Chapter ISG, Problem 8.6.2.2P

a.

To determine

To find: The value of $q$ for the given first equation.

Answer to Problem 8.6.2.2P

The value of $q$ for the given first equation is $16$ .

Explanation of Solution

Given information:

The equations are $2x^2+qx-12=6x$ and $3y^2+11y-q=0$ .

Calculation:

Consider the equation.

  $2x^2+qx-12=6x$ ……. (1)

Substitute $1$ for $x$ in Equation (1).

  $\begin{array}{l}2{\left(1\right)}^2+q\left(1\right)-12=6\left(1\right)\\ 2+q-12=6\\ q=6+12-2\\ q=18-2\\ q=16\end{array}$

Thus, the value of $q$ for the given equation is $16$ .

b.

To determine

To find: The value of $q$ for the given second equation.

Answer to Problem 8.6.2.2P

The value of $q$ for the given first second is $14$ .

Explanation of Solution

Given information:

The equations are $2x^2+qx-12=6x$ and $3y^2+11y-q=0$ .

Calculation:

Consider the equation.

  $3y^2+11y-q=0$ ……. (2)

Substitute $1$ for $y$ in Equation (2).

  $\begin{array}{l}3{\left(1\right)}^2+11\left(1\right)-q=0\\ 3+11-q=0\\ 14-q=0\\ -q=-14\\ q=14\end{array}$

Thus, the value of $q$ for the given equation is $14$ .

c.

To determine

To write: The given equations after substituting the value of $q$ .

Answer to Problem 8.6.2.2P

The given equations after substituting the value of $q$ are $2x^2+16x-12=6x$ and $3y^2+11y-14=0$ .

Explanation of Solution

Given information:

The equations are $2x^2+qx-12=6x$ and $3y^2+11y-q=0$ .

Calculation:

Consider the equation.

  $2x^2+qx-12=6x$ ……. (1)

Consider the equation.

  $3y^2+11y-q=0$ ……. (2)

Substitute $16$ for $q$ in Equation (1).

  $2x^2+16x-12=6x$

Substitute $14$ for $q$ in Equation (2).

  $3y^2+11y-14=0$

Thus, the given equations after substituting the value of $q$ are $2x^2+16x-12=6x$ and $3y^2+11y-14=0$ .

d.

To determine

To find: The value of $x$ and $y$ .

Answer to Problem 8.6.2.2P

The value of $x$ and $y$ is $x=-6,1$ and $y=1,-\frac{14}{3}$ .

Explanation of Solution

Given information:

The equations are $2x^2+qx-12=6x$ and $3y^2+11y-q=0$ .

Calculation:

Consider the equation.

  $\begin{array}{l}2x^2+16x-12=6x\\ 2x^2+16x-6x-12=0\\ 2x^2+10x-12=0\\ 2x^2+12x-2x-12=0\end{array}$

Further solve,

  $\begin{array}{l}2x\left(x+6\right)-2\left(x+6\right)=0\\ \left(x+6\right)\left(2x-2\right)=0\\ x+6=0\\ x=-6\end{array}$

Further solve,

  $\begin{array}{l}2x-2=0\\ x=1\end{array}$

Consider the equation.

  $\begin{array}{l}3y^2+11y-14=0\\ \left(3y^2-3y\right)+\left(14y-14\right)=0\\ 3y\left(y-1\right)+14\left(y-1\right)=0\\ \left(y-1\right)\left(3y+14\right)=0\end{array}$

Further solve,

  $\begin{array}{l}y-1=0\\ y=1\\ 3y+14=0\\ 3y=-14\\ y=-\frac{14}{3}\end{array}$

Thus, the value of $x$ and $y$ is $x=-6,1$ and $y=1,-\frac{14}{3}$ .