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Mathematics 15 Online
OpenStudy (anonymous):

what value of k makes the equation have no solution? 2x + 2 = kx - 3

OpenStudy (anonymous):

2

OpenStudy (anonymous):

can you explain how you get that answr?

OpenStudy (anonymous):

the only way to remove x totaly from the equation is to put k = 2

OpenStudy (anonymous):

better way of explaining would be to solve for x

OpenStudy (anonymous):

2x -kx = -3-2

OpenStudy (anonymous):

x(2-k) = -5

OpenStudy (anonymous):

x= -5/ ( 2-k)

OpenStudy (anonymous):

this solution for x is valid everywhere except k=2

OpenStudy (anonymous):

because you cant divide by zero

OpenStudy (anonymous):

wait why do you subract the kx and 2?

OpenStudy (anonymous):

2x + 2 = kx -3 this is the original right ?

OpenStudy (anonymous):

yah but i still dont quite understand how to do it... im not very good at math so can you explain it again

OpenStudy (anonymous):

substract 2 from both sides >>> 2x + 2 - 2 = kx -3 -2 2x = kx - 5

OpenStudy (anonymous):

now substract kx from both sides 2x - kx = kx - 5 - kx 2x - kx = 5

OpenStudy (anonymous):

then what

OpenStudy (anonymous):

then take x as a common factor from the left hand term x(2-k) = 5

OpenStudy (anonymous):

now devide both terms by (2-k) x(2-k) / (2-k) = 5 / (2-k)

OpenStudy (anonymous):

so x = 5 / (2-k)

OpenStudy (anonymous):

then where do I go from there

OpenStudy (anonymous):

u know how to devide a number by 0 ?

OpenStudy (anonymous):

no can you explain?

OpenStudy (anonymous):

you cannot devide any number by 0 thats mathimatically undefined

OpenStudy (anonymous):

so the only way to make x is undefined is to make it a number devided by 0

OpenStudy (anonymous):

again x = 5 / ( 2-k)

OpenStudy (anonymous):

so make 2-k = 0

OpenStudy (anonymous):

add k to both sides >>> 2 - k + k = 0 + k so k = 2

OpenStudy (anonymous):

got it ? :)

OpenStudy (anonymous):

the only part i'm confused is when you are at x =5/2-k and how you get to the next step.

OpenStudy (anonymous):

hmmm sorry i cannot really illustrate more

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