en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area x=8+2y,y=-1,y=3

Solution

area

x = 8 + 2 y, y = - 1, y = 3

Solution

No Solution

Solution steps

Isolate

y

for

x = 8 + 2 y : y = \frac{x - 8}{2}

f_{1} (x) = \frac{x - 8}{2}

f_{2} (x) = - 1

To find the intersection points solve

f_{i} (x) = f_{j} (x)

\frac{x - 8}{2} = - 1 : x = 6

\frac{x - 8}{2} = 3 : x = 14

- 1 = 3 :

False

Curves not bounded

= No Solution

Popular Examples

x^3(dy)/(dx)+5xy=x

x^{3} \frac{d y}{d x} + 5 x y = x

(dh)/(dt)=-0.006sqrt(19.6h)

\frac{d h}{d t} = - 0.006 19.6 h

tangent of y=1+sqrt(x),(4,3)tangent of

y = 1 + x, (4, 3)

derivative of-1/((3+x^2))

\frac{d}{d x} (- \frac{1}{( 3 + x ) ^{2}})

limit as x approaches-3 of (x^3)/(x^2-9)

x \to - 3 lim (\frac{x ^{3}}{x ^{2} - 9})

Frequently Asked Questions (FAQ)

What is area x=8+2y,y=-1,y=3 ?
The answer to area x=8+2y,y=-1,y=3 is No Solution