SOLUTION: The lines 2x+7y=19, 3x+2y=3 and kx+5y=17 are concurrent, find the value of k.

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Question 1102524: The lines 2x+7y=19, 3x+2y=3 and kx+5y=17 are concurrent, find the value of k.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
if the lines are concurrent, then they all intersect at the same point.

solve for x and y using the first 2 equations.

start with:

2x + 7y = 19
3x + 2y = 3

multiply the first equation by 3 and the second equation by 2 to get:

6x + 21y = 57
6x + 4y = 6

subtract the second equation from the first to get:

17y = 51

solve for y to get y = 3

replace y with either the first equation or the second equation and solve for x to get x = -1

if all 3 equations are concurrent then they all must intersect at the same point.

this means that the common solution to all 3 equations must be x = -1 and y = 3.

in the third equation of kx + 5y = 17, replace x with -1 and y with 3 to get:

-k + 15 = 17

solve for k to get k = -2.

that's your solution.

all 3 equations will intersect at x = -1 and y = 3 when k = -2.

your 3 equations that need to be solved simultaneously are:

2x + 7y = 19
3x + 2y = 3
-2x + 5y = 17

the following graph shows that the intersection for all 3 equations is equal to (-1,3).