Question

Percy said that any real number for k would cause the system of equations to have no solution. Explain the error in Percy’s statement.


6x + 4y = 14,

3x + 2y = k

Answer 1

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.

Answer 2

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

$3x + 2y = k \to 6x + 4y = 2k$

Subtract the new equation, from the first equation

$6x - 6x + 4y - 4y =14 -2k$

$0=14 -2k$

Collect like terms

$2k = 14$

Solve for k

$k = 7$

The above means that:

The system of equations would have infinite many solutions when k = 7

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