If 2 times an integer x is increased by 7, the result is always greater than 16 but less than 29. (a) Write down an inequality in x.
(b) Solve this inequality and write down the least possible integer value of x.

X=6

Step-by-step explanation:

Turning the question into a formulae

2X + 5 > 16; and 2X + 5 < 29

Let’s start by calculating the lower limit

2X + 5 = 16

X = (16–5)/2

X = 11/2 = 5.5

X > 5.5

The lowest value of X is infinitesimally close to 5.5, but greater than 5.5.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15