Question

Help I’m not sure with this one!! Urgent help

Answer 1

This system has exactly one solution, because the two equation are linearly independent (i.e. one is not a multiple of the other).

We can explicitly find the solutions: from the first equation we deduce

$x = \dfrac{c-by}{a}$

Plug this value for x in the second equation:

$\dfrac{a}{3}\cdot \dfrac{c-by}{a}+\dfrac{by}{2}=\dfrac{c}{6} \iff \dfrac{c-by}{3}+\dfrac{by}{2}=\dfrac{c}{6} \iff \dfrac{2(c-by)+3by}{6}=\dfrac{c}{6}$

Multiply both sides by 6 to get

$2(c-by)+3by=c \iff 2c-2by+3by=c \iff by=-c \iff y=-\dfrac{c}{b}$

Plug this value for y in the expression for x:

$x = \dfrac{c-by}{a} = \dfrac{c-b\cdot\frac{-c}{b}}{a}=\dfrac{2c}{a}$

Since both a,b are not zero, both solutions are well defined.