Each time x increases by 2 (eg: from x = -3 to x = -1), the value of y increases by 7 (eg: y = 9.5 to y = 16.5)

Change in y = 7
change in x = 2
Slope = (change in y)/(change in x) = 7/2 = 3.5

This all means that we have a linear equation

4 0

3 years ago

Type the correct answer in each box. Use numerals instead of words.

lubasha [3.4K]

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:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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

$x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}$

Finding value of y:

y = -1/2x

$y=-1/2(\frac{\pm2\sqrt{85}}{5})$

$y=\frac{\pm\sqrt{85}}{5}$

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:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}$

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

4 0

3 years ago

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