SOLUTION: Solve the system by the addition method
x/3-y/2=-5/6
x/5-y/3=-3/5
Algebra.Com
Question 4841: Solve the system by the addition method
x/3-y/2=-5/6
x/5-y/3=-3/5
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let's start by clearing those fractions. The LCD for the first equation is 6, for the second equation is 15, so multiply both sides of these equations by those numbers respectively:
= First Equation
= Second Equation
= First Equation
= Second Equation
2x - 3y = -5 First Equation
3x - 5y = -9 Second Equation
Now, eliminate either the x in these equations by getting a common number of 6 for the x coefficients. To do this, you must multiply the first equation by 3 and the second equation by -2. It looks like this:
3(2x - 3y ) = 3*(-5)
-2(3x - 5y ) = -2*(-9)
6x - 9y = -15
-6x + 10y = 18
Add the equations together, and you get:
y = 3
Substitute y = 3 in the first equation:
2x - 3y = -5
2x -3(3) = -5
2x - 9 = -5 Add +9 to each side of the equation:
2x - 9 + 9 = -5 + 9
2x = 4
x= 2
Check it by substituting both x and y in the second equation (NOTE: To be a valid check, you must substitute into the OTHER equation, not the one you just used!!)
3x - 5y = -9
3(2) - 5(3) = -9
6 - 15 = -9
Praise the Lord! It checks!!!
R^2 at SCC
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