Question

Which values, when placed in the box, would result in a system of equations with no solution? Check all that apply.

Answer 1

For this case we have the following system of equations:

$y = -2x + 4\\6x + 3y =$

Rewriting equation 1 we have:

$2x + y = 4$

Therefore, the equivalent system is:

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

The system will have no solution, if we write equation 2 as a linear combination of equation 1.

Therefore, since both lines have the same slope, they are parallel.

Parallel lines do not intersect when they have different cut points.

Therefore, there is no solution for:

-12, -4, 0, 4

The system has inifinites solutions for:

12

This is because the lines intersect at all points in the domain.

Answer:

The values, when placed in the box, would result in a system of equations with no solution are:

A: -12

B: -4

C: 0

D: 4

Answer 2

Answer:

Thus, option (a), (b) , (c) , (d) are correct.

The system will have no solution for all values except 12.

Step-by-step explanation:

 Given a system of equation y = –2x + 4  and 6x + 3y = ?

We have to check for which value the '?'  would result in a system of equations with no solution.

Consider a system of equation  $a_1x+b_1y+c_1=0 \\\\a_2x+b_2y+c_2=0$

For the system to have no solution the condition is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

For the given system of equation ,

Let unknown quantity be v.

y = –2x + 4   ⇒ 2x +y - 4 = 0  

and 6x + 3y = v ⇒  6x + 3y - v =0

On comparing, we get,

$a_1=2 , b_1=1,c_1=-4\\\\\a_2=6,b_2=3,c_3=-v$

Substitute the values in condition for no solution , we get ,

$\frac{2}{6}=\frac{1}{3}\neq \frac{-4}{-v}$

Consider second and third ratio, we get,

$\frac{1}{3}\neq \frac{-4}{-v}$

Solve for v , we get,

$\frac{1}{3}\neq \frac{-4}{-v} \\\\ \Rightarrow v \neq 12$  

Thus, for all values v except v = 12

The system will have no solution

at v = 12 the system will have infinite many solution.

Thus, option (a), (b) , (c) , (d) are correct.