Use the zero product property to find the solutions to the equation 6x^2-5x=56

6x² -5x -56 = 0 . . . . . subtract the constant

(3x +8)(2x -7) = 0 . . . .factor*

The zero product property says the product will be zero only when one (or both) of the factors is zero.

First Factor

... 3x +8 = 0

... x + 8/3 = 0 . . . . divide by the coefficient of x

... x = -8/3 . . . . . . subtract 8/3

Second Factor

... 2x -7 = 0 . . . . . . set the factor to zero

... x -7/2 = 0 . . . . . divide by the coefficient of x

... x = 7/2 . . . . . . . add 7/2

_____

*Comment on factoring

There are various ways to do this. In basic terms, we want to find numbers that are factors of the product 6·(-56) whose sum is -5. We found those numbers to be -21 and 16. Then we can rewrite the -5x term using these numbers and factor by grouping.

... 6x² -5x -56 = (6x² -21x) +(16x -56) = 3x(2x -7) +8(2x -7) = (3x +8)(2x -7)

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~