Question

16-(a+3)(a+2)=4-(6+a)(a-1)​

Answer 1

Ответ:

Объяснение:

16-(a^2+2a+3a+6)=4-(6a-6+a^2-a),  

16-a^2-5a-6=4-5a+6-a^2,  10=10, верно,  уравнение имеет

бесконечное множ-во решений

Answer 2

To solve the given equation $16 - (a+3)(a+2) = 4 - (6+a)(a-1)$, we will follow these steps:

1. Expand the expressions within the parentheses.
2. Simplify both sides of the equation.
3. Solve for the variable $a$.

Let's start:

1. Expand the expressions within the parentheses:

   • On the left side: $(a+3)(a+2) = a^2 + 2a + 3a + 6 = a^2 + 5a + 6$
   • On the right side: $(6+a)(a-1) = 6a - a + a^2 - 1 = a^2 + 5a - 1$

2. Substitute the expanded expressions into the equation: $16 - (a^2 + 5a + 6) = 4 - (a^2 + 5a - 1)$

3. Simplify both sides of the equation: $16 - a^2 - 5a - 6 = 4 - a^2 - 5a + 1$

   Simplifying the constants: $10 - a^2 - 5a = 5 - a^2 - 5a$

At this point, we notice that the terms involving $a$ on both sides are the same. Thus, the equation becomes:

$10 = 5$

Since $10$ is not equal to $5$, there's no solution for this equation. This means that the original equation doesn't have any solution for the variable $a$ that would satisfy the equation.