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Salsk061 [2.6K]
3 years ago
12

Hi how to solve this simultaneous equation

Mathematics
1 answer:
nalin [4]3 years ago
6 0

Answer:

\large \boxed{\sf \ \ x=\pm8 \ \ or \ \ x=\pm2  \ \ }

Step-by-step explanation:

Hello, please consider the following.

First of all, we can assume that x and y are different from 0, as we cannot divide by 0.

And from the first equation we can write y as a function of x as below.

y=\dfrac{16}{x}

And then, we replace it in the second equation to get.

\dfrac{x}{\frac{16}{x}}+\dfrac{\frac{16}{x}}{x}=\dfrac{17}{4}\\\\ \dfrac{x^2}{16}+\dfrac{16}{x^2}=\dfrac{17}{4}\\\\\text{*** We multiply by }16x^2 \text{ both sides ***}\\\\x^4+16*16=\dfrac{17*16}{4}x^2\\\\x^4-68x^2+3600=0\\\\\text{*** The product of the zeroes is 3600 = 64*4 and the sum is 64+4=68 ***}\\\\\text{*** So we can factorise *** }\\\\x^4-64x^2-4x^2+3600=x^2(x^2-64)-4(x^2-64)=(x^2-64)(x^2-4)=0\\\\x^2=64=8^2 \ \ or \ \ x^2=4\\\\x=\pm8 \ \ or \ \ x=\pm2\\

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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<em>putt</em><em>ing</em><em> </em><em>it</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>formu</em><em>la</em>

<em>p</em><em>=</em><em> </em><em>2</em><em>(</em><em>4</em><em>+</em><em>2</em><em>W</em><em>)</em><em> </em><em>+</em><em>2</em><em>W</em>

<em>but</em><em> </em><em>peri</em><em>meter</em><em> </em><em>was</em><em> </em><em>gi</em><em>ven</em><em> </em><em>as</em><em> </em><em>4</em><em>4</em><em>m</em>

<em>4</em><em>4</em><em>=</em><em> </em><em>8</em><em>+</em><em>4</em><em>w</em><em>+</em><em>2</em><em>w</em>

<em>4</em><em>4</em><em>=</em><em>8</em><em>+</em><em>6</em><em>w</em>

<em>4</em><em>4</em><em>-</em><em>8</em><em>=</em><em>6</em><em>w</em>

<em>3</em><em>6</em><em>=</em><em>6</em><em>w</em>

<em>divi</em><em>ding</em><em> through</em><em> </em><em>by</em><em> </em><em>6</em>

<em>w</em><em>=</em><em>3</em><em>6</em><em>/</em><em>6</em>

<em>w</em><em>=</em><em>6</em><em>m</em><em> </em>

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<em>L</em><em>=</em><em>1</em><em>6</em><em>m</em>

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<em />

<em>That's literally it- it won't show me the actual number xD</em>

<em />

<em>Hope this helps :]</em>

<em />

6 0
2 years ago
Read 2 more answers
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