Two solutions:x = 5 and y = 6, orx = 6 and y = 5.
xy = 20 x+y=9 x=9-y y(9-y) = 20 9y-y(squared) = 20 0= y(squared)-9y+20 0= (y-5)(y-4) y= 5 or 4, x= 5 or 4 The two numbers are 5 and 4.
Let x = one number = 2 or 2.5 Let y = the other number = 2.5 or 2 (x)(y)=5 x+y=4.5 y=4.5-x (x)(y)=5 (x)(4.5-x)=5 4.5x-x^2-5=0 I used the quadratic formula to get x. x=2 or x=2.5 (x)(y)=5 (2)(y)=5 y=5/2 y=2.5 (x)(y)=5 (2.5)(y)=5 y=5/2.5 y=2
The sum of x and y decreased by their product is (x + y)- xy.
What is the product of y and 10
if your question says 5(y+3) or 5 * (y+3) then its 5y+15
y - 5x
Two solutions:x = 5 and y = 6, orx = 6 and y = 5.
5 - 3y
The answer is y=3. barbrasydow@att.net.
let this integer be (xy) (xy) = 10x + y 10x+y = 2.x.y 10x + y - 2xy = 0 2x.(5-y) +y= 0 x= y / 2(y-5) when the integer provides this condition, it is equal to twice the product of its digits. And there is such only one integer. 36
W+5y
As an algebraic expression it is: 3y-5
xy = 20 x+y=9 x=9-y y(9-y) = 20 9y-y(squared) = 20 0= y(squared)-9y+20 0= (y-5)(y-4) y= 5 or 4, x= 5 or 4 The two numbers are 5 and 4.
x = 10, y = 25
Let x = one number = 2 or 2.5 Let y = the other number = 2.5 or 2 (x)(y)=5 x+y=4.5 y=4.5-x (x)(y)=5 (x)(4.5-x)=5 4.5x-x^2-5=0 I used the quadratic formula to get x. x=2 or x=2.5 (x)(y)=5 (2)(y)=5 y=5/2 y=2.5 (x)(y)=5 (2.5)(y)=5 y=5/2.5 y=2
5y-8 = 16 5y = 16+8 5y = 24 y = 24/5 = 4.8