Question

How can one thirdx − 2 = one fourthx + 11 be set up as a system of equations?

Answer 1

Answer:

3y-x= -6

4y-x=44 ....

Step-by-step explanation:

Let y= 1/3x-2

Let y= 1/4x+11

Now we are required to arrange them in standard form.

So,

y= 1/3x-2

Combine the variable terms:

-1/3x+y=-2

Multiply both sides by 3

3(-1/3x+y)=3* -2

-x+3y= -6

3y-x= -6 ---------equation 1

y= 1/4x+11

Combine the variable terms:

-1/4x+y = 11

Multiply both sides by 4

4(-1/4x+y) = 4*11

-x+4y=44

4y-x=44 -----------------equation 2

Therefore the system of equations is:

3y-x= -6

4y-x=44 ....

Answer 2

Answer: Hello there!

we have that "one thirdx − 2 = one fourthx + 11"

this means  (1/3)x - 2 = (1/4)x + 11

now, we also can write this as:

(1/3)x - 2 = y = (1/4)x + 11

and now we have a system of equations:

(1/3)x - 2 = y

(1/4)x + 11 = y

or

(1/4)x - y = -11

(1/3)x - y = 2