5x+5y=1 25x+ky=4 solve for k

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5x+5y=1 25x+ky=4 solve for k

Mathematics
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there is few ways of working this out. Elimination, Substitution and maybe multiplication
which way do you prefer?
i need my mind refreshed. whichever method works best is the method i choose.

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ill use substitution then :P x = \[x = y - \frac{ 1 }{ 5 }\] 25x + ky = 4
25(y - \[25(y - \frac{ 1 }{ 5 }) + ky = 4\]
\[25y - 5 + ky = 4\]
do you get it now?
howd u get y-(1/5)?
divide everything in the first equation by 5
there is no solution :)
ill use elimination now
times the first equation by -5 -25x - 25y = -5 25x + ky = 4
now add this equation vertically together. it would give you k=25
i probably shoulda phrased the question the way they have it on my hw. it says: Determine the value of k for which the system of linear equations has no solutions.
haha great :/ u say that now
you know why there is no solution? if you sub K = 25 it wont work
:(
yes?
because they equal different things
\[\Huge{\color{Purple}{\\{\ \boxed{\mathbb{\text{}{}}{\color{green}{\color{blue}T{\color{purple}h{\color{pink}a{\color{red}nk}}}}}}}}} \] \[\Huge{\color{Purple}{\\{\ \boxed{\mathbb{\text{}{}}{\color{blue}Y{\color{green}o{\color{red}u}}}}}}} \]
haha no, THANK YOU ! :)
ok i have another one lol this one wants the k to have an infinite amount of solutions
49x-49y=49 7x+ky=7
the answer is not 1,7,49,1/7, or 1/49.
sorry been bit busy
so do you have an idea where to start?
i would times the 2nd equation by 7
ok...
do you undertand this yet lol?
49x-49y=49 7x+ky=7 49x-49y=49 49x + 7ky = 49
dont know where to go from there
-49y - 7ky = 0 -7y(7 + k) = 0
how do we get the k from that
im bit confused :P you sure their is more than 1 answer??
its 1 answer
oh u said this one wants the k to have an infinite amount of solutions
1 K but infinite solutions we need the K
y(7 + k ) = 0 from this can you find K? im pretty sure you can :P
hint: you only need to look at this question to get K
ok its -7 lol
im feeling slooow ha
ok i g2g now :)
have fun with your maths
haha see ya.

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