Find 4 consectutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers.

Question

Find 4 consecutive even integers where the product of the two smaller numbers is 72 less than the product of the two larger numbers.
I'm never able to understand word problems...

Answer 1

Hey Veronica -- here's an intuitive approach ... since the difference is 72,

we need at least 10X8 ...

subtracting the 6x4 leaves 80-24=56 too small ... try the next set up

==> 12x10 minus 8x6= 120-48=72 ... 12,10,8,6  Best wishes, ma'am :)

Answer 2

First number: x

Second number: x+2

Third number: x+4

Fourth number: x+6

 

Product of two smaller numbers: (x)(x + 2) = x^2 + 2x

Product of two larger numbers: (x+4)(x+6) = x^2 + 6x +4x + 24 = x^2 +10x +24

 

product of the two smaller numbers is 72 less than the product of the two larger numbers:

x^2 + 2x = x^2 + 10x + 24 -72

So group your variables and you get: 

-8x = -48

x= 6

since you know x = 6, you know the even numbers that will come after it.

x = 16.5

 

 

Answer 3

x * (x +2) = (x + 4) * (x + 6) - 72

x² + 2x = x² + 10x + 24 - 72

subtract x² from both sides:

2x = 10x - 48

2x - 10x = -48

-8x = -48

x = 6

so, consecutive even integers are:

6, 8, 10, 12