10 times the first of 2 consecutive even integers is 8 times the second. Find the integers

10 times the first of 2 consecutive even integers is 8 times the second. Find the integers.

Let the first integer be x. Let the second integer be y. We're given:

1. 10x = 8y
2. We also know a consecutive even integer means we add 2 to x to get y. y = x + 2

Substitute (1) into (2):
10x = 8(x + 2)

Multiply through:
10x = 8x + 16

To solve for x, we type this equation into our search engine and we get:
x = 8

Since y = x + 2, we plug in x = 8 to get:
y = 8 + 2
y = 10

Now, let's check our work. Does x = 8 and y = 10 make equation 1 hold?
10(8) ? 8(10)
80 = 80 <-- Yes!