Please help me to solve this equation.
Here is the question:
a+b=c implies ac<+>bc = ab. Use this to create three reductions involving integers only. (E.g. 1+2=3, so 3<+>6=2.)
Heck, they gave you an example, so that makes it easy. Just pick any values for a,b,c:
9-3=6 ==> 9*6<+>-3*6=9*-3 or 54<+>-18=-27
Who cares what <+> really means?
Thanks Steve
To solve the equation a + b = c and use it to create three reductions involving integers only, we'll start by substituting the given equation into the equation ac <+> bc = ab:
Replacing c in the second equation with a + b, we have:
a(a + b) <+> b(a + b) = ab
Expanding and simplifying this expression, we get:
a^2 + ab + ab + b^2 = ab
Combining like terms, we have:
a^2 + 2ab + b^2 = ab
Now, let's create the three reductions using integers only:
1. First Reduction:
Let's substitute integers for a and b in the equation a^2 + 2ab + b^2 = ab to find values that satisfy the equation.
If we let a = 1 and b = 1, we have:
1^2 + 2(1)(1) + 1^2 = 1(1)
1 + 2 + 1 = 1
4 = 1, which is false.
We see that the first reduction does not yield a true equation.
2. Second Reduction:
Let's substitute different integers for a and b, again using the equation a^2 + 2ab + b^2 = ab.
If we let a = 2 and b = 3, we have:
2^2 + 2(2)(3) + 3^2 = 2(3)
4 + 12 + 9 = 6
25 = 6, which is also false.
Again, the second reduction does not give a valid equation.
3. Third Reduction:
Let's try one more substitution with integers for a and b, again using the equation a^2 + 2ab + b^2 = ab.
If we let a = 3 and b = 4, we have:
3^2 + 2(3)(4) + 4^2 = 3(4)
9 + 24 + 16 = 12
49 = 12, which is false as well.
Therefore, we can conclude that there are no valid reductions involving integers only that satisfy the equation ac <+> bc = ab, where a + b = c.