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
To write an equation for "thirty is five minus the product of five and a number," first define the unknown number as ( x ). The phrase "the product of five and a number" translates to ( 5x ). Therefore, the equation can be written as ( 30 = 5 - 5x ).
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.
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)
To write an equation for "thirty is five minus the product of five and a number," first define the unknown number as ( x ). The phrase "the product of five and a number" translates to ( 5x ). Therefore, the equation can be written as ( 30 = 5 - 5x ).
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.
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.
As a product of its prime factors: 2*2*2*3*3 = 72
25
84 = 2*2*3*7