Astudent is trying to solve the system of two equations given below:

equation p: y + z = 6
equation q: 8y + 7z = 1

which of the following is a possible step used in eliminating the y-term?
a (y + z = 6) ⋅ −8
b (y + z = 6) ⋅ 7
c (8y + 7z = 1) ⋅ 7
d (8y + 7z = 1) ⋅ 8

Option A

The possible step used in eliminating the y-term in Equation P: y + z = 6  Equation Q: 8y + 7z = 1 is (y + z = 6) ⋅ −8

SOLUTION:

Given that, A student is trying to solve the system of two equations given below:  

Equation P: y + z = 6  

Equation Q: 8y + 7z = 1  

we have to find which of the given steps is a possible step used in eliminating the y-term?

We know that, while solving the system of equations, if we want to eliminate any one of the variables then, co efficient of that variable must be same in both equations but should differ by sign.

Now, let us check it.

Here coefficient of y in P is less than in Q, so let us multiply P with coefficient of y in Q and with opposite sign.

Then, (y + z = 6) . -8

Hence, the correct option is A.

im alright what do you need

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!