Question

which input value produces the same output value for the two functions on the graph?
$f(x) = -\frac{2}{3}x + 1$
$g(x) = \frac{1}{3}x - 2$
graph of two lines, f(x) (red) and g(x) (blue), intersecting at x=3
options:
$x = -1$
$x = 1$
$x = 3$
$x = -3$

Explanation:

Step1: Set the two functions equal

To find the input value \( x \) where \( f(x) = g(x) \), we set \( -\frac{2}{3}x + 1=\frac{1}{3}x - 2 \).

Step2: Solve for \( x \)

First, add \( \frac{2}{3}x \) to both sides: \( 1=\frac{1}{3}x+\frac{2}{3}x - 2 \).
Simplify the right - hand side: \( 1 = x-2 \).
Then, add 2 to both sides: \( x=1 + 2=3 \).

We can also verify by plugging in the values:

• For \( x = - 1 \):

\( f(-1)=-\frac{2}{3}\times(-1)+1=\frac{2}{3}+1=\frac{5}{3} \), \( g(-1)=\frac{1}{3}\times(-1)-2=-\frac{1}{3}-2=-\frac{7}{3} \), \( f(-1)
eq g(-1) \).

• For \( x = 1 \):

\( f(1)=-\frac{2}{3}\times1 + 1=\frac{1}{3} \), \( g(1)=\frac{1}{3}\times1-2=-\frac{5}{3} \), \( f(1)
eq g(1) \).

• For \( x = 3 \):

\( f(3)=-\frac{2}{3}\times3+1=-2 + 1=-1 \), \( g(3)=\frac{1}{3}\times3-2=1-2=-1 \), \( f(3)=g(3) \).

• For \( x=-3 \):

\( f(-3)=-\frac{2}{3}\times(-3)+1 = 2 + 1=3 \), \( g(-3)=\frac{1}{3}\times(-3)-2=-1-2=-3 \), \( f(-3)
eq g(-3) \).

Answer:

\( x = 3 \) (the option with text "x = 3")