1st = x
2nd = 5 - x
x(5 - x) = -36
5x - x2 = -36
5x - x2 + 36 = -36 + 36
5x - x2 + 36 = 0
-1(- x2 + 5x + 36 = 0)
x2 - 5x - 36 = 0
(x + 4)(x - 9) = 0
(x + 4) = 0 or (x - 9) = 0
x = -4 or x = 9
Thus, the integers are -4 and 9
Call your number "x", write an equation, and solve it: 8x = x + 49 7x = 49 x = 7
You can find those by trial and error. You can also write an equation for the three consecutive integers, and solve it. If the first number is "n", the others are "n + 1" and "n + 2". By solving the equation for "n", you get the first of the three numbers.
As a product of its prime factors: 2*2*2*3*3 = 72
Let x be the first integer. Then the sum of the three consecutive integers is x + (x+1) + (x+2), which equals 3x + 3. We are given that this sum is 43, so we can write the equation 3x + 3 = 43. Solving this equation, we find that x = 13. Therefore, the three consecutive integers are 13, 14, and 15.
1.) Identify the names of the reactants and the product, and write a word equation. 2.) Write a formula equation by substituting correct formulas for the names of the reactants and the products. 3.) Balancing the formula equation according to the law of conservation of mass. 4.) Count atoms to be sure that the equation is balanced.
x + y = 41, where x and y are unknown integers.
Call your number "x", write an equation, and solve it: 8x = x + 49 7x = 49 x = 7
* n=15
3(x+4)
i dont now
How about: 2x = 16 meaning that x = 8
You can find those by trial and error. You can also write an equation for the three consecutive integers, and solve it. If the first number is "n", the others are "n + 1" and "n + 2". By solving the equation for "n", you get the first of the three numbers.
7x11=77
As a product of its prime factors: 2*2*2*3*3 = 72
That isn't possible; three consecutive integers, or three consecutive positive integers, always have a sum that is a multiple of 3. In general, you can solve this quickly by trial and error. In this case, you will quickly find that a certain set of three consecutive integers will give you a sum that is TOO LOW, while the next-higher even integers will give you a sum that is TOO HIGH. You can also write an equation and solve it: n + (n + 2) + (n + 4) = 32. If you solve it, you will find that the solution is fractional, not integral.
84 = 2*2*3*7
25