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3 years ago

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{-2x+y=0 5x+3y=-11 Solve the substitution

1 answer:

Rzqust [24]3 years ago

7 0

$\bf \begin{cases} -2x+y=0\\ \boxed{y} = 2x\\[-0.5em] \hrulefill\\ 5x+3y=-11 \end{cases}\qquad \qquad \implies \stackrel{\textit{substituting on the 2nd equation}}{5x+3\left( \boxed{2x} \right)=-11} \\\\\\ 5x+6x=-11\implies 11x=-11\implies x = \cfrac{-11}{11}\implies \blacktriangleright x = -1 \blacktriangleleft \\\\\\ \stackrel{\textit{we know that}}{y = 2x}\implies \blacktriangleright y = -2 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (-1~~,~~-2)~\hfill$

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Ksivusya [100]

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<h2>Step-by-step explanation:</h2>

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Let the length of R be: 2x

and the width of R be: 3x

i.e. The perimeter of R is given by:

$Perimeter\ of\ R=2(2x+3x)$

( Since, the perimeter of a rectangle with length L and breadth or width B is given by:

$Perimeter=2(L+B)$ )

Hence, we get:

$Perimeter\ of\ R=2(5x)$

i.e.

$Perimeter\ of\ R=10x$

Also, let " s " denote the side of the square region.

We know that the perimeter of a square with side " s " is given by:

$\text{Perimeter\ of\ square}=4s$

Now, it is given that:

The perimeters of square region S and rectangular region R are equal.

i.e.

$4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}$

Now, we know that the area of a square is given by:

$\text{Area\ of\ square}=s^2$

and

$\text{Area\ of\ Rectangle}=L\times B$

Hence, we get:

$\text{Area\ of\ square}=(\dfrac{5x}{2})^2=\dfrac{25x^2}{4}$

and

$\text{Area\ of\ Rectangle}=2x\times 3x$

i.e.

$\text{Area\ of\ Rectangle}=6x^2$

Hence,

Ratio of the area of region R to the area of region S is:

$=\dfrac{6x^2}{\dfrac{25x^2}{4}}\\\\=\dfrac{6x^2\times 4}{25x^2}\\\\=\dfrac{24}{25}$

6 0

3 years ago

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