Question

6(x-2)+4(x+1)<7 x^2+3x-4>0

Answer 1

6x-12+4x+4-7<0
x^2+3x-4>0
Смею предположить что это система, тогда.
1)10x-15<0       2) x^2+3x-4>0
    10x<15              D=9+16=25
     x<1.5                 x1=-3+5/2=1 x2=-8/2=-4
                                x⊂(-∞;-4) ∪ (1; +∞)
Находим их общее решение.
Ответ: x ⊂ (-∞;-4) ∪ (1;1.5)

Answer 2

To solve the inequalities, let's take them one at a time:

1. Solve the inequality 6(x - 2) + 4(x + 1) < 7:

Step 1: Distribute the terms inside the parentheses. 6x - 12 + 4x + 4 < 7

Step 2: Combine like terms. 10x - 8 < 7

Step 3: Add 8 to both sides to isolate the variable term. 10x < 15

Step 4: Divide both sides by 10 to solve for x. x < 1.5

So, the solution for the first inequality is x < 1.5.

2. Solve the inequality x^2 + 3x - 4 > 0:

Step 1: Factorize the quadratic expression. (x + 4)(x - 1) > 0

Step 2: Determine the critical points by setting each factor equal to zero and solving for x. x + 4 = 0 -> x = -4 x - 1 = 0 -> x = 1

Step 3: Create a sign chart to find the regions where the expression is positive (greater than zero).

scssCopy code
-4      1
|------|------|------|
(-)     (+)    (-)

Step 4: The expression is greater than zero (positive) in the intervals (-∞, -4) and (1, ∞).

So, the solution for the second inequality is x < -4 or x > 1.

To summarize:

1. 6(x - 2) + 4(x + 1) < 7: Solution is x < 1.5
2. x^2 + 3x - 4 > 0: Solution is x < -4 or x > 1