2x2 row and column equations

Question

Rules: No paper or notes, but a calculator is accepted.

Post the exact time min wise of how long it took.

And obviously give the answer.

Answer 1

Took one minute to solve.

This question has a reasonably easy solution if you can figure out the pattern in the the provided equations.

Look at the first row equation and the second column equation.
$a+b =8$
$b+c =8$

Notice that there is a common variable in the equations (the intersection of the equations in the picture denotes the common variable.)

Thus we can rewrite the set of equations as:
$a+b=8$
$c+b=8$

Subtracting these two equations can make us realize that $a=c$. Which means the top left box is equal to bottom right box.

The problem can be rewritten as

The values now seem fairly easy to compute.

$a+c=13$
$c-a=6$

By adding the above equations, the variable $c$ can be resolved. By evaluation of $c$ and substituting into either of the above equations, variable $a$ can be calculated. After which variable $b$ can be calculated by substituting variable $a$ in $a+b=8$.

Answer 2

Let a be the top - left square,
Let b be the top - right square,
Let c be the bottom - left square,
Let d be the bottom - right square,

a = 3.5,
b = 4.5,
c = 9.5,
d = 3.5

Method:

Equations:

a + b = 8
c - d = 6
Add to get: a + b + c - d = 14

a + c = 13
b + d = 8
Add to get: a + b + c + d = 21
Subtract equations to get: 2d = 7
d = 7/2 = 3.5
c - 3.5 = 6
c = 9.5
a + 9.5 = 13
a = 3.5
3.5 + b = 8
b = 4.5

Also ~ two minutes.

Answer 3

3.5 + 4.5 = 8
9.5 - 3.5 = 6

This took me two minutes to solve. There are four variables and four equations. Basic algebra rules the day.

Answer 4

Let squares be $a$, $b$, $c$, $d$. We have $a$ $+$ $b$ $+$ $c$ $+$ $d$ $=$ $21$, $a$ $+$ $b$ $+$ $c$ $-$ $d$ $=$ $14$, so $2d$ $=$ $7$, $d$ $=$ $3.5$. Got this bit in 5 seconds. Full square:

3.5 + 4.5 = 8
 +     +
9.5 - 3.5 = 6
 =     =
 13    8

$20$ seconds in total - roughly

Answer 5

Here is how to solve in your mind (I did it in this way in 2-3 minutes to solve and confirm):

if you sum all equations you will be summing up the squares two times but the square with opposite signs (the forth square) cancels and you left with summing up squares with + signs two times. So 2x(first+second+third squares) = 8 + 8 + 6 + 13 = 35, hence first+second+third = 17.5, you know first + second = 8 therefore third = 9.5. Then it follows.

Answer 6

Pretty much the same as everybody else, with a slightly different twist:

Add the columns: \begin{align}a + c = 13\\+\qquad b + d = ~~8\\\hline\llap{\text{Add to get:}\quad}a + b + c + d = 21\\\llap{\text{Subtract:}\qquad}-\qquad a+b=~~8\\\hline\llap{\text{to get:}~~\qquad\qquad\qquad}c+d=13\end{align} So we have $c+d=13$ and $c-d=6$, so $c$ must be the average of $13$ and $6$.  We can calculate this as \begin{align}c+d&=13\\+\qquad c-d&=~~6\\\hline\llap{\text{Add to get:}\quad\qquad}2c~\phantom{+d}&=19\\[2ex]\llap{\text{So,}\quad\qquad\qquad}c~\phantom{+d}&=19\rlap{/2}\\[2ex]\llap{\implies\qquad\qquad}c&=~9\rlap{.5}\end{align} but I didn't need to do that.  I saved a few seconds by realizing that, since $c$ is equally distant from $13$ and $6$, it must be halfway between $13$ and $6$, i.e., their average.

Then (the same as everybody else), it's a simple matter to solve for the other three:

$c+d=13\implies 9.5+d=13\implies\boxed{d}=13-9.5=\boxed{3.5}$
$a+c=13\implies a+9.5=13\implies\boxed{a}=13-9.5=\boxed{3.5}$
$b+d=~8~\implies\,b+3.5=~8~\implies\boxed{b}=~~8-3.5=\boxed{4.5}$
Interestingly, I didn't notice that $a=d$ until after I had solved for their values.

Results: $\qquad a=3.5\qquad b=4.5\qquad c=9.5\qquad d=3.5$

So the filled-in grid is:
$$\begin{array}{c}\Large{3.5}&+&\Large{4.5}&=&\Large{8}\\+& &+\\\Large{9.5}&-&\Large{3.5}&=&\Large{6}\\||& &||\\\Large{13}&&\Large{8}\end{array}\hskip1.1in$$

I didn't exactly time myself, but I'm pretty sure that I did this in under a minute.  In deference to rand al'thor, I'll stipulate that I took at least 45 seconds.

^{P.S. For convenience, I copied some of the equations from greenturtle3141's answer.}

Answer 7

This took me a minute using brute force.

I instinctively started with the bottom row first because it seemed to have a narrower range of reasonable answers. After plugging 8,9,10 in the bottom left corner it was pretty obvious that a whole number wasn't going to work - my answers for the top row were bracketing 8. Once I reached that conclusion the problem was basically solved.

I can see from other answers that there is a more theoretical way to solve the problem but I figured it would take me longer to work out the theory than to solve it with brute force.