Question

5x-5>2x+1 X^2-x-12<0

Answer 1

Объяснение:

$\left \{ {{5x-5 > 2x+1} \atop {x^2-x-12 < 0}} \right.\ \ \ \ \left \{ {{3x > 6\ |:3} \atop {x^2-4x+3x-12 < 0}} \right.\ \ \ \ \left \{ {{x > 2} \atop {x*(x-4)+3*(x-4) < 0}} \right. \ \ \ \ \left \{ {{x > 2} \atop {(x-4)*(x+3) < 0}} \right. .$

-∞__+__-3__-__4__+__+∞         ⇒

x∈(-3;4).          

$></p> <p><strong>Ответ: x∈(2;4).</strong></p> </div> <div class=$ 0 0

Answer 2

To solve the inequalities:

1. 5x - 5 > 2x + 1

Let's simplify and solve for x:

5x - 2x > 1 + 5

3x > 6

Dividing both sides by 3:

x > 2

Therefore, the solution for this inequality is x > 2.

2. x^2 - x - 12 < 0

Let's factorize the quadratic equation:

(x - 4)(x + 3) < 0

Now, we can analyze the sign of the expression for different intervals:

Interval (-∞, -3): For x < -3, both factors are negative, so the expression is positive.

Interval (-3, 4): For -3 < x < 4, the factor (x + 3) becomes positive, while (x - 4) remains negative. Therefore, the expression is negative in this interval.

Interval (4, +∞): For x > 4, both factors are positive, so the expression is positive.

Combining these intervals, we find that the expression is negative when -3 < x < 4.

Therefore, the solution for this inequality is -3 < x < 4.