Here's the problem: Jim and Tim are sharing money. If I square Jim’s money and add on Tim’s, I get £10,050. If I square Tim’s money and add on Jim’s, I get £2,600. How much do they each have? From this the equations are as follows $x^2 +y=10050$ and $y^2 + x= 2600$ This is a grade 10 math problem and want to solve it w/o using differentiation, integration etc.
Simultaneous equation $x^2 +y=10050$ and $y^2 + x= 2600$ solution
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systems-of-equations
diophantine-equations
symmetric-polynomials
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0$x^2=(2600-y^2)^2=10050-y$ ... hint $(y-50)$ is a factor. – 2017-02-25
2 Answers
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Hint: assuming whole numbers, $y^2 \le 2600 \implies y \le 50\,$ and $x^2 \le 10050 \implies x \le 100\,$. But then $x^2 = 10050-y \ge 10050-50 = 10000\,$ so $x=100\,$.
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$\text{Jim} = x$, $\text{Tim} = y$
$$\begin{cases} x^2 +y=10050\\
y^2 + x= 2600\end{cases}$$
Subtracting the equations,
$$x^2 – y^2 +y – x = 10050-2600= 7450
\\( x-y) (x+y-1)= 7450 = 149 \times 50$$
Since $149$ is prime and $x\ge y$,
$$\begin{cases} x- y = 50 \\
x + y -1 = 149\end{cases}$$
After solving $x= 100$ and $y = 50$.
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0You should also include the hypothesis of $x$ and $y$ being non-negative integers (which is not to be taken for granted, since English pounds famously admit decimal values). – 2017-04-06
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2`Since 149 is prime ...` You need to consider all the other possibilities as well, like $x-y = 1\,$, $x-y = 2 \cdot 149\,$, $x-y = 5 \cdot 149\,$ etc. – 2017-04-06