Question 280764
Using the substitution method {x/8-y/5=1, x/6-y=5}


{{{x/8 - y/5 = 1}}}
{{{x/6 - y = 5}}}


I would first get rid of the denominators.  For the first equation, multiply everything by 40 (which I found by multiplying the denominator 8 by the denominator 5)



40{{{(x/8 - y/5)}}} = 40{{{(1)}}}  
5x - 8y = 40  This is the new first equation.  


I would get rid of the denominator 6 in the 2nd equation by multiplying everything by 6. 


6{{{(x/6 - y)}}} =6{{{(5)}}}
x -6y = 30  This is the new second equation.


Here are the two equations now:



5x - 8y = 40
x -6y = 30 <----- now let's take this equation and solve for "x"


x - 6y = 30
x = 6y + 30 (added 6y to both sides to isolate the x)
Now that we know that "x" is equal to 6y + 30, let's "plug" that into the first equation.  


5x - 8y = 40  First equation
5(6y + 30) -8y =40 Plugged in 6y + 30 for the "x" variable.
30y + 150 - 8y =40 Distributed 5 to 6y and 5 to 30.
22y + 150 = 40 Combined like terms:  30y - 8y = 22y
22y = 40 - 150 Subtracted 150 from both sides to begin to isolate the y
22y = -110 Found -110 by determining 40 - 150.
y = -5 Divided both sides by 22 to isolate the y.  -110 divided by +22 equals -5.


Now we know that y = -5.  Let's plug that into our 2nd equation.


x -6y = 30   Second equation
x -6(-5) = 30 Plugged -5 in for the "y" variable.
x + 30= 30  Determined 30 because -6 times -5 = 30
x = 30 - 30 Subtracted 30 from both sides to isolate the "x"
x = 0.


Now we have x = 0 and we have y = -5.


Let's check:


Original equations:


{{{x/8 - y/5 = 1}}} First original equation

{{{0/8 - (-5)/5 = 1}}}  Plug in the 0 for the X variable and -5 for the y variable.
{{{0 + 5/5 = 1}}}  
{{{5/5 = 1}}}
{{{1 = 1}}} This is correct.  Yay.



{{{x/6 - y = 5}}}  Second original equation
{{{0/6 - (-5) = 5}}} Plug in 0 for the x variable and -5 for the y variable.
{{{0 + 5 = 5}}}
{{{5 = 5}}} This is also correct.  Yay.


We have proved the answers.  I hope this helps you. :-)