Answer:
[tex]6y-x=-30[/tex]
[tex]5y-x=10[/tex].
Step-by-step explanation:
Given
[tex]\frac{1}{6}x-5=\frac{1}{5}x+2[/tex]
Let, [tex]\frac{1}{6}x-5=y[/tex]
[tex]\frac{x-30}{6}=y[/tex]
By cross multiply we get
[tex]x-30=6y[/tex]
By subtraction property of equality
[tex]-30=6y-x[/tex]
[tex]6y-x=-30[/tex]
Let ,[tex]\frac{1}{5}x+2=y[/tex]
[tex]\frac{x+10}{5}=y[/tex]
By cross multiply we get
[tex]x+10=5y[/tex]
By subtraction property of equality we get
[tex]5y-x=10[/tex]
Hence, [tex]6y-x=-30, 5y-x=10[/tex].
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 = -42*5 = -210
And this is the value of x we wanted to find.
