Let the numbers equal x and y.... So you have the two equations x+y=77 and 6x=y Using substitution.... x+6x=77 7x=77 x=11 Plugging x back into our original equation.... 11+y=77 y=66 So your two numbers are 11 and 66
54
45
57
a=58.4615 b=5.2308 c=41.3076
= The sum of two numbers is -42 the first number minus the second number is 52 Find the numbers? =
54
45
57
= The sum of two numbers is -42 the first number minus the second number is 52 Find the numbers? =
a=58.4615 b=5.2308 c=41.3076
Take two numbers. Throw (the first number) into a box (the second number) times. The total amount in the box is then the product of the two numbers.
Add the first to itself.
four times a number is 6 more than three times a second number while 8 times the first number is 22 less than 7 times the second number. find the 2 numbers
y = 3x+8 x + y = 72 x = 16 y = 56
The numbers are 14, 16 and 18.
25 and 14
This problem asks us to find 2 numbers, n1 and n2, with the following relations between them: * n2 = 4 n1 * n1 + n2 = 45 Substituting the first equation into the second one gives us: * n1 + 4n1 = 45 which gives us * n1 = 9. We can now use this solution to find n2 with the first equation * n2 = 4 n1 = 36 So the first number is 4, the second number is 36.