Type the correct answer in each box. use numerals instead of words.
consider the systems of equations below.

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17   eq(1)

y = -1/2x            eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

a = 5/4, b =0 and c = -17

Finding value of y:

y = -1/2x

System A has 4 real solutions.

System B

y = x^2 -7x + 10    eq(1)

y = -6x + 5            eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

a= 1, b =-1 and c =5

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9    eq(1)

8x - y = -17        eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17    

8(-2) - y = -17    

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

System A has 2 real solutions.

System B has 0 real solutions.

System C has 1 real solutions.

Step-by-step explanation:

If the graph of system of equation intersect each other at n points then the system of equation has n real solutions.

System A:

              .... (1)

            .... (2)

Plot the graph of these equations.

From graph (1) it is clear that the graph of equation (1) and (2), intersect each other at two points, (-3.688,1.844) and (3.688,-1.844).

Therefore, System A has two real solutions.

System B:

              .... (3)

            .... (4)

Plot the graph of these equations.

From graph (2) it is clear that the graph of equation (3) and (4), never intersect each other.

Therefore, System B has 0 real solutions.

System C:

              .... (5)

            .... (6)

Plot the graph of these equations.

From graph (3) it is clear that the graph of equation (5) and (6), intersect each other at one points, (-2,1).

Therefore, System C has 1 real solutions.

the least value out of the data shown is 5, maybe if you post a picture with it i could a little more.

step-by-step explanation: