two numbers have the sum of 54 and a product of 200. What are the two numbers?

x + y = 54

xy = 200

y = 54 - x

x(54 - x) = 200

54x - x^2 = 200

x^2 - 54x + 200 = 0

x = (54 +/- sqrt(2916 - 800)) / 2

x = (54 +/- sqrt(2116)) / 2

x = (54 +/- 46) / 2

x = 4 and x = 50

4 + y = 54

y = 50

50 + y = 54

y = 4

So it's 50 & 4.

I hope this information was very helpful.

a + b = 54

a x b = 200

b = 200/a

Combine the equations

a + 200/a = 54

a^2 + 200 = 54a

a^2 - 54a + 200 = 0

(a - 50) (a - 4) = 0

So a must be 50 or 4

a + b = 54

if a is 50 then b is 4

is a is 4 then b is 50

So the numbers are 4 and 50

50 + 4 =54

50 x 4 =54

50+4 = 54

50*4 = 200

The two numbers are 50 and 4

50+4 = 54

50*4 = 200

hope this helps :)

x+y=54----(1)

x*y=200----(2)

From (1), x=54-y

Then (2) becomes, (54-y)y=200

54y-y^2=200

y^2-54y+200=0

(y-50)(y-4)=0

ie, y=50 or 4

Put these y in (1)

if y=50 then x+50=54==> x=4

if y=4 then x+4=54==> x=50

So the numbers are 50 and 4

50+4=54 and 50x4=200

x + y = 54

xy = 200

solve simultaneously (substitution is the best method here)

You will get, as stated above, 50 and 4.

x^2-54x+200=0

x's=[27+&-(27^2-200)^1/2]=50 & 4

God bless you.