What is the solution to this system: 4x+9y=28 - 4x-y=-28

Mathematics

1 answer:

Degger [83]3 years ago

6 0

Answer:

x = 7

y = 0

Step-by-step explanation:

→To solve this, you can use the elimination method. To do this, you must have one set of variables that can cancel each other out. In the problem given, we already have positive 4x and -4x, making them cancel out:

4x + 9y = 28

-4x - y = -28

__________

8y = 0

y = 0

<u>→Plug in 0 for y, into an equation:</u>

-4x - 0 = -28

-4x = -28

x = 7

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PLEASE HELP.

solniwko [45]

recall that the radius is the distance from the center of a circle to any point on the circle.

we know the center is at -16,-14, and we know that -8,-8 is a point on the circle, so the distance between both must be the radius.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{-16}~,~\stackrel{y_1}{-14})\qquad
(\stackrel{x_2}{-8}~,~\stackrel{y_2}{-8})\qquad \qquad
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
\stackrel{radius}{r}=\sqrt{[-8-(-16)]^2+[-8-(-14)]^2}
\\\\\\
r=\sqrt{(-8+16)^2+(-8+14)^2}\implies r=\sqrt{8^2+6^2}
\\\\\\
r=\sqrt{100}\implies \boxed{r=10}
\\\\[-0.35em]
\rule{34em}{0.25pt}$

$\bf \textit{equation of a circle}\\\\
(x- h)^2+(y- k)^2= r^2
\qquad
center~~(\stackrel{-16}{ h},\stackrel{-14}{ k})\qquad \qquad
radius=\stackrel{10}{ r}
\\\\\\\
[x-(-16)]^2+[y-(-14)]^2=10^2\implies \blacktriangleright (x+16)^2+(y+14)^2=100 \blacktriangleleft$