If you mean: (x+3)^2 = (x+2)^2 +13 Then the value of the number x works out as 4
(x + 3)^2 - 13 = (x + 2)^2 x^2 + 6x + 9 -13 = x^2 + 4x + 4 x^2 + 6x - 4 = x^2 + 4x + 4 6x - 4x = 4 + 4 2x = 8 x = 4
x + (13-10)
Let x be the number you seek. 6(x+2) = 5x + 13 Solve for x: 6x + 12 = 5x + 13 x = 1 Verify: 6(3) = 5(1) + 13 18 = 18
3(x + 7) > 5x - 13 3x + 21 > 5x - 13 21 + 13 > 5x - 3x 34 > 2x 17 > x The solution set: {x| x < 17}
[(x+3)^2] - [(x+2)^2] = 13 ==> 2x = 8 ==> x = 4. the number is 4!
If you mean: (x+3)^2 = (x+2)^2 +13 Then the value of the number x works out as 4
Let the number be x and so if x+13 = 75 then the value of x must be 62
It is: 13+x+7x or as 13+8x
x+10>13 it wouldn't be x+10-13 ??
Let the first number be x and the second number is 27 − x. [As the sum of both the numbers is 27] Therefore, their product = x (27 − x) It is given that the product of these numbers is 182. x (27 – x) = 182 x2 + 27x - 182 = 0 Changing the signs on both sides we get,x2 - 27x + 182 = 0 Factorizing we get , 13 and 14 are the numbers whose sum is 27 and product is 182 x2 – 13x – 14x + 182 = 0 = x(x – 13) – 14 (x – 13)= 0 = (x – 13) (x – 14) = 0 Either x – 13 = 0 or x − 14 = 0 i.e., x = 13 or x = 14 If first number = 13, then Other number = 27 − 13 = 14 If first number = 14, then Other number = 27 − 14 = 13 Therefore, the numbers are 13 and 14.
call the first number x. The second number would be x + 2 (as they're consecutive odd numbers). The sum of both numbers would be x + x + 2 = 28, or 2x + 2 = 28. This can be evaluated to 2x = 26, x = 13. The first number would be 13, the second 15, 13 + 15 = 28.
(x + 3)^2 - 13 = (x + 2)^2 x^2 + 6x + 9 -13 = x^2 + 4x + 4 x^2 + 6x - 4 = x^2 + 4x + 4 6x - 4x = 4 + 4 2x = 8 x = 4
x + (13-10)
Le the "number" be x. The sum is x + 6.
Let x be the number you seek. 6(x+2) = 5x + 13 Solve for x: 6x + 12 = 5x + 13 x = 1 Verify: 6(3) = 5(1) + 13 18 = 18
Let 'N' be the number. The sum of the number and 13 is (N+13).The square of the sum of the number and 13 is (N+13)2N2 + 26 N + 169.It's probably easier to add 13 to the number and then square that answer ... less arithmetic.