In the system of equations below, k is a constant and x and y are variables. For what value of k will the system of equations have no solution? \(kx-3y=4\\4x-5y=7\) A) \(\dfrac{12}{5}\) B) \(\dfrac{16}{7}\) C) \(-\dfrac{16}{7}\) D) \(-\dfrac{12}{5}\)
`t` suppose to be y ?? right?
ye
in order to get no solution, the coefficients of both variables should be the same but opposite sign. like in this example \[\large\rm 3x -4y=9\]\[\large\rm -3x+4y=7\] 3x cancels out same with y variable left with 0=16
so first i would try to get same coefficients for y variable. then it would be easy to find k
how would I do that?
multiply first equation by 5 and equation by -3
2nd e quation by -3*
\(5(kx-3y=4)\) I multiply k by 5 too or no?
yes multiply all terms by 5
\(5kx-15y=20\) \(-12x+15y=-21\)
looks good to me now you can try each option to find value for k \(\rm 5kx-15y=20\) \(\color{Red}{-12x}+15y=-21\) remember our goal is to cancel out both variables so which option will give you 12 when you multiply by 5 ??
D?
\[\rm 5(\frac{-12}{5})x-15y\] you will get \[\large\rm -12x -15y=20\] and the 2nd e quation is \[\large\rm -12x -15y=20\]\[\large\rm -12x+15y=-21\] would you get no solution ?
there is one solution
yes try other option \[\large\rm -12x \cancel{-15y}=20\]\[\large\rm -12x\cancel{+15y}=-21\] like i said the sign should be opposite -12x-12x is not equal to 0
try A?
try it do you get 0 ??
did you substitute k for option A ??
\(5(\dfrac{12}{5}x-15y=20\\12x-15=20\) ?
yes that's correct now what's the 2nd equation ? and i guess there is a typo you forgot the y variable
\(12x-15y=20\\-12x+15y=-21\)
right solve it
0 = -1 ?
right so that means no solution.
so A is my answer then?
thats what we get so yes
oke ty ^_^
yw
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