Chapter 11: Q. 55 (page 773)

Graph the equations given in Example $4$ .

Short Answer

Expert verified

The graph for the system of equations $\{\begin{cases} x^{2} + x + y^{2} - 3 y + 2 = 0 \\ x + 1 + \frac{y^{2} - y}{x} = 0 \end{cases}$ is,

Step by step solution

Step 1 The equations in the example 4 are,

$\{\begin{cases} x^{2} + x + y^{2} - 3 y + 2 = 0 \\ x + 1 + \frac{y^{2} - y}{x} = 0 \end{cases}$

Consider the first equation and simplify it.

$x^{2} + x + y^{2} - 3 y + 2 = 0 x^{2} + x + y^{2} - 3 y = - 2 x^{2} + x + {(\frac{1}{2})}^{2} + y^{2} - 3 y + {(\frac{3}{2})}^{2} = - 2 + {(\frac{1}{2})}^{2} + {(\frac{3}{2})}^{2} {(x + \frac{1}{2})}^{2} + {(y - \frac{3}{2})}^{2} = - 2 + \frac{1}{4} + \frac{9}{4} {(x + \frac{1}{2})}^{2} + {(y - \frac{3}{2})}^{2} = \frac{1}{2}$

Step 2 The obtained equation takes the form of x-h2+y-k2=r2, which is the equation of a circle.

So, the circle has radius $\frac{1}{2}$ and the center $(- \frac{1}{2}, \frac{3}{2})$ .

Now consider the second equation and simplify it.

$x + 1 + \frac{y^{2} - y}{x} = 0 x (x + 1) + y^{2} - y = 0 x^{2} + x + y^{2} - y = 0 x^{2} + x + {(\frac{1}{2})}^{2} + y^{2} - y + {(\frac{1}{2})}^{2} = 0 + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} {(x + \frac{1}{2})}^{2} + {(y - \frac{1}{2})}^{2} = \frac{1}{2}$

So, the second equation also takes the form of the equation of the circle with radius $\frac{1}{2}$ and the center $(- \frac{1}{2}, \frac{1}{2})$ .

Step 3 Now sketch a graph for the given equations.

From the second equation we can see that $x$ cannot be zero. To indicate this, there are holes at points where the circle touches the $y$ -axis.

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Graph the equations given in Example $4$ .

Short Answer

Step by step solution

Step 1 The equations in the example 4 are,

Step 2 The obtained equation takes the form of x-h2+y-k2=r2, which is the equation of a circle.

Step 3 Now sketch a graph for the given equations.

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Graph the equations given in Example 4.

Short Answer

Step by step solution

Step 1 The equations in the example 4 are,

Step 2 The obtained equation takes the form of x-h2+y-k2=r2, which is the equation of a circle.

Step 3 Now sketch a graph for the given equations.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Decision Maths

Statistics

Mechanics Maths

Theoretical and Mathematical Physics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Graph the equations given in Example $4$ .