[02.03]
compare and contrast: two equations are listed below. solve each equation and
compare the solutions. choose the statement that is true about both solutions.
equation 1 equation 2
15x + 6 = 41 2x + 13 = 28
(2 points)
1) equation 1 has more solutions than equation 2
2) equation 1 and equation 2 have the same number of solutions.
3) equation 2 has more solutions than equation 1.
4) the number of solutions cannot be determined.

2)  Equation 1 and Equation 2 have the same number of solutions.

Step-by-step explanation:

The two given equations are

1)  15x + 6 = 41                and                   2) 2x + 13 = 28

Solving both equations, we get

Solving (1) :   15x + 6 = 41 ⇒   15x  = 41 -  6 = 35

or, x = 35/15 ⇒ x = 7/3

Solving (2) :   2x + 13 = 28⇒   2x  = 28 -  13  = 15

or, x = 15/2 ⇒ x = 15/2

So, from above solutions we can say that Equation 1 and Equation 2 have the same number of UNIQUE solution.

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!