Solve the system of equations two ways: 3x+y=5 and x-2y=4 A) using substitution B) using elimination

1 answer:

5 0

A. solve for 1 variable
let's solve for x in 2nd equation
add 2y to both sides
x=2y+4
sub 2y+4 for x in other equation
3(2y+4)+y=5
6y+12+y=5
7y+12=5
minu12 both sides
7y=-7
divide 7
y=-1
sub back
x=2y+4
x=2(-1)+4
x=-2+4
x=2
(2,-1)

B. eliminate
eliminate y's
multiply first equation by 2 and add to first

6x+2y=10
x-2y=4 +
7x+0y=14

7x=14
divide by 7
x=2
sub back
x-2y=4
2-2y=4
minus 2
-2y=2
divide -2
y=-1
(2,-1)

(2,-1) is answer

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Step-by-step explanation:

The equation for a circle is $(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is the vertex of the circle and $r$ is the radius. Immediately, since the center of the circle is given, we know what $h$ and $k$ are. $h$ is 230 and $k$ is 220.

The only thing we need to find is the radius, which will just be the distance from the center (230,220) to a point on the circumference (170,170). The distance between them can be calculated using the distance formula, which is really just the Pythagorean Theorem rearranged. The formula states that $d=\sqrt{x^2+y^2}$ , where $x$ is the change in the x-coordinates of the two points and $y$ is the change in the y-coordinates of the two points. Plug-in $x$ for 60 and $y$ for 50 to get $d=\sqrt{50^2+60^2}$ . Solve for $d$ , arriving at $\sqrt{6100}$ . Therefore, the radius of the circle is $\sqrt{6100}$ .