y = cx + d
2y = 10
In the system of equations above, cand d are constants.
For which of the following values of cand d does the
system of equations have no real solutions?
A) c= -6, d = 6
B) C=-5, d=4
C) c= 6, d = 4
D) c = 6, d= 5

Answer :

Answer is a hope this helps

The system of equations will have no real solutions for c = -5, d = 4, i.e. option B.

What is system of equations?

System of equations is a finite set of equations for which common solutions are sought.

We have,

y = cx² + d     .....(i)

And,

2y = 10

y = 5     .....(ii)

So,

Now,

Comparing equation (i) and (ii),

We get,

cx² + d = 5

Now,

According to the question,

For,

c = -6, and d = 6

i.e.

cx² + d = 5

-6x² +6 = 5

We get,

-6x² = -1

i.e. [tex]x^2 = \frac{1}{6}[/tex]

That means equation have real solution.

Now,

For,

c = -5, and d = 4

i.e.

cx² + d = 5

-5x² + 4 = 5

We get,

-5x² = 1

i.e. [tex]x^2 = \frac{1}{-5}[/tex]

So, For these values of c and d, equation will not have any real solution, because square root can not be negative.

Hence, we can say that the system of equations will have no real solutions for c = -5, d = 4, i.e. option B.

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