From the problem you know two things. The two numbers have a quotient of 9 and a difference of 56. Step 1: Write these facts as equations. x/y = 9 x - y = 56 Step 2: Take one equation and solve for one of the variables. x/y = 9 (Multiply both sides by y) x = 9y Step 3: Substitute this result into the other equation and solve for variable. x - y = 56 9y - y = 56 8y = 56 y = 7 Step 4: Plug this result into either equation and solve for other variable. x - y = 56 x - 7 = 56 x = 63 Step 5: Test your answer in both equations. Is the result correct? x = 63; y = 7 63 - 7 = 56? YES 63/7 = 9? YES The two numbers are 63 and 7.
63 and 7
56 and 7
63/7 = 9 63 - 7 = 56 The two numbers are therefore 7 and 63.
There are no two consecutive primes whose sum, difference, product or quotient is 56.However, there may be some non-standard binary operation such that two primes can be combined to make 56.
x/y = 9 = quotient x - y = 56 = difference x = 9y and x = 56 + y therefore 9y = 56 + y and 9y - y = 56 so 8y = 56 therefore y = 56/8 = 7 and x =9y = 9x7 = 63 so x = 63 and y = 7 your two numbers are 63 and 7.
The two numbers will be 48 and 8. 48 + 8 = 56 48 / 8 = 6
The numbers are: 76 and 20
The quotient of 56 and 7 is 8.
Let the numbers be a and b. Then a + b = 56 And a - b = 18.........adding these two equations together to eliminate b gives : 2a + b - b = 56 + 18 : 2a = 74 : a = 37 Substituting for a in the first equation gives : 37 + b = 56 : b = 56 - 37 = 19 The two numbers are 37 and 19.
11, 12 -11, -12
(+7 and +8) and (-7 and-8)
56/b