Question

How many solutions does the following equation have?

Answer 1

Answer:

B. Exactly One Solution

Step-by-step explanation:

Given:

$7(y+3)=5y+8$

Expand the brackets:

$\implies 7y + 21 = 5y + 8$

Subtract 21 from both sides:

$\implies 7y + 21-21 = 5y + 8-21$

$\implies 7y=5y-13$

Subtract 5y from both sides:

$\implies 7y-5y=5y-13-5y$

$\implies 2y=-13$

Divide both sides by 2:

$\implies \dfrac{2y}{2}=\dfrac{-13}{2}$

$\implies y=-\dfrac{13}{2}$

Therefore, the equation has exactly one solution.

Answer 2

B. Exactly one solution

7y +21 = 5y +8
7y +13=5y
13=-2y
-6.5=y