# SOLUTION: Using the substitution method {x/8-y/5=1, x/6-y=5}

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 Question 280764: Using the substitution method {x/8-y/5=1, x/6-y=5}Answer by PRMath(123)   (Show Source): You can put this solution on YOUR website!Using the substitution method {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 = 40 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 =6 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: First original equation Plug in the 0 for the X variable and -5 for the y variable. This is correct. Yay. Second original equation Plug in 0 for the x variable and -5 for the y variable. This is also correct. Yay. We have proved the answers. I hope this helps you. :-)