1) 2x + 5y = 7
2) 3x + 6y = 3
I'm going to use the method of elimination to solve for x and y.
Multiply 1) by 3 and 2) by 2 to allow the x's to be eliminated.
1) 6x + 15y = 21
2) 6x + 12y = 6
Now subtract line 2 from line 1.
0x + 3y = 15
---- ----
3 3 divide both sides by 3 to get y by itself.
y =5.
Substitute into 1) to calculate x.
2x + 5(5) = 7
2x + 25 = 7
2x + 25 -25 = 7 - 25
2x = -18
---- ----- divide both sides by 2 to get x by itself
2 2
x = -9
Check by plugging in answer into the other equation, in this case 2)
3 (-9) + 6(5) = 3
-27 + 30 = 3
3 = 3
We did it correctly and checked to prove that we did it right.
Good luck.
Paul
SOURCE: analytic geometry
assuming the question is what is the circle equation?
and if (-2,2) is the center of the circle
the equation should look like this: (x+2)^2+(Y-2)^2=R^2
And now only R is needed.
given 2x-5y+4=0 equation of line perpendicular
we can rearange the equation to be y=(2x+4)/5
from that we can see that the slope of the line is 2/5
And from the fact of perpendicular line we can say that the slope
of the radius line is -2/5.
The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular
For that we would calculate the radius line equation and compare it to the equation of line perpendicular
As mentioned earlier the slope of the radious line is -2/5.
So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates
2= - (2/5)*(-2)+b ------> b=2-4/5=1.2
radius equation is y=-(2/5)x+1.2
Now the cross point is calculated by comparing the equations:
-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1
So the cross point is (1/2,1).
The distance between the points is calculated by the following
Formula:
R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=
SQR(7.25)
Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25
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