There are 120 solutions.
Assuming that you are reffering to something like this: (x - h)(x - k) = 0 x = h, x = k This is the fundamental theorem of algebra which states that is given a polynomial (multiple terms raised to positive powers ex) x^3 + 2x + 1), then the number of solutions to that polynomial is equal to the degree (or highest exponent) in the polynomial. The factorization in the beginning was dealing with a quadratic equation - when foiled out it equals x^2 - hx - kx + hk. The highest exponent in the quadratic is two and therefore there are two solutions. You can even think back to the factorization again: if x = h then the whole equation is 0, if x = k then the whole equation is 0.
To solve the equation "x times y minus z equals 13," we need to set up the equation as follows: xy - z = 13. Since we are looking for three unknowns, there are multiple solutions to this equation. One possible solution could be x = 5, y = 3, and z = 2, as 5 times 3 minus 2 equals 13.
First we can solve for y by factoring it out: y(x+1)+x=30, so y=(30-x)/(x+1)= -1+31/(x+1). Let's start by assuming x and y are positive integers and look for a logical contradiction. Since y is positive, the right side (30-x)/(x+1) must be positive. Since x (and therefore x+1) is positive, 30-x must be positive. Therefore x is less than 31. But hold on! Since y is an integer, y+1=31/(x+1) is an integer. Since 31 is only divisible by 1 and 31, y+1 is an integer implies that (x+1) is 1 or 31, making x either 0 or 30. However, x is positive and less 30, which is impossible! There it cannot be the case that x and y are both positive integers.AnswerIt's not as complicated as that. Just add 1 to both sides and factorise, and you get: (x+1)(y+1) = 31. Since 31 is prime, one of its factors must be either 1 or -1. So we'd get x+1=1 (so x=0) or else x+1=-1 (so x=-2), or else y=0 or y=-2. But the question says x and y have to be positive.
The question is suppose to read: Find the equation of the line tangent to y=(x²+3x)²(2x-2)³, when x=8
give an example of two fractions whose product equals 1
This question cannot be answered because there are no operations defined, such as +, -, x, /, and = .
The person or program that solves the equation does.
If x = y, then the equation is true for any integer pair. Otherwise, the equation is not equivalent and is impossible.
Negative.
A negative integer. Every time.
The quadratic equation will have two solutions.
You will get a positive integer. If you subtract a negative number, you will be adding it. I think of it like 2 minuses equals a plus. :P
This equation describes all the points on the unit sphere. There is an infinite number of solutions. Some quick integer solutions would be (1,0,0) and (0,1,0) and (0,0,1) which are the one the axes.
There are no solutions because the discriminant of this quadratic equation is less than zero
6
it cant a negative + negative always equals a positive -Posted By Anonymous 7th Grader it cant be a negative plus a negative equals a negative always -Posted BY Anonymous 7th Grader♥
Set up an equation using x for the integer. The equation would be: x/(-4+6)=-15 Firstly find the sum of -4 and 6. It would be 2. The equation is now: x/2=-15 Multiply both sides by 2. This means that: x=-30 Plug it back into the equation to check: -30/(-4+6)=-15 -30/2=-15 -15=-15 It works therefore -30 is the integer that when divided by the sum of negative 4 and positive 6 equals negative 15.