Question

How can one third x − 2 = one fourth x + 11 be set up as a system of equations?

Answer 1

Answer: (x-2) = (x+11)

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3 4

Step-by-step explanation:

First part is x-2 divided by on third. So the x-2 acts a the 1 so you put x-2 over 3. And you get:

(x-2)

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3

For the other side of the equation it is x +11 divide by 1/4. So the x+11 will act a the 1 in the fraction. And you will get:

(x+11)

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4

Answer 2

For this case we have the following expression:

$\frac {1} {3} (x-2) = \frac {1} {4} (x + 11)$

So, we have two equations of the form:$y = \frac {1} {3} (x-2)\\y = \frac {1} {4} (x + 11)$

Rewriting equation 1:

$y = \frac {1} {3} (x-2)\\3y = x-2\\3y-x = -2$

If we multiply by 3 on both sides of the equation we have:

$9y-3x = -6$

Rewriting equation 2:

$y = \frac {1} {4} (x + 11)\\4y = x + 11\\4y-x = 11$

If we multiply by 4 on both sides of the equation:

$16y-4x = 44$

Answer:

$9y-3x = -6\\16y-4x = 44$