Question

find whether p=2and p=3 are solution for equation 7(p+2)=5 (p+4)? justify?​

Answer 1

Answer:

$\huge\boxed{\text{only}\ p=3}$

Step-by-step explanation:

$7(p+2)=5(p+4)\qquad|\text{use the distributive property}\\\\(7)(p)+(7)(2)=(5)(p)+(5)(4)\\\\7p+14=5p+20\qquad|\text{subtract}\ 14\ \text{and}\ 5p\ \text{from both sides}\\\\2p=6\qquad|\text{divide both sides by 2}\\\\p=3$

Answer 2

Answer:

No, p=2 and p=3 are not the solution

Step-by-step explanation:

Let's assume p = 2

.'. 7 (2+2) =5 (2+4)

.'. 7×4 = 5×6

.'. 28 = 30

LHS is not equal to RHS

Let's assume p = 3

.'. 7 (3+2) = 5 (3+4)

.'. 7 × 5 = 5 × 12

.'. 35 = 60

LHS is not equal to RHS