The equations 4 x minus 5 y = 5, 3 x + 10 y = 20, 4 x minus 5 y = negative 5, and 3 x + 10 y = negative 20 are shown on the graph below. On a coordinate plane, there are 4 lines. Green line goes through (0, 1) and (2.5, 3). Orange line goes through (0, negative 1) and (2.5, 1). Blue line goes through (0, 2) and (3.5, 1). Pink line goes through (0, negative 2) and (3.5, negative 3). Which system of equations has a solution of approximately (–2.7, –1.2)? 4 x minus 5 y = negative 5 and 3 x + 10 y = negative 20 4 x minus 5 y = 5 and 3 x + 10 y = 20 4 x minus 5 y = 5 and 3 x + 10 y = negative 20 4 x minus 5 y = negative 5 and 3 x + 10 y = 20

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!