# How do you write two different addition equations that have 12 as the solution?

Jun 23, 2017

See below

#### Explanation:

I don't really get your question, but here is my answer nonetheless

Just pick two of the equations from the ones I have listed below, they all equal 12.

$1 + 11$
$2 + 10$
$3 + 9$
$4 + 8$
$5 + 7$
$6 + 6$

$- 1 + 13$
$- 2 + 14$
$- 3 + 15$
$- 4 + 16$
$- 5 + 17$
$- 6 + 18$
$- 7 + 19$
$- 8 + 20$
$- 9 + 21$
$- 10 + 22$
$- 11 + 23$
$- 12 + 24$
$- 13 + 25$... and it goes indefinitely

There are infinitely many equations, so I will not list them all.

Jun 23, 2017

#### Answer:

$x - 1 = 11$ and $x - 2 = 10$
[See logic below]

#### Explanation:

Let's first limit the question to linear equations with integer coefficients of the form: $a x + b = c$ $\left\{a , b , c \in \mathbb{Z}\right\}$

Actually, this restriction is not necassary to answer the question but, since there are an inifinite number of these, it will not limit us.

So, we are asked to find two different equations of the form $a x + b = c$ where $x = 12$

To simplify matters still further, let's say $a = 1$
[We still have an infinite number of possible equations]

$\therefore x + b = c$

But we must make $x = 12$, so: $c - b = 12 \to c = b + 12$

To generate our equations we can set $b$ to any number $\in \mathbb{Z}$

$b = - 1 : \to c = 11 \to x - 1 = 11$ Is our first example

$b = - 2 : \to c = 10 \to x - 2 = 10$ Is our second example

and so on ....

From which we can see that there is an infinite number of equations that satisfy the condition even after the restrictions we have placed.