How many sets of natural number solutions does the equation 2 X + Y + Z = 8 have?
The total being a small number one can write down all possible combinations. if the total is large and only number of combinations is required,one can use the divider method.The divider method is suitable for non negative numbers( natural numbers and zero)but we can modify the method.let x=a+1,y=b+1 and z=c+1 in which case a,b,c can take zero values also.Substituting we get 2a+b+c=4(1)Consider a=0 b+c=4 we can consider 4 dots on a line and one more dot as a divider between them.any dot can be considered as divider.If leftmost is chosen as divider b=0 and c=4.if 3rd dot is chosen as divider b=2 and c=2. so the problem is choosing one dot (as divider) out of 5 dots. It is [math]^5C_1 [/math]= 5 ways.(2)a=1 ; b+c=2; 2+1=3 dots out of which one is to be chosen. So [math]^3C_1[/math]=3 ways(3) a=2 b+c=0 total dots=0+1;there is only one way to choose.So totally we have 5+3+1 =9 ways
Which numbers below belong to the solution set of the inequality?
Let's start by solving for x. 8x >= 64 Divide by 8 on both sides to get x alone. x >= 8 So, the numbers that are 8 or above would be .. A. 12 C. 9 D. 10 F. 8
How do you solve equations over a set of complex numbers?
It just means to solve the problem, even if the answers are complex numbers with an imaginary part, like 3 + 4i or some such non-real number. (i stands for the square root of -1, which is why it's "imaginary.") Your first example, 6x^2-5x+3=0, has such answers. If you apply the quadratic formula, you get x = (5 +- √(25- 4*6*3))/2*6 x = (5+- √(25-72)) / 12 x = (5 +- √-47) /12 x = 5/12 +- (√47/12)*i You want any complex answers to be expressed in the form a + bi, where a and b are both real numbers, so your two answers are: x = 5/12 + (√47/12)*i and x = 5/12 - (√47/12)*i
How can I solve this, The number of solutions of the equation x+y+z = 10 in positive integers x,y, z is equal to (A) 36 (B) 55 (C) 72 (D) 45?
Imagine 10 as being 10 balls on a line. To find set of all three positive non zero integers which adds up to 10, all you have to do is to trifurcate those 10 balls into 3 parts. For this purpose, you can imagine 2 bars which can be placed in between the balls. There exactly are 9 places where the 2 bars can be placed to divide balls into 3 different sets. All the possible division of balls in three sets will add up to 10. So, the problem of finding three positive integers which add up to 10 is equivalently transformed into the problem of finding no of ways in which two bars can be placed at 9 possible locations. Applying combinatorics, this comes out to be 36 (9C3=36).To learn basics- look stars and bars article at wiki.
How many solutions does the equation x1+x2+x3=11 have where x1, x2, and x3 are Natural Numbers?
We can solve this problem in a case where ``natural number" is meant to include zero (see Is zero a natural number? Why or why not?) Otherwise, we can just say that [math]x_1 = 1 + y_1[/math], etc., and solve the same problem with the sum on the right-hand side being smaller by 3.There is a general way to find the number of solutions to [math]x_1 + x_2 + ... + x_m = n[/math]. Suppose that we have [math]n[/math] balls and [math]m - 1[/math] dividers (all indistinguishable). Then there is a one-to-one mapping between solutions to the equation above and arrangements between balls and dividers (by taking [math]x_1[/math] to be the number of balls to the left of the first divider, [math]x_2[/math] the number of balls between the first and second, etc.) The number of such arrangements is just the number of ways to choose [math]n[/math] elements from a set of [math]n + m - 1[/math], which is [math]n + m - 1\choose m - 1[/math]. In this case, it would be [math]13 \choose 2[/math] or 78 (if we exclude zero, it would be [math]10 \choose 2[/math] or 45).
Solving equations in the indicated number set...?
It's asking you to find x. In the first set, since the domain is real, it just means your answer can be anything from lower than -99999 to 999999 etc. But if your domain, for example, is (0,3], then your answer must be greater than 0 (when there is a round bracket you don't include the point) and less than or equal to 3 (you include the point). So, 6-3x= -31 -3x = -31 - 6 -3x = -37 x = -37/-3 x = 37/3 or if you want it in decimal form, x= 12.333
How do you solve the equation in the complex number system: x^4=16? And what is the solution set?
I am glad you asked “How do you solve” and not just “What is the answer”. This shows you really want to understand and learn something!This method will work beautifully in all cases like this one.x^4 = 16The fundamental Theorem of Algebra says that this equation has four solutions.You can find these solutions using De Moivre’s Theorem as follows:Let x = rcis(θ)
What is the solution set of the equation | 2x | = 6?
2x=6 =x=6/2=3