Question #230380

You are given the system of linear equations

2x+ky=5,x+3y=7,2x+ky=5, \quad x+3y=7,

where k is a constant.


The system above has no solution when k= _________________




1
Expert's answer
2021-08-31T14:09:08-0400

The given system of linear equations is:

2x+ky=5x+3y=72x+ky=5\\x+3y=7

The given system is of the type:

a1x+b1y+c1=0, a2x+b2y+c2=0a_1x+b_1y+c_1=0,\space a_2x+b_2y+c_2=0

which has no solution if:

a1a2=b1b2c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}


Therefore, the given system can be written as:

2x+ky5=0x+3y7=02x+ky−5=0\\x+3y−7=0

which has no solution if:

21=k35721=k3k=6\frac{2}{1}=\frac{k}{3}≠\frac{−5}{−7}\\⇒\frac{2}{1}=\frac{k}{3}⇒\\k=6

so, the given system has no solution for k=6.


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