The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:

b² + 2b + 1 = 2b + 5
b² + 2b + 1 - 5 = 2b
b² + 2b - 2b - 4 = 0
b² - 4 = 0

Next, factor the quadratic equation:

b² - 4 = 0
(b - 2)(b + 2) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (b - 2) or (b + 2) must equal zero:

If (b - 2) = 0, b must equal 2
If (b + 2) = 0, b must equal -2

So the solution is that b = 2 or -2

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the < sign and the answer on the other.

-2c - 3 < \( \frac{c}{9} \)
9 x (-2c - 3) < c
(9 x -2c) + (9 x -3) < c
-18c - 27 < c
-18c - 27 - c < 0
-18c - c < 27
-19c < 27
c < \( \frac{27}{-19} \)
c < -1\(\frac{8}{19}\)

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c²) is equal to the sum of the two perpendicular sides squared (a² + b²): c² = a² + b² or, solved for c, \(c = \sqrt{a + b}\)

To solve this equation, isolate the variable for which you are solving (c) on one side of the equation and put everything else on the other side.

5c + y = -c - 3y + 4
5c = -c - 3y + 4 - y
5c + c = -3y + 4 - y
6c = -4y + 4
c = \( \frac{-4y + 4}{6} \)
c = \( \frac{-4y}{6} \) + \( \frac{4}{6} \)
c = -\(\frac{2}{3}\)y + \(\frac{2}{3}\)

You need to find the value of b so solve the first equation in terms of y:

8b + y = 1
y = 1 - 8b

then substitute the result (1 - 8b) into the second equation:

-7b - 3(1 - 8b) = 2
-7b + (-3 x 1) + (-3 x -8b) = 2
-7b - 3 + 24b = 2
-7b + 24b = 2 + 3
17b = 5
b = \( \frac{5}{17} \)
b = \(\frac{5}{17}\)