5
8 + y
8+y is the sum of 8 and y
abs(y^3 - 4) < 8*y
Sum of digits x+y=8 Two digit no assume as 10x+y It is given that 10x+y+18=10y+x 9x-9y+18=0 x-y=-18/9 x-y= -2 Equate both x+y-8=x-y+2 2y=10 y= 5 Substitute this y in x+y x+5=8 x=3 So 35 is original two digit no Reversed new no is 53
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
It is an algebraic equation.
The expression "twice the quotient of 50 and the sum of a number y and 8" can be written mathematically as ( 2 \times \frac{50}{y + 8} ). This represents taking the sum of ( y ) and 8, finding the quotient of 50 divided by that sum, and then multiplying the result by 2.
y = x +8 where x is the number
Y = 4 (x+8)
8 + y
8+y is the sum of 8 and y
The word sum is the clue. It means the problem will be an addition problem.y + 3(8) = ?y + 24 = ?
The algebraic expression that represents fifteen less than the sum of ( y^3 ) and seven is ( (y^3 + 7) - 15 ). This simplifies to ( y^3 - 8 ).
In an addition sum, if the missing number is the first number, for example, x + 5 = 10, then to find x, perform the sum 10 - 5, producing the solution x = 5. In a subtraction sum, if the missing number is the second number, for example, 7 - y = 4, then to find y, perform the sum 7 - 4 = 3, production the solution y = 3.
y + 3
abs(y^3 - 4) < 8*y
Let x be the unknown number and let y be the ten's digit,then x = 10y + y - 3 = 11y - 3,and x = 8 + 6(y + y - 3) = 8 + 6(2y - 3) = 8 + 12y - 18 = 12y -10,therefore 12y - 10 = 11y - 3.Subtracting 11y and adding 10 to each side, y = 7.Since the ten's digit is 7, the unit's digit must be 4, and the number is 74.