The sum of twice a number and five is greater than three times the number minus three.
2 answers:
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Answer:
n = , n =
Step-by-step explanation:
6n² - 5n - 7 = - 8 ( add 8 to both sides )
6n² - 5n + 1 = 0 ← in standard form
Consider the product of the factors of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term
product = 6 × 1 = 6 and sum = - 5
The factors are - 3 and - 2
Use these factors to split the n- term
6n² - 3n - 2n + 1 = 0 ( factor the first/second and third/fourth terms )
3n(2n - 1) - 1(2n - 1) = 0 ← factor out (2n - 1) from each term
(2n - 1)(3n - 1) = 0 ← in factored form
Equate each factor to zero and solve for n
3n - 1 = 0 ⇒ 3n = 1 ⇒ n =
2n - 1 = 0 ⇒ 2n = 1 ⇒ n =