2x - 6x + 3x = -412x - 3x = -41-1x = -41x = 41
44
x=5
Let the number be x and so if 3x+5 = 41 then x = 12
3x = 24, x = 8
2x - 6x + 3x = -412x - 3x = -41-1x = -41x = 41
3x-x44 = -41
44
41
It is -41.
x=5
Let the number be x and so if 3x+5 = 41 then x = 12
Presumably you mean: -2x2-3x+41 If so then the discriminant is: (-3)2-4*-2*41 = 337 The expession is: i/2+i
3x = 24, x = 8
3x-28-5x=5x+13 Let us first solve the variables -2x-28=5x+13 Transposing -28 to Right hand side -2x=5x+13+28 -2x=5x+41 Transposing 5x to left hand side -2x-5x=41 -7x=41 equation 1 x=41/-7 x=-41/7 Hence the required answer is -41/7. You can verify it by substituting the value of x in the given equation. If Right hand side is equal to Left hand side the answer is right. Let us verify Substituting the value of x=-41/7 in the given equation1 -7(-41/7)=41 -7 X -41/7 =41 41=41 LHS=RHS Hence Verified
The consecutive integers are 39, 40, and 41. To solve this algebraically, note that consecutive numbers can be indicated by the variables x, x+1, and x+2. For the sum x + (x+1) + (x+2) = 120, 3x +3 = 120 3x= 117 x= 39, followed by 40 and 41 39+40+41 = 120
39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43