Ms. Jones wrote the expression below on the chalkboard for her class. 4(3x + 5y + 2z) + 3(x – z)

She asked the students to write an equivalent expression using no more than one set of parentheses.

Tom wrote 12x + 20y + 8z
Jenna wrote 15x + 20y +5z
Chris wrote 40xyz

Which, if any, of the three students wrote an expression that is equivalent to Ms. Jones’ expression? Justify by showing your work!

12x + 20y + 8z + 3(x -z)

12x + 20y + 8z + 3x - 3z

15x + 20y + 5z

15x + 20y + 5z

the answer is:15x + 20y + 5z

(w – 2.5)(w + 2.5)

(–4v – 9)(–4v + 9)

step-by-step explanation:

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!