Chapter 0.4, Problem 7QR

a.

To determine

The given point lie or does not lie on the graph of the equation.

Answer to Problem 7QR

The point $\left(1,1\right)$ lies on the graph of the equation $2x^{2}y\text{+}{y}^2\text{=}3$ .

Explanation of Solution

Given:

Given data: The given equation is $2x^{2}y\text{+}{y}^2\text{=}3$ and the point $\left(1,1\right)$ .

Method Used:

A point $\left(p,q\right)$ lies on the graph of an equation $y=f\left(x\right)$ , if and only if the point satisfies the equation

Calculations:

The given equation is $2x^{2}y\text{+}{y}^2\text{=}3$ .

The given point is $\left(1,1\right)$.

Substitute $x=1,y=1$ in the equation,

  $\begin{array}{c}2x^{2}y\text{+}{y}^2\text{=}3\\ 2{\left(1\right)}^2\left(1\right)\text{+}{\left(1\right)}^2\text{=}3\\ 2+1=3\\ 3=3\end{array}$

Thus, the point $\left(1,1\right)$ satisfies the equation, therefore it lies on the graph of the equation

b.

To determine

The given point lie or does not lie on the graph of the equation.

Answer to Problem 7QR

The point $\left(-1,-1\right)$ does not lie on the graph of the equation $2x^{2}y\text{+}{y}^2\text{=}3$ .

Explanation of Solution

Given data: The given equation is $2x^{2}y\text{+}{y}^2\text{=}3$ and the point is $\left(-1,-1\right)$

Method used:

Same as in part (a).

Calculations:

The given equation is $2x^{2}y\text{+}{y}^2\text{=}3$ .

The given point is $\left(-1,-1\right)$.

Substitute $x=-1,y=-1$ in the equation,

  $\begin{array}{c}2x^{2}y\text{+}{y}^2\text{=}3\\ 2{\left(-1\right)}^2\left(-1\right)\text{+}{\left(-1\right)}^2\text{=}3\\ -2+1=3\\ -1\ne 3\end{array}$

Thus, the point $\left(-1,-1\right)$ does not satisfy the equation, therefore it does not lie on the graph of the equation.

c.

To determine

The given point lie or does not lie on the graph of the equation.

Answer to Problem 7QR

The point $\left(1/2,-2\right)$ lie on the graph of the equation $2x^{2}y\text{+}{y}^2\text{=}3$ .

Explanation of Solution

Given:

Given data: The given equation is $2x^{2}y\text{+}{y}^2\text{=}3$ and the point is $\left(1/2,-2\right)$ .

Method used:

Same as in part (a).

Calculations:

Calculation:

The given equation is $2x^{2}y\text{+}{y}^2\text{=}3$ .

The given point is $\left(1/2,-2\right)$ .

Substitute $x=1/2,y=-2$ in the equation,

  $\begin{array}{c}2x^{2}y\text{+}{y}^2\text{=}3\\ 2{\left(\frac{1}{2}\right)}^2\left(-2\right)\text{+}{\left(-2\right)}^2\text{=}3\\ -1+4=3\\ 3=3\end{array}$

Thus, the point $\left(1/2,-2\right)$ satisfies the equation, therefore it lies on the graph of the equation.