Chapter 11: Q. 49 (page 774)

Solve the system. Use any method you wish.
$\{\begin{cases} y^{2} + y + x^{2} - x - 2 = 0 \\ y + 1 + \frac{x - 2}{y} = 0 \end{cases}$

Short Answer

Expert verified

The solution of system of equations are $(0, 1), (0, - 2), (2, - 1)$ .

Step by step solution

Step 1. Given information.

Consider the given question,

$\{\begin{cases} y^{2} + y + x^{2} - x - 2 = 0 . . . . . . (i) \\ y + 1 + \frac{x - 2}{y} = 0 . . . . . . (i i) \end{cases}$

Multiply equation (ii) with y,

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

Subtract equations (i) and (iii),

role="math" localid="1646976142501" $(y^{2} + y + x^{2} - x - 2) - (y^{2} + y + x - 2) = 0 x^{2} - 2 x = 0 x (x - 2) = 0 x = 0, 2$

Step 2. Substitute the values in equation (iii).

Substitute $x = 0$ in equation (iii),

$y^{2} + y + 0 - 2 = 0 y^{2} + y - 2 = 0 (y - 1) (y + 2) = 0 y = 1, - 2$

Substitute $x = 2$ in equation (iii),

$y^{2} + y + 2 - 2 = 0 y (y + 1) = 0 y = 0, - 1$

$y = 0$ cannot be considered as in equation (ii) y is in the denominator.

Therefore, the solution sets arerole="math" localid="1646976404925" $(0, 1), (0, - 2), (2, - 1)$ .

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Solve the system. Use any method you wish.
$\{\begin{cases} y^{2} + y + x^{2} - x - 2 = 0 \\ y + 1 + \frac{x - 2}{y} = 0 \end{cases}$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Substitute the values in equation (iii).

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Solve the system. Use any method you wish.y2+y+x2-x-2=0y+1+x-2y=0

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Substitute the values in equation (iii).

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Discrete Mathematics

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Solve the system. Use any method you wish.
$\{\begin{cases} y^{2} + y + x^{2} - x - 2 = 0 \\ y + 1 + \frac{x - 2}{y} = 0 \end{cases}$