Question

Find the number x and y if:-
a.) (2x-1),y+2)=(1,2)
(Specially find x)

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Answer 1

Answer:

Step-by-step explanation:

et’s do this one using middle school math. We have  f(x,y)=0  where

f(x,y)=x2–2x+4y2+16y+1  

Let’s zoom into a point on the curve  (r,s)  

f(x,y)=f(r+(x−r),s+(y−s))  

=(r+(x−r))2−2(r+(x−r))+4(s+(y−s))2+16(s+(y−s))+1  

=r2+2r(x−r)+(x−r)2−2r−2(x−r))4s2+8s(y−s)+(y−s)2+16s+16(y−s)+1  

=r2−2r+4s2+16s+1+(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2  

f(x,y)=f(r,s)+(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2  

Since  (r,s)  is assumed on the curve,  f(r,s)=0  

f(x,y)=(2r−2)(x−r)+(8s+16)(y−s)+(x−r)2+(y−s)2  

The tangent at  (r,s)  is the best linear approximation at  (r,s) . We simply drop the second degree terms and set the whole thing to zero:

0=(2r−2)(x−r)+(8s+16)(y−s)  

For the no slope case (vertical line) the coefficient on y is zero:

8s+16=0  

s=−2  

0=f(r,−2)=r2–2r+4(−2)2+16(−2)+1  

0=r2–2r−15  

(r−5)(r+3)=0  

r=5,r=−3  

Points with vertical tangents:  (5,−2)  and  (−3,−2)  

plot 0= x^2 – 2x + 4y^2 + 16y + 1 , x=5, x=-3

hope it helps

:)

Answer 2

Answer:

2X-1=1

=>2x=1+1

=>X=2/2

=>X=1.