What Is The Coordinates Of X And Y Of The Point Intersection Of The Lines Help

Finding the coordinates of the point of intersection of these lines? Please help!?

The point of intersection is the x & y values that you get when you solve both equations simultaneously.
3x + 4y = 10
5x - 6y = 23

Multiply every term of the first equation by 1.5 so that we can eliminate the y term:
4.5x + 6y = 15
5x - 6y = 23

9.5x = 38
x = 4

Now plug in the x value into one of the original equations & solve for y:
3x + 4y = 10
3(4) + 4y = 10
12 + 4y = 10
4y = -2
y = -1/2

So, the point of intersection is (4, -1/2)

Find the coordinates of the points of intersection of the line y+2x=11 and the curve xy=12 ??

Find the coordinates of the points of intersection of the line y+2x=11 and the curve xy=12

i find an answer for this question but i am not sure and i dont have the answer key.
Thanks for answering

Plx help!! find the corrdinates of the point of intersection of the lines y = 3x + 1 and x + 3y = 6?

1) The coordinates of the point of intersection are found where the values of both lines are equal. So

y = 3x + 1 and x + 3y = 6

Choosing one of the two equations, solve for one variable, then the other:

x + 3y = 6

Since, from the first equation y = 3x + 1, substitute that into the second equation

x + 3y = 6 == > x + 3(3x + 1) = 6

x + 9x + 3 = 6

10x = 6 - 3

x = 3/10

Then substituting this in for x in the first equation

y = 3x + 1

y = 3(3/10) + 1

y = 9/10 + 1 = 19/10

So x = 3/10, y = 19/10



2) The line L is parallel to Y = -2x+ 1 and passes through the point 5,2.

Two or more parallel lines have equal slopes. Since line Y has a slope of -2, so does the new line. Since you have both a coordinate pair and a slope, we may proceed to checkout....

Use the point-slope form of the equation:

y - y1 = m(x - x1)

y - 2 = -2(x - 5) [Where m = -2, (x1,y1) = (5,2)]

y - 2 = -2x + 10

y = -2x + 12

Which is the equation for the new line. All that remains is to find the x- and y-intercepts.

To find the y-intercept, set x = 0

y-int = -2(0) + 12

Similarly, to find the x-intercept set y = 0

0 = -2x + 12

2x = 12

x-int = 6

So the x- and y-intercepts are (6,0) and 0,12) respectively.

How to find the x-coordinates of the points of intersection? PLEASE HELP, 10 POINTS?

The equation of a curve C is y=2x^2-8x+9 and the equation of a line L is x+y=3. Find the x-coordinates of the points of intersection of L and C, and show that one of these points is also the stationary point of C.

How can I solve this problem? Please help, 10 points??

How do u find the coordinates of the points of intersection of the parabola y= -x2+9x-3 and the line y=3x?

Substituting Y = 3x in the parabola Y = -x^2 + 9x - 3 gives
3x = -x^2 + 9x - 3 and x^2 - 6x + 3 =0.

Therefore, x = [-(-6) + or - sqrt of {(-6)^2 - 4*3}]/2. In this x will have two answers.
x = [6 + sqrt of (36 - 12)]/2 or [6 - sqrt of (36 - 12)]/2
x = [6 + sqrt of (24)]/2 or [6 - sqrt of (24)]/2
x = [3 + sqrt of (6)] or [3 - sqrt of (6)]

Substituting the above two x values in y = 3x, you will get the corresponding values for y.

Find the coordinates of the points of intersection of the parabola y=x^2-ax and line y=bx-ab plz show steps?

You need to know where one equation equals the other, so:

x²-ax = bx -ab
x² - (a+b)x +ab = 0

(x-a)(x-b) = 0, x= a or b, so plug those in to the other equation:

y = ab-ab =0 when x = a, and
y = b²-ab when x = b, so:

(a, 0) and (b, b²-ab)