An integer is 3 less than 5 times another. If the product of the two integers is 36, then find the integers.

let

x  = one integer

y = another integer

x = 5y - 3

If the product of the two integers is 36, then find the integers.

x * y = 36

(5y - 3) * y = 36

5y² - 3y = 36

5y² - 3y - 36 = 0

Solve the quadratic equation using factorization method

That is, find two numbers whose product will give -180 and sum will give -3

Note: coefficient of y² multiplied by -36 = -180

5y² - 3y - 36 = 0

The numbers are -15 and +12

5y² - 15y + 12y - 36 = 0

5y(y - 3) + 12 (y - 3) = 0

(5y + 12) (y - 3) = 0

5y + 12 = 0      y - 3 = 0

5y = - 12           y = 3

y = -12/5

The value of y can not be negative

Therefore,

y = 3

Substitute y = 3 into x = 5y - 3

x = 5y - 3

x = 5(3) - 3

= 15 - 3

= 12

x = 12

Therefore,

(x, y) = (12, 3)

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3, 12

Step-by-step explanation:

Et x and y be the required integers.

Case 1: x = 5y - 3...(1)

Case 2: xy = 36

Hence, (5y - 3)*y = 36

Hence, the required integers are 3 and 12.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.