Question

​How many ordered pairs $(x,y)$ satisfy the system of equations $x=2y+5$ and $y=(2x-3)(x+9)$?

Answer 1

Step-1: Find the number of ordered pairs $\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$:

Given the system of equations $x=2y+5$ and $y=(2x-3)(x+9)$.

To solve a system of equations with two variables, we find the expression for one of the variables in terms of the other by using one of the equations and then substituting it in the other equation.

Given two solutions are in the form of the general quadratic equation: $a{x}^2+bx+c=0$.

Use the quadratic formula: $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
(i) lf $b^2-4ac>0$, then there are two distinct solutions.
(ii) If $b^2-4ac=0$, then there are two solutions, each equal to $\frac{-b}{2a}$.

Step-2: Substitute one equation into another.

Form an equation involving only one of the variables, either $x$or $y$.

Substitute the value of $x=2y+5$ into the equation $y=(2x-3)(x+9)$:

$\begin{array}{rcl}y& =& (2\left(2y+5\right)-3)(2y+5+9)\\ & \Rightarrow & y=\left(4y+10-3\right)(2y+14)\\ & \Rightarrow & y=\left(4y+7\right)(2y+14)\\ & \Rightarrow & y=8y^2+56y+14y+98\\ & \Rightarrow & y=8y^2+70y+98\\ & \Rightarrow & 0=8y^2+70y-y+98\\ & \Rightarrow & 0=8y^2+69y+98\end{array}$

Step-3: Check the condition to find the number of ordered pairs:

Substitute $a=8,b=69\mathrm{and}\mathrm{c}=98$ in $b^2-4ac$:

$$\begin{gathered} {69}^2-4\times 8\times 98=4761-3136 \\ =1625 \end{gathered}$$

The value of $1625$ is greater than $0.$

So, it satisfies the condition $b^2-4ac>0$ and will have two distinct solutions.

Step-4: Find the ordered pairs.

Use the quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to find the ordered pairs.

Substitute $a=8,b=69\mathrm{and}\mathrm{c}=98$ in above formula:

$\begin{array}{rcl}y& =& \frac{-69\pm \sqrt{{69}^2-4\times 8\times 98}}{2\times 8}\\ & =& \frac{-69\pm \sqrt{4761-3136}}{16}\\ & =& \frac{-69\pm \sqrt{1625}}{16}\\ & =& \frac{-69\pm 40.3113}{16}\\ & =& -1.793\mathrm{or}-6.832\end{array}$

Substitute $\begin{array}{rcl}y& =& -1.793\end{array}$ and $-6.832$ in the equation $x=2y+5$ to find $x$ values:

$\begin{array}{rcl}x& =& 2(-1.793)+5\\ & =& -3.586+5\\ & =& 1.414\end{array}$

And,

$\begin{array}{rcl}x& =& 2(-6.832)+5\\ & =& -13.664+5\\ & =& -8.664\end{array}$

So, the ordered pairs are $(1.414,-1.793)$ and $(-8.664,-6.832)$.

Hence, when satisfying the system of equations $\mathit{x}\mathbf{=}\mathbf{2}\mathit{y}\mathbf{+}\mathbf{5}$ and $\mathit{y}\mathbf{=}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{9}\mathbf{)}$ the number of ordered pairs $\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$ are $\mathbf{2}$.