Question 1085977
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<pre>
1.  One way to solve it is using determinant.


2.  The other way is  <U>THIS</U>:

    the system has INFINITELY MANY solutions if and only if the two equations are equivalent.

    In other words, if and only if

    {{{1/(k+1)}}} = {{{(k+1)/9}}} = {{{5/(8k-1)}}}.    (1)

    (the coefficients and the right side terms are proportional with the same proportionality coefficient).


3.   From  {{{1/(k+1)}}} = {{{(k+1)/9}}}  you have  {{{(k+1)^2}}} = 9  ====>  k+1 = +/-3  ====>  a) k = -1 + 3 = 2,  and  b) k = -1 - 3 = -4.



     Now check k = 2 for this proportion: {{{(k+1)/9}}} = {{{5/(8k-1)}}}.

     You have {{{(2+1)/9}}} = {{{1/3}}} (left side)  and  {{{5/(8*2-1)}}} = {{{5/15}}} = {{{1/3}}}  (right side).

     So, the value k = 2 satisfies (1).



     Now check k = -4 for this proportion: {{{(k+1)/9}}} = {{{5/(8k-1)}}}.

     You have {{{(-4+1)/9}}} = {{{-1/3}}} (left side)  and  {{{5/(8*(-4)-1)}}} = {{{5/(-33)}}}  (right side).

     So, the value k = -4 does not satisfy (1).
</pre>

<U>Answer</U>. The value of "k" under the question is  k = 2.



See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/coordinate/lessons/Geom-interpret-of-the-lin-system-of-two-eqns-with-two-unknowns.lesson>Geometric interpretation of the linear system of two equations in two unknowns</A> 

in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic "<U>Systems of two linear equations in two unknowns</U>".