Distributive Property

$ab-cd+ac-bd$ factors into $(a-d)(b+c)$ . Using substitution, we get $(a-d)(b+c)=4\cdot 15=\boxed{60}$ .

ab - cd + ac -bd = a(b+c) - d(c+b) = (b+c) (a-d) = 15*4 = 60

ab-cd+ac-bd = ab+ac-cd-bd = a(b+c)-d(c+b) = a(b+c)-d(b+c) = (b+c)(a-d) = 15X4 = 60

ab-cd+ac-bd or, ab-bd+ac-cd or, b(a-d)+c(a-d) or, (b+c)(a-d) or, 15*4 [this value is given] so, 60 (ANS)

ab-cd+ac-bd then it is equal to a(b+c)-d(b+c) by taking common =(b+c)(a-d) =15 * 4 [ given that b+c = 15 & a-d = 4] =60 [solved]

Rearranging the terms so that it is easier to factorise the equation, $\begin{aligned} ab - cd + ac - bd &= ab + ac - bd - cd \\ &= a(b + c) - d(b + c) \\ &= (a - d)(b + c) \end{aligned}$ Now substituting $b + c = 15$ and $a - d = 4$ gives: $\begin{aligned} (a - d)(b + c) &= (4)(15) \\ &= \boxed{60} \end{aligned}$ Therefore, the answer is 60 .

(ac-cd)+(ab-bd)=c(a-d)+b(a-d)=(c 4)+(b 4)=4(c+b)=4*15=60

b+c=15. a-d=4. (b+c)(a-d)=15*4. ab-db+ac-dc=60. simple!!

from given we can assume a=4+d;b=15-c substitute in ab-cd+ac-bd v will get 60

ab+ac-cd-bd=a(b+c)-d(b+c)=(a-d)(b+c)=4 * 15=60

Given that, $b+c=15$ ....(i) and $a-d=4$ ....(ii)

Multiplying (i) and (ii), we get,

$(b+c)(a-d)=15\times 4$

$\implies ab-bd+ac-cd=60 \implies ab-cd+ac-bd=\boxed{60}$

ab - cd + ac - bd = a(b + c) -d(b + c) = (a - d)(b + c) = 15 x 4 = 60

ab-bd+ac-cd = b(a-d) + c(a-d) =(a-d)(b+c) =(4)(15) =4x15 =60

ab - bd + ac - cd = b(a-d) + c(a-d) = (b+c)(a-d) = 15 x 4 = 60. ANSWER :60

b+c =15 a-d = 4 ab-cd+ac-bd or a(b+c)-d(b+c) (a-d)(b+c) = 15 x 4 = 60

ab+ac-bd-cd = a(b+c)d(b+c)

= (b+c)(a-d)

= (15)(4)

=60

(ab-bd) +(ac-cd)= b(a-d)+c(a-d) ,a-d=4 @ 4b+4c=4(b+c)=4x15=60

(b+c).(a-d) = ab-cd+ac-bd= 15x4=60

ab-cd+ac-bd =ab+ac-bd-cd =a(b+c)-d(b+c)=(b+c)(a-d)=15*4=60 [as b+c=15; a-d=4]

(b+c) * (a-d) = 15*4

ba-bd+ac-dc = 60

ab-cd+ac-bd = 60.

Hence prooved

ab-cd+ac-bd= (b+c)(a-d)=15*4=60

ab-cd+ac-bd=ab+ac-(bd+cd)=a(b+c)-d(b+c)=(b+c)(a-d)=15*4=60

15 X 4

a(b+c)-d(b+c); (b+c)(a-d)=15*4=60.

(b+c)(a-d) = (15)(4) ab-cd+ac-bd= 60

(b+c)(a-d)=15*4 ab-cd+ac-bd=60

1. Combine all similar terms ab+ac-cd-bd and then transpose -cd-bd .... ab+ac=cd+bd
2. Factorization.. a(b+c)=d(b+c)
3. Substitute the given values... a(15)=d(15) a(15)-d(15)= x 15(a-d) =x 15(4) = x 60=x

ab-cd +ac -bd given: b+c = 15 and a - d = 4........ first re-arrange the given equation in the form => ab + ac - bd - cd => a(b + c) - d(b + c) .....(eqn 1) put the value in the eqn......1 => a . 15 - d . 15 => 15 (a - d) => 15 . 4 => 60 ans...

We Have b+c=15 And a-d=4...Simply Multiply Both Equations (b+c)(a-d)=15 Times 4=60

ab-cd+ac-bd = a(b+c)-d(b+c)=(a-d)(b+c)=15\times 4 = 60

a(b+c) - d(b+c) =(a-d)(b+c) sub in the statement: b+c=15, a-d=4; =4(15) =60

a(b+c)-(b+c)d =a(15)-(15)d =15(a-d) =15(4) =60

15 x 4 = 60

ab-cd+ac-bd =a(b+c)-d(b+c) =(a-d)(b+c) =4*15= 60

We can rearrange the equation to $ab + ac - cd - bd$ . Then we factor out the a and d to get $a(b+c) - d(b+c)$ . That is the same as $(a-d) + (b+c)$ We can replace the with the values of $b+c$ and $a-d$ to get $15*4$ , which is $60$

ab + ac - cd -bd = a(b + c) -d(b + c) = (b + c) (a - d) =(15)(4) = 60

Multiplying (a-d) (b+c)=15 4 ab-ac-bd-cd=60

Therefore ab-cd+ac-bd=60

(b+c)x(a-d)=ab-cd+ac-bd 15x4=60

By multiplying the two equations together and expanding, we get $ab-cd+ac-bd=60.$