10 Answers

(8) + (15) = 23
and
(8) (15) = +120
1) Form equations and solve to get the answer.
Here, the eqns are x + y = 23 and xy = 120
y = 23 x
Sub in the other as x(23x) = 120
23xx^2 = 120
x^2 + 23x + 120 = 0
Solution for this is
x = [ 23 + sqrt{23^2  4(1)(120)} ]/2 and
x = [ 23  sqrt{23^2  4(1)(120)} ]/2
Solve and u'll get 8 and 15
2) Trial & Error
120 = 12 * 10
= 4 3 5 * 2
= 4 ( 3 5) * 2 > Combine suct that their sum is 23.
= (42) (35)
= 8 * 15 >8+15=23
Since the req. 2 nos. add to give 23, ur answer should be negative.
Therefore, 8 and 15
It is known that multiplication of 2 negative nos. yields a positive number.

Many of these types of problems are trial and error.
First, what two numbers add up to 23 that are fairly big numbers (since they have to make a large number: 120).
After trial and error, I figured out it is 8 and 15.
Hope that helps!

x + y = 23
xy = 120
x( 23  x) = 120
23x  x^2 = 120
0 = x^2 + 23x + 120
0 = (x + 15)(x + 8)
x = 15 or x = 8

x+y=23 ==> y=x23
xy=120
x(x23)=120
x^2+23x+120=0
Solve for quadratic equation:
[23+/ sqrt(23^24(1)(120))]/2
x = 15, y = 8

What Two Numbers

X + Y = 23
X times Y = 120
Use this system of equation to solve.
you will get 15 and 8

8 and 15

15 and 8

x= first number
y= second number
1st equation x+ y = 23 or y= 23x
2nd equation x*y= 120
Substituting 1 in 2
.... x times (23x)= 120
Now please solve the quadratic equation and find the x. If nobody does iyt for you and you cant emaill me

x+y=23
x=23y
Also
xy=120
so substitute the 23y for x and you get
y(23y) =120
y^223y=120
Then use the quadratic formula.
x is plus or minus 11.5,