#maths 2

[isehs]

A number exceeds 4 times its reciprocal by 3. Find the number.

•Let the number be x "A number exceeds 4 times its reciprocal..." x = 1/x *4 x = 4/x "...by 3..." x = 4/x + 3 •Now multiply all the terms by x so that the x below 4 can be canceled. x^2 = 4 + 3x x^2 - 3x - 4 = 0 •You can now solve this sum using the quadratic equation method. x^2-3x-4=0 Product=1*-4=-4 Sum=-3 The numbers which give the product of -4 and a sum of -3 are -4 and +1. x^2-3x-1x-4=0 x(x+4)-1(x+4)=0 (x-1)(x+4) So the possible answers are: x-1=0 x=1 OR x+4=0 x=-4 Both of these answers are correct and you can check them with this equation that we formed earlier: x=4/x+3

[isehs]

The sum of the digits of a two-digit number is 14. If the digits are reversed, the number is 18 less than the original number. Find the original number.

Let x = the 10's digit Let y = the units : 10x + y = the original two digit number : The sum of the digits of a two-digit number is 14. x + y = 14 : "If the digits are reversed, the number is 18 less than the original number." Rev no. = orig no. - 18 10y + x = 10x + y - 18 10y - y = 10x - x - 18 9y = 9x - 18 simplify, divide by 9 y = x - 2 : In the sum equation replace y with (x-2) x + (x-2) = 14 2x = 14 + 2 2x = 16 x = 8 obviously y = 6 : 86 is the original number : : Check solution in the statement: "If the digits are reversed, the number is 18 less than the original number." 68 = 86 - 18