=44,bc=112$ .

Adding, $(a+c)b=156$ . But, $a+c=13$ , so, $13b=156$ . Which gives us $b=12$

Given:

 ab = 44

 bc = 112

 a+c = 13

Solution:

 ab = 44

 since a+c = 13, therefore a = 13-c

 (13-c) * b = 44

 (13*b) - (c*b) = 44

 (13*b) - 112 = 44

 (13*b) = 44 + 112

 13*b = 156

  b = 156 / 13

  b = 12

a+c=13 so the value of a is 13-c putting the value of a in ab=44 (13-c)b=44 13b-cb=44 cb=112 13b-112=44 13b=156 b=12

ab+bc=44+112=156; b(a+c)=156; b=156/(a+c)=156/13=12.

ab = 44 bc = 112 a + c = 13

so, ab + bc = 44 + 112, b(a+c) = 156, b(13) = 156, b = 156/13, so ans 12

ab+bc=44+112 b(a+c)=156 b*13=156 (because a+c=13) b=156/13 b=12

a=44/b

c=112/b

a+c=44/b+112/b=156/b

156/b=13 >> b=12

ab = 44 , bc = 112 and a+c=13

So, $ab+bc=44+112 \implies b(a+c)= 156 \implies 13b=156 \implies b=\boxed{12}$

a=13-c (13-c)*b=44 13b-bc=44 13b-112=44 b=12

ab+bc=156 b(a+c)=156 b(13)=156 b=156/13 b=12

ab=44, cb=112 ab+cb=156,take out b b(a+c)=156, a+c= 13 so b= 156/13= 12

we know a=44/b, and c=112/b, hence putting these in a+c=13 =>44/b+112/b=13 =>(44+112)/13=b =>b=12

we have : ab = 44 ac = 112 => b = 44/a b = 112/a <=> 44/a = 112/a

=> a + c = 13 112a - 44c = 0 => a = 11/3, c = 28/3 => b = 12

from ab=44, you'll get a=44/b (eq1).. then from bc=112, you'll get c=112/b (eq2)... then substitute eq1 and eq2 to a+c=13, which is 44/b + 112/b = 13.. then solve for b, and my answer is 12...

consider ab=44 as eguation 1 and bc=112 as equation 2 and a+c=13 as equation 3. add equations 1 and 2 and take 'b' common from these equations and then put the value of a+c from equation 3 in new equation.then divide 156 by 13 and get the answer...

ad+bc=44+112->b(a+c)=156->b(13)=156->b=156/13=12

12

ab= 44 a=44/b

bc= 112 c=112/b

a+c= 44/b + 112/b =13 = 156/b = 13 = 156 = 13b b= 156/13 b= 12

ab / bc = 44 / 112 <=> a/c = 11/28 <=> a = 11/28 c a + c = 13 11/28 c + c = 13 39/28 c = 13 c = 28/3

bc = 112 28/3 b = 112 b = 12

ab = 44 then a = 44/b....... eqn (i)

bc = 112 then c = 112/b..... eqn (ii)

substitute eqn i and ii into a+c=13 then, 44/b + 112/b = 13 (44+112)/b = 13 156 = 13b b = 12

therefore b = 12. ANS

ab=44 bc=112 add them , b(a+c)=112+44=156 (a+c)= 13 so, b=12. :)

(a+c)*b=44+112; =>b=156/13=12

ab=44, a=44/b...............(1) bc=112, c=122/b.............(2) => a+c=13, from...(1) & (2)..we get that...... 44/b + 112/b=13, 44 +112=13b, 4(11+28)=13b, 4 * 39=13b, b=(39/13) * 4, b=3 * 4, b=12.

a= b/44 ; c= b/112; a+c= (44+112)/b = 13 ; b=12

a+c=13 multiplying both sides with b (a+c)b=13b ab+bc=13 44+112=13b (by substituing the corespnding given values) 13b=156 b=156/13

b=12

a+c=13 a=13-c Substitute a into "ab=44" Then (13-c)b=44 13b-cb=44

From bc=112 b=\frac{112}{c} Substitute "b" into "13b-cb=44" Then 13(112/c)-c(112/c)=44 Hence, (1456/c)-112=44 (1456/c)=156 C=9.333

Last, substitute C into "bc=112" b(9.333)=112 Then B=12

We know that ab=44, bc=112 and a+c=13. From that we know : b=44/a and c=13-a. Plugging those into the equation bc=112, we get (44/a)(13-a)=112. 13(44/a)=572/a;-a(44/a)=-44. We now have 572/a - 44=112. Getting a common denominator we have 572 - 44a all over a =112. Multiplying a to both sides we get 572-44a=112a. 572=156a (+44a). a = 572/156 . a = 11/3. Since we know ab=44 : (11/3)b=44. 11b/3=44; 11b=132; b=12.