Solving system of equation?

4x² + 4y² = 64
x² + y² = 16

x - y = 1
y = x - 1

x² + (x-1)² = 16
x² + x² - 2x + 1 = 16
2x² - 2x - 15 = 0

x = (2±√(2²+4*2*15))/(2*2)
= (2±√(4+120))/(2*2)
= (2±2√31)/(2*2)
= (1±√31)/(2)

x = (1+√31)/(2)
y = (√31-1)/(2)

OR

x = (1-√31)/(2)
y = (-1-√31)/(2)

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6x² - x < 20
6x² - x - 20 < 0

If
6x² - x - 20 = 0
then
x = (1±√(1²+4*6*20))/(2*6)
= (1±√481)/12

x < (1-√481)/12
OR
x > (1+√481)/12

x - y = 1
y = x - 1

4x² + 4y² = 64
x² + y² = 16
x² + (x -1)² = 16

expand and add together. Use formula: x= (- b +- √b² 4ac ) /2a to find out x value.



b.
6x² - x < 20
6x² - x - 20 < 0

Assume it is equal to zero.
6x² - x - 20 = 0

Use formula: x= (- b +- √b² 4ac ) /2a to find out x value.

x = (1±√(1²+4*6*20))/(2*6)
= (1±√481)/12

x < (1-√481)/12 or x > (1+√481)/12

In the first question, isolate x for the x-y=1 equation. Doing that will give x=y+1. Since x=y+1, throw in y+1 wherever you see an x in the first equation. Let's review:

x - y = 1 is the same as x = y + 1 because I added y to both sides to get that. Now, substitute y+1 into x:

4(y+1)^2 + 4y^2 = 64

4(y^2 + 2y + 1) + 4y^2 = 64

4y^2 + 8y + 4 + 4y^2 = 64

8y^2 + 8y + 4 = 64

8y^2 + 8y + 4 - 64 = 0

8y^2 + 8y - 60 = 0

4(2y^2 + 2y - 15) = 0

2y^2 + 2y - 15 = 0

(Use quadratic formula to solve for y)

Just a thought: Make sure the original question was typed in correctly.

After you get y, you can solve for x:

x = y + 1 where y is the answer you got from above.

And now for the second question:

6x^2 - x < 20

Subtract 20 from both sides and you get:

6x^2 - x - 20 < 0

(Use quadratic formula to get x)

Graph it. The answer is going to be -m <= x <= n where m,n are the answers from the quad formula.

for a. make x = y+ 1

then substitute it

so it becomes

4(y+1)^2 + 4y^2 = 64