Mr. Martin’s math test, which is worth 100 points, has 35 problems. Each problem is worth either 5 points or 2 points. Let x = the number of questions worth 5 points. Let y = the number of questions worth 2 points. x + y = 35, 5x + 2y = 100 How many problems of each point value are on the test? 10 problems worth 5 points and 25 problems worth 2 points 15 problems worth 5 points and 20 problems worth 2 points 20 problems worth 5 points and 15 problems worth 2 points 25 problems worth 5 points and 10 problems worth 2 points

10 problems worth 5 points and 25 problems worth 2 points

Step-by-step explanation:

Let x = the number of questions worth 5 points.

Let y = the number of questions worth 2 points

x+y = 35  since there are 35 problems

5x + 2y = 100  since there are 100 points

Multiply the first equation by -5 so we can eliminate x

-5(x+y) = 35*-5

-5x -5y =-175

Add this to the second equation

-5x -5y =-175

5x + 2y = 100

    -3y = -75

Divide each side by -3

-3y/-3 = -75/-3

y = 25

There are 25 2 point questions

Now find x

x+y = 35

x+25 = 35

x = 35-25

x =10

There are 10 5 point questions

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!