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Reptile [31]
4 years ago
15

5x+4y=25 - (5x+2y=3)

Mathematics
1 answer:
olchik [2.2K]4 years ago
6 0

Answer:

T1he values for x and y of the equations is y = 11, and x = -19/5.

Step-by-step explanation:

To solve this question, we need to rearrange the expression:

5x+4y=25 - (5x+2y=3)

Look, if we subtract one equation to the other, then:

5x+4y=25 [1]

-(5x+2y=3) [2]

Which is the same as:

5x+4y=25

-5x-2y=-3

Subtract them:

5x+4y=25

-5x-2y=-3

---------------

      2y =22

y = 22/2 = 11

Then, y = 11.

To find x, we can substitute<em> y</em> in either equation [1] or [2].

Let us use [1]

5x+4y=25

5x+4(11)=25

5x+44=25

5x=25-44

5x=-19

x = -19/5

Then, the values for x and y of the equations is y = 11, and x = -19/5.

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Check the picture below.  so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}

(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}

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