OpenStudy (exoexoexo):

solve the following system of linear equation algebraically ax=by 3ax-2by=ab

1 year ago
OpenStudy (tkhunny):

Have you considered attempting the solution?

1 year ago
OpenStudy (exoexoexo):

I have tried to but it ended up confusing me more

1 year ago
OpenStudy (michele_laino):

by substitution of the first equation into the second one, we get: \[\begin{gathered} 3by - 2by = ab \hfill \\ \hfill \\ by = ab \hfill \\ \end{gathered} \]

1 year ago
OpenStudy (michele_laino):

now, we have to distinguish two cases, namely: 1) \(b \neq 0\) so we can divide both sides by \(b\), and we get: \[\frac{{by}}{b} = \frac{{ab}}{b}\] please continue

1 year ago
OpenStudy (michele_laino):

what is \(y\)?

1 year ago
OpenStudy (exoexoexo):

y=a?

1 year ago
OpenStudy (michele_laino):

perfect!

1 year ago
OpenStudy (michele_laino):

now I substitute \(y=a\) into the first equation, and I get: \[ax = ab\]

1 year ago
OpenStudy (michele_laino):

again I have to consider two cases, namely: 1) \(a \neq 0\) 2) \(a=0\) case 1) being \(a \neq 0\) I can divide by \(a\), so I get: \[\frac{{ax}}{a} = \frac{{ab}}{a}\] what is \(x\) ?

1 year ago
OpenStudy (exoexoexo):

x=b

1 year ago
OpenStudy (michele_laino):

correct! so we have the first solution of the system, like below: \[\left\{ \begin{gathered} y = a \hfill \\ x = b \hfill \\ \end{gathered} \right.,\quad a \ne 0,b \ne 0\]

1 year ago
OpenStudy (michele_laino):

next we have to consider the case 2) being \(a=0\), how many solution has this equations: \[ax = ab\]

1 year ago
OpenStudy (michele_laino):

solutions*

1 year ago
OpenStudy (exoexoexo):

1?

1 year ago
OpenStudy (michele_laino):

if \(a=0\), we can write: \(0\;x=0\) so any number satisfies such equation, right?

1 year ago
OpenStudy (exoexoexo):

yes

1 year ago
OpenStudy (michele_laino):

so such equation has infinite solutions, and the second solution of the system is: \[\left\{ \begin{gathered} y = a \hfill \\ x = k \hfill \\ \end{gathered} \right.,\quad a = 0,b \ne 0,k \in \mathbb{R}\]

1 year ago
OpenStudy (michele_laino):

now we have to consider this case: 2) \(b=0\)

1 year ago
OpenStudy (michele_laino):

in that case I can rewrite the above equation: \[by = ab\] in this way: \(0\;y=0\) so how many solutions \(y\) has such equation?

1 year ago
OpenStudy (exoexoexo):

infinite amount of solutions

1 year ago
OpenStudy (michele_laino):

correct!

1 year ago
OpenStudy (michele_laino):

we can write this: \[y = h,\quad h \in \mathbb{R}\]

1 year ago
OpenStudy (michele_laino):

now, I substitute \(y=h\) into the original first equation: \[ax = bk = 0\] since \(b=0\) then we have: \(ax=0\) Now, as befor, I have to distinguish two cases: 1) \(a \neq 0\) 2) \(a=0\) now, I consider the case 1): \(a \neq 0\) so I can divide by \(a\), and I get: \[\frac{{ax}}{a} = \frac{0}{a}\] what is \(x\) ?

1 year ago
OpenStudy (michele_laino):

before*

1 year ago
OpenStudy (exoexoexo):

\[x=\frac{ 0 }{ a }\]

1 year ago
OpenStudy (michele_laino):

which is equal to \(0\), so the third solution of the system is: \[\left\{ \begin{gathered} y = h \hfill \\ x = 0 \hfill \\ \end{gathered} \right.,\quad a \ne 0,b = 0,h \in \mathbb{R}\]

1 year ago
OpenStudy (michele_laino):

next I have to consider the case \(a=0\) now I can rewrite the equation: \[ax = 0\] like below: \(0\;x=0\) so, how many solutions \(x\) has such equation

1 year ago
OpenStudy (exoexoexo):

infinite amount of solutions

1 year ago
OpenStudy (michele_laino):

that's rright! so we can write the fourth solution of the system as below: \[\left\{ \begin{gathered} y = h \hfill \\ x = m \hfill \\ \end{gathered} \right.,\quad a = 0,b = 0,\quad h,m \in \mathbb{R}\]

1 year ago
OpenStudy (michele_laino):

oops.. that's right*!

1 year ago
OpenStudy (michele_laino):

and we have completed your exercise

1 year ago