Find the equations of circles passing through (1,-1),touching the lines 4x+3y+5=0 and 3x-4y-10=0

Negative x plus 3y = 9....the slope is opposite

This is best written as two separate equations:

8x+3y = -23 and 34x+ 5y = -23
Solving the first one for x:
8x = -23-3y
x = -23/8 - 3/8y
Substituting this value for x into the second equation:
34(-23/8 - 3/8y) + 5y = -23
-97.75 - (34)(.375)y + 5y = -23
-97.75 - 12.75y + 5y = -23
-97.75 -7.75y = -23
-7.75y = 97.75-23=74.75
y = -74.75/7.75 = -9.645161
Substitution back into the equation for x:
x = -23/8 - 3/8(-9.645161)
x = -2.875 + 3.616935
x =.741935

To find the solution, first find the value of y for each equation.
Then substitue one equation into the other so that you only the x variable left.
Then just solve for x.
Once you have a value for x, then you can easily solve for y.


So for the first equation:

3y - 6x = -3
3y = 6x - 3

y = 2x - 1


Now for the second equation:

2y + 8x = 10
2y = -8x + 10

y = -4x + 5

Since both equations equal y, they also equal each other, therefore:

2x - 1 = -4x + 5

Now just solve for x:

2x + 4x = 5 + 1
6x = 6

x=1

Now substitute x=1 into either original equation:

y = 2x - 1
y = 2 (1) - 1
y = 2 - 1

y = 1


Therefore the solution is x=1 and y=1



Good luck, I hope that helps.


Joe.

Price of bag of chips s X and price of box of pretzels is Y.
Now you can write following equations:

2X+3Y=413
X+2Y=239

From second equation you know that X is 239-2Y, and you put that in first equation:

2*(239-2Y)+3Y=413

or

-4Y+3Y=413-2*239

Finally we have for Y: Y=65 and X=239-2Y=109.

This should start wit X=something and Y=something, sorry I'm not an human algebra calculator....

You can solve it with following method.

5x+3y=6 2x-4y=5
So 5x=6-3y so 2[(6-3y)/5]-4y=5
So x=(6-3y)/5 so 12-6y-20y=25
so -26y=25-12
so -26y=13
so y= -(1/2)
2x-4y=5
so 2x=5+4y
so 2x=5+4(-1/2)
so 2x=(10-4)/2
so 2x=6/4
so x =3/2

The value of x=3/2 and value of y= -1/2

Let me know if you need further assistance.
Thanks for using FixYa.

The two numbers are 9 and 27.

Solution:
Suppose the two numbers are x and y, then,

x + y = 36 ---------1
and
x = 3y -------------2

Solving equation 1 and 2...

3y+y=36 => 4y=36 => y = 9
x + 9 = 36 => x = 27

Please rate and comment if my answer is right.

Hello,
I am not lecturing you but I would rather you understand how to do the manipulations involved in isolating a variable.
You want to isolate y, Ok
Start stripping it of all that is not y.

3y + 4x =6. (Addition is commutative, I can change the order of addition)
The term with y is added to 4x. If I want the term in y alone on one side I perform the inverse operation of addition, a substraction. I subtract 4x from both sides.
3y+4x-4x=6-4x. But 4x-4x=0, and we are left with

3y= - 4x+6

This operation is sometimes summarized as make one term change side while changing its sign

It would do no harm to put the right side of the foregoing equation in parentheses. I do that to avoid errors)

3y= (-4x+6).

Now y is multiplied by the number 3. To isolate y I have to perform the inverse operation of the multiplication, namely the division by 3

3y/3 =(-4x+6)/3. The left hand side is just y

y= (-4x+6)/3.

While result is correct, I can also open the parentheses
y= -4x/3 +6/3
y= -(4/3)*x +2.

Hope it helps.

Go to your matrix button and enter a "3x4" matrix.

Then enter it as follows:

-3 4 5 7
4 3 2 9
-5 5 3 -10

Then exit out and go to "2nd->matrix->math->rref(". Then press enter.

Your screen should look like this:

rref(

Then go to matrix and select your 3x4 matrix, press enter and close it with a parathesis. Your screen should look like this:

rref([A])

Press enter and the screen should say this:

1 0 0 2
0 1 0 -3
0 0 1 5

So,
x=3
y=-3
z=5

Hope this cleared up the confusion!

SJ_Sharks

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