To calculate: The solution of a(by−2)≥b(2y+5).

Question

Brittany Patton · Accepted Answer

Formula used:
The addition principle for inequalities:
For any real numbers a, b and c:
$a < b$ is equivalent to $a + c < b + c$
$a > b$ is equivalent to $a + c > b + c$
Similar statement holds for $\leq and \geq$ .
The multiplication principle for inequalities:
For any real numbers a and b, and for any negative number c,
$a < b$ is equivalent to $a c < b c$ .
$a > b$ is equivalent to $a c > b c$ .
Similar statement holds for $\leq and \geq$ .
The distributive property is sometimes called the distributive law of multiplication and division.
Calculation:
Consider, the equation $a (b y - 2) \geq b (2 y + 5)$ .
Apply distributive law on inequality, $a (b y - 2) \geq b (2 y + 5)$ .
$a b y - 2 a \geq 2 b y + 5 b$
Subtract 2by on both the sides of the inequality $a b y - 2 a \geq 2 b y + 5 b$ .
$a b y - 2 b y - 2 a \geq 2 b y - 2 b y + 5 b$
$a b y - 2 b y - 2 a \geq 5 b$
Add 2a on both the sides of the inequality $a b y - 2 b y - 2 a \geq 5 b$ .
$a b y - 2 b y - 2 a + 2 a \geq 2 a + 5 b$ .
$a b y - 2 b y \geq 2 a + 5 b$
$y (a b - 2 b) \geq 2 a + 5 b$
Divide $(a b - 2 b)$ into both the sides of the inequality $y (a b - 2 b) \geq 2 a + 5 b$ .
$y \frac{(a b - 2 b)}{(a b - 2 b)} \geq \frac{2 a + 5 b}{(a b - 2 b)}$
$y \geq \frac{2 a + 5 b}{(a b - 2 b)}$
The inequality symbol will not change when we divided $(a b - 2 b)$ to the inequality because $a > 2 and b > 0 t h e n (a b - 2 b) < 0$ .
Hence, the solution to inequality ${y ∣ y \geq \frac{2 a + 5 b}{(a b - 2 b)}}$ .

To calculate: The solution of a(by-2) \geq b(2y+5).

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