Question

Can you answer these please many thanks

Answer 1

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$.

To find: The expanded form of the given expressions.

(c) $4(m+n)=4m+4n$       (using $a(b+c)=ab+ac$)

(d) $3(5-q)=15-3q$           (using $a(b-c)=ab-ac$)

(e) $5(2c+1)=10c+5$          (using $a(b+c)=ab+ac$)

(f) $3(2x-5)=6x-15$          (using $a(b-c)=ab-ac$)

(g) $7(4b-1)=28b-7$          (using $a(b-c)=ab-ac$)

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒$3(2x+y-5)=3(2x+y)-15$         (using $a(b-c)=ab-ac$)

⇒$3(2x+y-5)=6x+3y-15$            (using $a(b+c)=ab+ac$)

(i) $2(6a-4b+3)=2((6a-4b)+3)$

⇒$2(6a-4b+3)=2(6a-4b)+6$         (using $a(b+c)=ab+ac$)

⇒$2(6a-4b+3)=12a-8b+6$            (using $a(b-c)=ab-ac$)

(j)$6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$           (using $a(b+c)=ab+ac$)

⇒ $6(m+n+p)=6m+6n+6p$            (using $a(b+c)=ab+ac$)

(k) $y(y+2)=y^2+2y$          (using $a(b+c)=ab+ac$)

(l) $g(g-3)=g^2-3g$           (using $a(b-c)=ab-ac$)

(m) $n(4-n)=4n-n^2$        (using $a(b-c)=ab-ac$)

(n) $a(b+c)=ab+ac$           (using $a(b+c)=ab+ac$)

(o) $s(3s-4)=3s^2-4s$        (using $a(b-c)=ab-ac$)

(p) $2x(x+5)=2x^2+10x$     (using $a(b+c)=ab+ac$)

(q) $4y(x-3)=4xy-12y$      (using $a(b-c)=ab-ac$)

(r) $5a(2b-5)=10ab-25a$     (using $a(b-c)=ab-ac$)

(s) $4a(3b+2c)=12ab+8ac$    (using $a(b+c)=ab+ac$)

(t) $5p(4p-5q)=20p^2-25pq$    (using $a(b-c)=ab-ac$)