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TiliK225 [7]
3 years ago
11

I need help in learning how to solve this equation using substitution:

Mathematics
1 answer:
Tomtit [17]3 years ago
4 0

Answer:

x=3, z=-1, y=6

Step-by-step explanation:

you need to make the coefficients of each variable equal to each other then subtract 2 equations.

this one: multiply -4 in 1st

4x+4y+4z=32

-4x+4y+5z=7

_____________

8x+0-z=25

we have 2 equations with 2 variables now

2x+2z=4

8x-z=25

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Write the equation of the circle with center (−1, −3) and (−7, −5) a point on the circle.
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So, we know the center is at -1, -3, hmmm what's the radius anyway?

well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus

\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ -1 &,& -3~) 
%  (c,d)
&&(~ -7 &,& -5~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2}
\\\\\\
r=\sqrt{36+4}\implies r=\sqrt{40}\\\\
-------------------------------

\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad 
radius=\stackrel{\sqrt{40}}{ r}
\\\\\\\
[x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40
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Step-by-step explanation:

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