Find all real numbers x such that 3x-7<5x+9 in interval notation THE LESS THAN SIGN IS LESS THAN EQUAL TO

3x-7 ≤ 5x+9 subtract 5x from both sides

-2x - 7 ≤ 9 add 7 to both sides

-2x ≤ 16

Divide both sides by -2 and reverse the inequality sign [ since we are dividing an inequality by a negative ]

x ≥ -8

If it's in interval notation, wouldn't it be \([-8, \infty)\) ?

Correct!!!!....I forgot about the interval notation.......!!!!