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3 years ago

What is the solution to 2x+y=1 and x-2y=3

Mathematics

1 answer:

Masteriza [31]3 years ago

5 0

Answer:

Step-by-step explanation:

2x+y=1---------> y=-2x+1

x-2y=3

2) x-2(-2x+1)=3

x+4x-2=3

5x=5

x=1

3) 2x+y=1 when x=1

2(1)+y=1

2+y=1

y=1

the answer is (1,1)

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(i) Yes. Simplify $f(x,y)$ .

$\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}$

Now compute the limit by converting to polar coordinates.

$\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0$

This tells us

$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1$

so we can define $f(0,0)=1$ to make the function continuous at the origin.

Alternatively, we have

$\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2$

and

$\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2$

Now,

$\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0$

$\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0$

so by the squeeze theorem,

$\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0$

and $f(x,y)$ approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

$\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x$

$f(0,y)$ and $f(x,0)$ are undefined, so there is no way to make $f(x,y)$ continuous at (0, 0).

(iii) No. Similarly,

$\dfrac{x^2 + y}y = \dfrac{x^2}y + 1$

is undefined when $y=0$ .

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Answer:

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