Question

PLEASE HELP!!!How can one thirdx − 2 = one fourthx + 11 be set up as a system of equations?3y − x = −64y − x = 443y + x = −64y + x = 443y − 3x = −64y − 4x = 443y + 3x = −64y + 4x = 44

Answer 1

3y - x = -6

4y - x = 44

Step-by-step explanation:

The given equation is

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

We can say that these expressions are equal because both are equal to y, so we have the following equations

`\begin{gathered} y=(1)/(3)x-2 \\ \\ y=(1)/(4)x+11 \end{gathered}`

Now, we can rewrite the first equation as

`\begin{gathered} y=(1)/(3)x-2 \\ \\ 3y=3((1)/(3)x)-3(2) \\ \\ 3y=x-6 \\ 3y-x=x-6-x \\ 3y-x=-6 \end{gathered}`

In the same way, we can rewrite the second equation as

`\begin{gathered} y=(1)/(4)x+11 \\ \\ 4y=4((1)/(4)x)+4(11) \\ \\ 4y=x+44 \\ 4y-x=x+44-x \\ 4y-x=44 \end{gathered}`

Therefore, the system of equations is

3y - x = -6

4y - x = 44