# Given the two functions f(x)=2x-1 and g(x)=-4/3x+9 how do you determine the point that lies in the solution set for both f(x) and g(x)?

Jan 7, 2017

Two find the point in the solution set for both these functions equation the two functions ($f \left(x\right) = g \left(x\right)$) and solve for $x$. See full process below.

#### Explanation:

Two find the point in the solution set for both these functions equation the two functions ($f \left(x\right) = g \left(x\right)$) and solve for $x$.

Therefore:

$2 x - 1 = - \frac{4}{3} x + 9$

We can now solve for $x$:

$2 x - 1 + \textcolor{red}{\frac{4}{3} x} + \textcolor{b l u e}{1} = - \frac{4}{3} x + 9 + \textcolor{red}{\frac{4}{3} x} + \textcolor{b l u e}{1}$

$2 x + \textcolor{red}{\frac{4}{3} x} - 1 + \textcolor{b l u e}{1} = - \frac{4}{3} x + \textcolor{red}{\frac{4}{3} x} + 9 + \textcolor{b l u e}{1}$

$2 x + \textcolor{red}{\frac{4}{3} x} - 0 = 0 + 9 + \textcolor{b l u e}{1}$

$2 x + \textcolor{red}{\frac{4}{3} x} = 10$

$\left(\frac{3}{3} \times 2\right) x + \textcolor{red}{\frac{4}{3} x} = 10$

$\frac{6}{3} x + \textcolor{red}{\frac{4}{3} x} = 10$

$\frac{10}{3} x = 10$

$\frac{\textcolor{red}{3}}{\textcolor{b l u e}{10}} \times \frac{10}{3} x = \frac{\textcolor{red}{3}}{\textcolor{b l u e}{10}} \times 10$

cancel(color(red)(3))/cancel(color(blue)(10)) xx color(blue)(cancel(color(black)(10)))/color(red)cancel(color(black)(3))x = color(red)(3)/cancel(color(blue)(10)) xx color(blue)(cancel(color(black)(10)

$x = 3$ is the point which lies in the solution set for both functions.