Question

Solve the system using graphing, substitution or elimination. If needed round soulutions to the nearest tenth

Answer 1

Given

The system of equations,

$\begin{gathered} 9x+y=45\text{ \_\_\_\_\_\lparen1\rparen} \\ x^3-3x^2-25x+93=y\text{ \_\_\_\_\_\_\lparen2\rparen} \end{gathered}$

To find the solution.

Explanation:

It is given that,

$\begin{gathered} 9x+y=45\text{ \_\_\_\_\_\lparen1\rparen} \\ x^3-3x^2-25x+93=y\text{ \_\_\_\_\_\_\lparen2\rparen} \end{gathered}$

From (1),

$y=45-9x$

Substitute y in (2).

Then,

$\begin{gathered} x^3-3x^2-25x+93=45-9x \\ x^3-3x^2-25x+9x+93-45=0 \\ x^3-3x^2-16x+48=0 \\ x^2(x-3)-16(x-3)=0 \\ (x-3)(x^2-16)=0 \\ (x-3)(x^2-4^2)=0 \\ (x-3)(x-4)(x+4)=0 \end{gathered}$

That implies,

$\begin{gathered} x-3=0,x-4=0,x+4=0 \\ \text{ }x=3,\text{ }x=4,\text{ }x=-4 \end{gathered}$

Therefore, for x=3,

$\begin{gathered} y=45-9\times3 \\ =45-27 \\ =18 \end{gathered}$

For x=4,

$\begin{gathered} y=45-9\times4 \\ =45-36 \\ =9 \end{gathered}$

For x=-4,

$\begin{gathered} y=45-(9\times-4) \\ =45+36 \\ =81 \end{gathered}$

Hence, the solution set is (3,18), (4,9), (-4,81).