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Which two numbers when added together yield $16$, and when multiplied together yield $55$.

I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't.

In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong?

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    You can see the motivation for the answers below by drawing the graph of both functions see that they intercept. Having that visualization the algebra is obvious. Once you have the algebra you can see that you are merely "constraining" one equation using the other. Then you don't need the picture, and go on to more complicated simultaneous constraints.2018-04-24

9 Answers 9

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Our two equations are: $x + y = 16 \tag{1}$ $xy = 55\tag{2}$

Rewriting equation (1) in terms of just $y =$ something, we get:

$y = 16-x$

Substituting this into equation (2) leaves us: $x(16-x) = 55$ $16x-x^2=55 \implies x = 5 \ \ \text{or} \ \ 11$

which can be easily seen by factoring or using the quadratic formula. It follows that $y=11|x=5$ and $y=5|x=11$.

Thus your solutions in terms of $(x,y)$ are $(5,11)$ and $(11,5)$.

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    @HenningMakholm Simple. You know the factorized quadratic will look like $(x-X)(x-Y)$, and $X$ and $Y$ satisfy... oh crap.2015-01-19
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We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general.

Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$. In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\pm 6$.

Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$.

The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$.

Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get $x(16-x)=55$. Simplification gives $x^2-16x+55=0$. The quadratic factors as $(x-5)(x-11)$, so our equation becomes $(x-5)(x-11)=0$, which has the solutions $x=5$ and $x=11$.

But we cannot necessarily rely on there being such a straightforward factorization. So in general after we get to the stage $x^2-16x+55=0$, we would use the Quadratic Formula. We get $x=\frac{16\pm\sqrt{(-16)^2-4(55)}}{2}.$
Compute. We get the solutions $x=5$ and $x=11$. The corresponding $y$ are now easy to find from $x+y=16$.

Remarks: $1,$ Recall that the Quadratic Formula says that if $a\ne 0$, then the solutions of the quadratic equation $ax^2+bx+c=0$ are given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

Your approach was along reasonable lines, but things went wrong in the details. From $xy=55$ we get $x=\frac{55}{y}$. Substituting in the formula $x+y=16$, we get $\frac{55}{y}+y=16.$ A reasonable strategy is to multiply through by $y$, getting $55+y^2=16y$, or equivalently $y^2-16y+55=0$. Now we have reached a quadratic equation which is basically the same as the one we reached above.

$2.$ The first approach that we used (presented as an algorithm, and stripped of algebraic notation) goes back to Neo-Babylonian times. The "standard" problem was to find the dimensions of a door, given its perimeter and area.

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    Approach 1 is the most elegant answer. (To see how you'd think of using $x-y$, you get to it by looking at the graph of $xy=55$ and $x+y = 16$ plotted on the same axes, and seeing that it's symmetrical about the line $y=x$. So it's natural to change variables and think of the graph as plotted on the axes $y=x$, $y=-x$.)2014-07-31
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Here is another method: suppose you are told that two numbers, $x$ and $y$, have a certain sum $x+y=S$, and a certain product $xy=P$. How to find $S$ and $P$?

We can use the fact that we know how to solve quadratic equations. Notice that $(t-x)(t-y) = t^2 - (x+y)t + xy = t^2 - St + P.$

That means that $x$ and $y$ are precisely the solutions to $t^2 - St + P = 0.$

In your specific case, $S=16$ and $P=55$. So we want to find the solutions to $t^2 - 16t + 55 = 0.$

The quadratic formula gives $t = \frac{16 \pm\sqrt{256 - 220}}{2} = 8 \pm\frac{1}{2}\sqrt{36} = 8\pm\frac{6}{2} = \left\{\begin{array}{l} 11\\ 5 \end{array}\right.$ So the two numbers are $5$ and $11$.

(Of course, we often solve quadratic equations $t^2 + at + b=0$ by figuring out by eyeballing two numbers whose product is $b$ and whose sum is $-a$, but we can always use the quadratic formula to take the guessing out of it.)

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The average of x and y is 16/2 = 8, their product is xy = 55 therefore: x, y = 8 plus or minus sqrt of 8 square minus 55 = 8 +/- sqrt of 9 = 8 +/- 3, x, y = 11, 5

by: GeorgeB reference: Vedic Book

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let the sum be S and the product be P

you can get the two numbers x and y using the following formular

x = (S + (S^2 - 4P)^-2)/2

y = (S - (S^2 - 4P)^-2)/2

I've done the math just like above

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should be

x = (S + ((S^2 - 4P)^(1/2)))/2

y = (S - ((S^2 - 4P)^(1/2)))/2

i.e. raising to the power of a half is taking the square root. raising to the power of -2 is the reciprocal of the square.

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    [Here's a MathJax tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) :)2014-07-31
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$ x + y = 16 ... (1*) $

$ x * y = 55 ... (2*) $

Use identity: $ ( x-y)^2 = = ( x+y)^2 - 4 *x*y = ... (3*) $

$ x - y = 256 - 4 * 55 = \sqrt{36} \rightarrow x -y = \pm \, 6 ...(4*) $

Find half sum and half difference of equations (1*) and (2*):

$ (x,y) = (11,5) , \, (5,11) .. (*5) $

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My favorite way

Square the sum to get $S^2=a^2+2ab+b^2$.

Get the squared difference $D^2=a^2-2ab+b^2=S^2-4P$ by deducting four times the product.

Then $a,b=\dfrac{S\pm D}2$.

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    Mentally, $D^2=256-220=6^2$, then $a,b=5,11$.2015-05-07