1. 3x - 2y = 0 => y = 3/2x
4x + 2y = 14<=> 4x + 2*(3/2x) = 14<=> 7x = 14<=> x = 2 & y = 3
2.3p + q = 7=> q = 7 - 3p
2p - 2q = -6<=> 2p - 2*(7-3p) = -6<=> 2p - 14 + 6p = -6<=> 8p = -6 + 14 = 8<=> p = 1 & q = 4
3.3x - 2y = 1=> x = (1+2y)/3
8x + 3y = 2<=> 8*(1+2y)/3 + 3y = 2<=> 8*(1+2y)/3 - 2 = -3y<=> 3*(8*(1+2y)/3 - 2) = -3*(3y)<=> 8*(1+2y) - 6 = -9y<=> 8 + 16y - 6 = -9y<=> 2 = -25y<=> y = -2/25 & x = 7/25
Answer:
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
Answer:
74 hope this works if not I'm sorry
Answer:
I think Its C
Step-by-step explanation:
I did it on unit test so idk if its right or wrong