If f(x)=(x-2)(x+3), then f(1)=

-4

Step-by-step explanation:

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The numbers are:  "30" and "32" .

The two (2) consecutive even numbers are:  "x" and "(x+2)" .
We are asked to find these numbers.  
If we can solve for "x" (one of these numbers); we can solve for "(x+2)" by adding "2" to "x".

  →   (⅔) x + (¼) (x + 2) = 28 ;  Solve for "x" ;

Note the distributive property of multiplication:

 a(b+c) = ab + ac ;
 
 a(b − c) = ab − ac ;

As such:  " + (¼) (x + 2) = (¼) x  + (¼) (2)  =  (¼) x + ½ ;

Rewrite the equation:  →   (⅔) x + (¼) (x + 2) = 28 ; 

as:  " →   (⅔) x + (¼) x + ½  = 28 ;

Multiply the entire equation by "12" ; to get rid of the fractions:

      → 12 * { (⅔) x + (¼) x + ½  = 28 } ;

      → (24/3) x + (12/4)x + (12/2) = 28*(12) ;

            →  8x + 3x + 6 = 336

            →  11x + 6 = 336 ;

Subtract "6" from each side of the equation:

            →  11x + 6 − 6 = 336 − 6 ; 

            →  11x = 330 ;

Divide each side of the equation by "11" ; 
   to isolate "x" on one side of the equation; & to solve for "x" ;

            → 11x / 11 = 330 / 11 ;

             → x =  30 ; 

The next consecutive even integer is: 
 "(x + 2)" ; which equals:  "30 + 2 = 32" .

The numbers are:  "30" and "32" .

Let us check our answer, by plugging the value for "x" in the original equation; to see if it holds true:

  →   (⅔) x + (¼) (x + 2) = 28 ; 

  →   (⅔) (30) + (¼) (30 + 2) = ? 28 ? ;

  →  (60/3)  +  (¼)*(32) = ? 28 ? ;

      →   20 + 8 = ? 28 ? ;  Yes!