Respuesta :
Percy is wrong since the system will have no solution for any real numbers except k = 7, because, at k = 7, both lines become the same, thus having infinite solutions.
Given that
The given system of the equation;
[tex]\rm 6x + 4y = 14\\\\ 3x + 2y = k[/tex]
Percy said that any real number for the value of k would cause the system of equations to have no solution.
We have to determine
Explain the error in Percy’s statement.
According to the question
The given system of the equation;
[tex]\rm 6x + 4y = 14\\\\ 3x + 2y = k[/tex]
When both the lines(represented by both the equations) are parallel and not coincident, then that system of linear equations has no solutions as there is no common point on both lines.
Thus, if two lines are not in any of the above cases, they intersect at a single point, and in that case, the considered system of the equation has a unique single solution.
The first equation is;
[tex]\rm 6x+4y = 14\\ \\ Dividing \ by \ 2 \ on \ both \ side\\ \\ 3x+2y = 7\\ \\ [/tex]
On comparing with the second equation;
K = 7
Thus, there are infinite solutions for all values of k except 7.
Hence, Percy is wrong since the system will have no solution for any real numbers except k = 7, because, at k = 7, both lines become the same, thus having infinite solutions.
To know more about System of Equation click the link given below.
https://brainly.com/question/4125955
Answer:
Sample Response: k = 7 proves the equations are equivalent and, therefore, have infinitely many solutions. That is illustrated by the fact that multiplying 2 by 3x +2y = 7 will create the equivalent of 6x + 4y =14.
Step-by-step explanation:
