3x2-4x-15 = 0
(3x+5)(x-3) = 0
x = -5/3 or x = 3
5x = 15 so x = 15/5 = 3
x+15=16x=16-15x=1
If: 4x = x+15 Then: x = 5
3x + 5y = -1 2x - 5y = 16 Add the two equations: 5x = 15 or x = 3 Substitute for x in the first equation: 3*3 + 5y = -1 15 + 5y = -1 5y = -10 so y = -10/5 = -2 The solution is (x, y) = (3, -2)
x = 13
It is: x = 5/3 and x = -3
What is the solution set of 2x2 + x = 15
no 3x2=6 then 4x-15= -60 so 6 added with - 60 = -54
3x2 + 4x - 15 = 0, solve by using the quadratic formula. a = 3, b = 4, c = -15 x = [-b ± √(b2 - 4ac)]/(2a) x = [-4 ± √(42 - 4*3*-15)]/(2*3) = [-4 ± √(16 + 180)]/6 = (-4 ± √(196)/6 = (-4 ± 17)/6 x = (-4 +17)/6 = 13/6 or x = (-4 - 17)/6 = -21/6 = -7/2 The solution set: {-7/2, 13/6}.
the empty set.
(2x3)+(3x5)-(3x2)= 2x3=6 3x5=15 3x2=6 So..... 6x25-6= 6x25=150 150+6=156
Factors: (2x + 3)( x - 5) so x = -1.5 or 5
5x = 15 so x = 15/5 = 3
x = -5
15-b=7.
-5
Y=-6