Choose h and k such that the system has​ (a) no​ solution, (b) a unique​ solution, and​ (c) many solutions.
x1 +hx2 = 3
5x1 + 15x2 = k
a. Select the correct answer below and fill in the answer​ box(es) to complete your choice. ​(Type an integer or simplified​ fraction.)
A. The system has no solutions only when k nothing and h is any real number.
B. The system has no solutions only when h nothing and k nothing.
C. The system has no solutions only when h nothing and k is any real number.
D. The system has no solutions only when h nothing and k nothing.
E. The system has no solutions only when h nothing and k nothing.
F. The system has no solutions only when k nothing and h is any real number.
G. The system has no solutions only when h nothing and k nothing.
H. The system has no solutions only when h nothing and k is any real number.
b. Select the correct answer below and fill in the answer box(es) to complete your choice. (Type an integer or simplified fraction.)
A. The system has a unique solution only when h and k
B. The system has a unique solution only when h =-and k is any real number.
C. The system has a unique solution only when k | and h is any real number.
D. The system has a unique solution only when h-U and k = 1

Answer:

a) C) The system has no solutions only when h=3 and k is any real number.

b) D) The system has a unique solution when [tex]h=(-\infty,3)U(3,\infty)[/tex] and k is any real number.

c) The system has may solutions when h=3 and k=15

Step-by-step explanation:

a) In order to determine when the system will have no solution, we can start by solving the equation by substitution. We can solve the first equation for x1:

[tex]x_{1}+hx_{2}=3[/tex]

so

[tex]x_{1}=3-hx_{2}[/tex]

Next we can substitute this into the second equation so we get:

[tex]5(3-hx_{2})+15x_{2}=k[/tex]

We distribute the 5 into the first parenthesis so we get:

[tex]15-5hx_{2}+15x_{2}=k[/tex]

and group like terms:

[tex]-5hx_{2}+15x_{2}=k-15[/tex]

we factor x2 so we get:

[tex]x_{2}(-5h+15)=k-15[/tex]

and solve for x2:

[tex]x_{2}=\frac{k-15}{-5h+15}[/tex]

this final answer is important because it tells us what value the system of equations is not valid for. That answer will not ve vallid if the denominator is zero, so we can set the denominator equal to zero and solve for h, so we get:

[tex]-5h+15= 0[/tex]

and solve for h:

[tex]-5h= -15[/tex]

[tex]h=\frac{-15}{-5}[/tex]

[tex]h= 3[/tex]

so it doesn't really matter what value k gets since all that matters is that the denominator of the answer isn't zero.

b)

For part b we need to know when the system of equations will have infinitely many answers. Generally, this will happen when both equations are basically the same, so we need to make sure to simplify the second equation so it looks like the first equation, compare them and determine the respective coefficients.

So we take the second equation and factor it:

[tex]5x_{1}+15x_{2}=k[/tex]

we start by factoring a 5 from the left side of the equation so we get:

[tex]5(x_{1}+3x_{2})=k[/tex]

Next, we divide both sides of the equation into 5 so we get:

[tex]x_{1}+3x_{2}=\frac{k}{5}[/tex]

we now compare it to the first equation:

[tex]x_{1}+hx_{2}=3[/tex]

[tex]x_{1}+3x_{2}=\frac{k}{5}[/tex]

In this case, every coefficient of the two equations must be the same for us to get infinitely many answers, so we can see that h=3 and [tex]\frac{k}{5}=3[/tex]

when taking the second condition and solving for k we get that:

[tex]k=3(5)[/tex]

so

k=15

Anything else than the specific combination h=3 and k=15 will give us unique solutions, so for b, the answer is:

D) The system has a unique solution when and k is any real number.

c)

We have already solved part c on the previous part of the problem, so the answer is:

The system has many solutions when h=3 and k=15

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