SOLUTION: How do you factor these equations: x(squared) + 3x -4 = y(squared) + 13y - 48= And how would you solve these equations: (x-3) (x-7)= 0 b(squared) + 3b

SOLUTION: How do you factor these equations: x(squared) + 3x -4 = y(squared) + 13y - 48= And how would you solve these equations: (x-3) (x-7)= 0 b(squared) + 3b - 4= 0 n(squared)

Question 138045: How do you factor these equations:
x(squared) + 3x -4 =
y(squared) + 13y - 48=
And how would you solve these equations:
(x-3) (x-7)= 0
b(squared) + 3b - 4= 0
n(squared) + n - 12= 0
5x (squared) + 18x = 8
3a (squared) + 4a = 2a(squared) - 2a - 9

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How do you factor these equations:
x(squared) + 3x -4 = (x+4)(x-1)
y(squared) + 13y - 48= (y+16)(y-3)
----------------------
And how would you solve these equations:
(x-3) (x-7)= 0
x = 3 or x=7
----------------------
b(squared) + 3b - 4= 0
(b-4)(b+1)=0
b=4 or b=-1
------------------
n(squared) + n - 12= 0
(n+4)(n-3) = 0
n = -4 or n = 3
-------------------
5x (squared) + 18x = 8
5x^2 + 18x -8 = 0
5x^2 +20x-2x -8 = 0
5x(x+4) -2(x+4) = 0
(x+4)(5x-2) = 0
x = -4 or x = 2/5
-------------------------
3a (squared) + 4a = 2a(squared) - 2a - 9
3a^2 + 4a = 2a^2 -2a-9
a^2 + 6a + 9 = 0
(a+3)^2 = 0
x = -3 with multiplicity two
===========================
Cheers,
Stan H.

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