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How can one sixthx _ 5 = one fifthx + 2 be set up as a system of equations?

6y _ x = _30

5y _ x = 10


5y + x = _30

5y + x = 10


6y _ 6x = _30

5y _ 5x = 10


5y + 5x = _30

5y + 5x = 10

Verified Answer :

lublana

Answer:

\(6y-x=-30\)

\(5y-x=10\).

Step-by-step explanation:

Given

\(\frac{1}{6}x-5=\frac{1}{5}x+2\)

Let, \(\frac{1}{6}x-5=y\)

\(\frac{x-30}{6}=y\)

By cross multiply we get

\(x-30=6y\)

By subtraction property of equality

\(-30=6y-x\)

\(6y-x=-30\)

Let ,\(\frac{1}{5}x+2=y\)

\(\frac{x+10}{5}=y\)

By cross multiply we get

\(x+10=5y\)

By subtraction property of equality we get

\(5y-x=10\)

Hence, \(6y-x=-30, 5y-x=10\).

Answer:

6y - x = -30

5y - x = 10

Step-by-step explanation:

one sixth x - 5 = one fifth x  + 2 is equal to (1/6)x - 5 = (1/5)x + 2. Isolating x, we get:

(1/6)x - 5 = (1/5)x + 2

(1/6)x - (1/5)x = 2 + 5

(-1/30)x = 7

x = -210

Now we want to isolate x from the next system of equations

6y - x = -30  (eq. 1)

5y - x = 10 (eq.  2)

To do so, we follow the next steps. First multiply eq. 2 by 6/5

(6/5)(5y - x) = (6/5)10

6y - (6/5)x = 12 (eq. 3)

Next, subtract eq. 3 to eq. 1

6y - x = -30

-

6y - (6/5)x = 12

----------------------

- x + (6/5)x = -30 - 12

(1/5)x = -42

x = -425 = -210

And this is the value of x we wanted to find.

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