Dude, stop trying to cheat on your own maths enrichment task loser.
3x + 6y = 47 6y = -3x + 47 y = -3x/6 + 47/6 y = (-3x + 47)/6 Since we are looking for positive integer, 3x (x > 0) cannot be greater than 47, the possibilities of the numerator are 42, 36, 24, 18, 12, and 6, so that - 3x must be -5, -11, -23, -29, -35, and -41. Since any of them are not multiples of -3, we can not find any pair of positive integers that satisfy the equation. So x and y cannot be both positive integers.
The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11
x = -6 or 65x2 - 136 = 445x2 = 180x2 = 36
3
why will the equations x+14=37 and x-14=37 have different solutions for x
There are 120 solutions.
There are 6 such triples.
x^2 + x = 110. x(x + 1) equals x squared + x.
The quadratic equation will have two solutions.
This question cannot be answered because there are no operations defined, such as +, -, x, /, and = .
The rule in dividing integers is to divide the absolute values. Two positive integers or two negative integers equals positive product. If one integer is positive and the other is negative, the product is negative.
6
There are no solutions because the discriminant of this quadratic equation is less than zero
The first three positive integers, 1, 2, and 3, satisfy this condition.
using the t-table determine 3 solutions to this equation: y equals 2x
Negative times negative equals positive.
A positive... in basic math (integers).