how do u solve this quadratic system equation: 2x^2-y^2=1 & x+y=12?

Question

This is from my pre calc class and i was hoping that someone out there can help me with this problem. I'm trying to figure out what x and y equals. I know how to do it using the substitution method, but the quadratic throws me off. Please Help.

Answer 1

2x^2 - y^2 = 1
x + y = 12 So y = 12 - x
So 2x^2 - (12 - x)^2 = 1
ie 2x^2 - 144 + 24x - x^2 = 1
ie x*2 + 24x - 145 = 0
ie (x + 29) (x - 5) = 0
x = 5. -29

when x = 5 y = 7 Check 2x^2-y^2 = 50 - 49 = 1

when x = -29 y = -41 Check 2x^2-y^2 = 1682 - 1681 = 1

Answer 2

Try the substitution method:
y=-x+12 and plug it into 2x^2-y^2=1

2x^2-(-x+12)^2=1
2x^2-(x^2-24x+144)=1
2x^2-x^2+24x-144=1
x^2+24x-145=0
(x+29)(x-5)=0
x=-29 and 5

Take these numbers and plug it into x:
y=-5+12
y=7

y=-(-29)+12
y=41

The answer is x=-29 and 5 and y=7 and 41.

Check:
2(5)^2-(7)^2=1
50-49=1
1=1

2(-29)^2-(41)^2=1
1,682-1,681=1
1=1

7=-5+12
7=7

41=-(-29)+12
41=41

I hope this helps!

Answer 3

substitute y=12-x in the first equation and solve for x.
then subtract x from 12 and that = y.

Oops, didnt see the rest of your question.

Plot both curves on a TI 83. The curves will cross at the solutions.

Answer 4

2x^2 - y^2 = 1
x + y = 12
y = 12 - x
2x^2 - (12 - x)^2 = 1
2x^2 - (144 - 24x + x^2) = 1
2x^2 - 144 + 24x - x^2 = 1
x^2 + 24x - 145 = 0
x = -12 ± √(144 + 145)
x = -12 ± 17
x = (5,-29)
y = 12 - x
y = (7,41)

2x^2 - y^2 = 1
2*25 - 49 = 1
2*841 - 1681 = 1

Answer 5

2x^2-y^2=1 & x+y=12
x=12-y substitute
2(12-y)^2-y^2-1=0
2(144-24y+y^2)-y^2-1=0
288-48y+2y^2-y^2-1=0
y^2-48y+287=0
(y-41)(y-7)=0
y=41 x=12-y=12-41=-29
y=7 x=12-y=12-7=5

Answer 6

there is a way to do this with algebraic formulas,but use substitution method ,it will be easier and gives you the same answer! good luck,

Answer 7

That looks hard. Wow I don't even know how to read that!