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solve the following… - QuestionCove
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