Answer:
Except k=7, any real number for k would cause the system of equations to have no solution.
Step-by-step explanation:
In general a system of equations can be represented as ax+by=c and dx+ey=f. In order this system of equations to have NO SOLUTIONS a/d=b/a≠c/f. In our example a=6, b=4, c=14, d=3, e=2 and f=k. To apply the formula above, 6/3=4/2≠14/k. Hence k≠7. It can be concluded that except k=7, any real number for k would cause the system of equations to have no solutions.
Just for information, if k=7 the system will have infinitely many solutions.
The error in Percy's statement is that, the system of equations would have infinite many solutions when k = 7
The system of equations is given as:
6x + 4y = 14,
3x + 2y = k
Multiply the second equation by 2
[tex]3x + 2y = k \to 6x + 4y = 2k[/tex]
Subtract the new equation, from the first equation
[tex]6x - 6x + 4y - 4y =14 -2k[/tex]
[tex]0=14 -2k[/tex]
Collect like terms
[tex]2k = 14[/tex]
Solve for k
[tex]k = 7[/tex]
The above means that:
The system of equations would have infinite many solutions when k = 7
Read more about system of equations at:
https://brainly.com/question/14323743
