Mathematics formulae

THE QUADRATIC EQUATION FORMULAE_

X={-b±√(b^2-4ac)}÷2a

Where;
a= first term of the quadratic
b= second term of the quadratic
c= third term of the quadratic

The sum of a two digit number is 15. When the digits are reversed, the new number formed exceeds the original number by 9. What is the number?

SOLUTION

Let the number be YZ
Therefore, Y+Z=15....(1)

YZ reversed=ZY
Give the first digit the power of 10
(10Z+Y)-(10Y+Z)=9
Open the brackets
10Z+Y-10Y-Z=9
Collect like terms
10Z-Z-10Y+Y=9
9Z-9Y=9....(2)

Y+Z=15...(1)
9Z-9Y=9...(2)

USING SUBSTITUTION METHOD,
Y+Z=15
Y+Z-Z=15-Z
Y=15-Z
SO,
9Z-9(15-Z)=9
9Z-135+9Z=9
9Z+9Z=9+135
18Z=144
DIVIDE BOTH SIDES BY 18
Z=8

IF Z=8. BUT, Y+Z=15
THEN, Y+8=15
Y+8-8=15-8
Y=7

BUT THE NUMBER WAS YZ
REPLACING THE LETTERS WITH THE INTEGERS,
YZ=78

Mathematical mind will always give you a good idea