Finding the numbers that satisfies the boxes

Question

How do you solve this question I represented the unknown numbers that add to give 8 and subtract to give 6 by x and y. I solved it and had 7 and 1. But they don't match with the 13 and 8 below the boxes

I'm stuck

$$\begin{array}{ccccc} \Box&+&\Box&=&8\\ +&&+\\ \Box&-&\Box&=&6\\ =&&=\\ 13&&8\end{array}$$

Answer 1

As a linear system, assuming $x_1$ is the lower-left box and proceeding counterclockwise, we have $$\begin{bmatrix} 1&-1&0&0\\ 0&1&1&0\\ 0&0&1&1\\ 1&0&0&1\end{bmatrix}\mathbf x=\begin{bmatrix}6\\8\\8\\13\end{bmatrix}$$ From this we get the unique solution $$\begin{array}{ccccc} 3.5&+&4.5&=&8\\ +&&+\\ 9.5&-&3.5&=&6\\ =&&=\\ 13&&8\end{array}$$

Answer 2

You can set up four equations from this. I will assign variables for each corner. Top left is a. Top right is b. Bottom left is c, and bottom right is d.

So these are the equations that are able to be set up.

$a+b=8$, $a+c=13$, $b+d=8$, $c-d=6$

So, let us combine the first two equations.

$ \ \ \ \ \ a+b=8$

$-(a+c=13)$

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$b-c=-5$

Now, let us combine the last two equations.

$ \ \ \ \ \ b+d=8$

$+(c-d=6)$

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$b+c=14$

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We now use the two derived equations to solve for either b or c.

$ \ \ \ \ \ b-c=-5$

$+(b+c=14)$

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$2b = 9$

$\therefore b = 4.5$

With this being said, plug $b$ back into the equation and get $c$. $b+c=14$ So, $4.5+c=14$.

$\therefore c=9.5$

Plug $b$ and $c$ back into the rest of the equations to get the variables.

$a+b=8 \therefore a=3.5$

$c-d=6 \therefore d=3.5$

Therefore, the answers to the question is: Top Left: 3.5, Top Right: 4.5, Bottom Left: 9.5, Bottom Right: 3.5