To solve [tex]$e^{4-3x}=\frac{4}{3}x+9$[/tex] by graphing, which equations should be graphed?

A. [tex]y=0[/tex]
B. [tex]y=\frac{4}{3}x+9[/tex]
C. [tex]y=e^{4-3x}-\frac{4}{3}x+9[/tex]
D. [tex]y=\frac{4}{3}x+9+e^{4-3x}[/tex]
E. [tex]y=e^{4-3x}[/tex]



Answer :

To solve the equation [tex]\( e^{4 - 3x} = \frac{4}{3} x + 9 \)[/tex] by graphing, we need to graph the individual components of the equation and find their intersection points. Specifically, we will graph the following equations:

1. [tex]\( y = \frac{4}{3} x + 9 \)[/tex] — This is a linear equation representing the right-hand side of the original equation.

2. [tex]\( y = e^{4 - 3x} \)[/tex] — This is an exponential equation representing the left-hand side of the original equation.

By graphing these two equations on the same coordinate plane, the solutions to the original equation [tex]\( e^{4 - 3x} = \frac{4}{3} x + 9 \)[/tex] are where these graphs intersect.

So, the equations that should be graphed are:

[tex]\[ y = \frac{4}{3} x + 9 \][/tex]
[tex]\[ y = e^{4 - 3x} \][/tex]

These graphs will intersect at points whose [tex]\( x \)[/tex]-coordinates are the solutions to the equation [tex]\( e^{4 - 3x} = \frac{4}{3} x + 9 \)[/tex].