you will desire to pass the fractions so which you have: (5)(n+3) - (6)(n+2) = (n+3)(n+2) Then amplify: 5n+15 - 6n+12 = n^2 + 5n + 6 Then simplify: -n + 3 = n^2 + 5n + 6 Then positioned each little thing on an identical area to = 0: n^2 + 6n + 3 = 0 Now as you won't be able to factorise, you are able to desire to apply the quadratic equation (the link shows it extra advantageous), x = b +or- sq. root of b^2 - 4ac/2a a is n^2 that's 1n so a = one million b is 6n so b = 6 and c = 3 so which you wind up with: x = -6 + v/ 36 - 12 / 2 which simplifies to: x = -6 + v/24 / 2 AND x = -6 - v/ 36 - 12 / 2 which simplifies to: x = -6 - v/24 / 2 remedy the two a kind of (simplified equations) and you wind up with the two values for x: *Angel*
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Verified answer
1 + √6-n = 3
√6-n = 2 {subtracting 1 from both sides}
6-n = 4 {squaring both sides}
6-4 = n {subtracting 4 and adding n to both sides}
n = 2
so, I pick...
d.)
;)
you will desire to pass the fractions so which you have: (5)(n+3) - (6)(n+2) = (n+3)(n+2) Then amplify: 5n+15 - 6n+12 = n^2 + 5n + 6 Then simplify: -n + 3 = n^2 + 5n + 6 Then positioned each little thing on an identical area to = 0: n^2 + 6n + 3 = 0 Now as you won't be able to factorise, you are able to desire to apply the quadratic equation (the link shows it extra advantageous), x = b +or- sq. root of b^2 - 4ac/2a a is n^2 that's 1n so a = one million b is 6n so b = 6 and c = 3 so which you wind up with: x = -6 + v/ 36 - 12 / 2 which simplifies to: x = -6 + v/24 / 2 AND x = -6 - v/ 36 - 12 / 2 which simplifies to: x = -6 - v/24 / 2 remedy the two a kind of (simplified equations) and you wind up with the two values for x: *Angel*
1 + √(6 - n) = 3
√(6 - n) = 3 - 1
6 - n = 2^2
n = 6 - 4
n = 2
(answer d)