Find 2 numbers such that their sum is 9 times their difference and twice the smallest number exceed the largest by 21.

Question

I please want the calculations to be able to also understand, especially where they say twice the number exceed the larger number by 21.

Answer 1

let the 2 numbers be S and L, with S=the smaller number and L = the larger number

S+L = 9(L-S)

2S= L+21

L=2S -21

S+2S-21 = 9(2S-21 -S)

3S = 9S -189+21 = -168

6S = 168

S = 168/6 = 28

L =56-21 = 35

L+S = 63 = 9(45-28) = 9(7)

Answer 2

Let a and b be the two numbers such that a > b.

Their sum is 9 times their difference can be written as the equation: (a+b) = 9(a-b)

For the second part, twice the smaller number exceeds the larger number by 21 means 2 times the smaller number is 21 more than the the larger number or 2b = a + 21.

So now you have a system of equations with 2 equations and 2 unknown.

So just solve for a and b.