(x/2)+(y/3)=-1
(x/2)+2y=4
Can someone please explain how to get the solutions set. What do I start with? Where do the numbers come from and why?
Please and Thank you!
+23254 First, I would get rid of fractions:
--- multiply the first equation by 6: 6(x/2) + 6(y/3) = 6(-1) ---> 3x + 2y = -6
--- multiply the second equation by 2: 2(x/2) + 2(2y) = 2(4) ---> x + 4y = 8
Now, combine these: 3x + 2y = 6
x + 4y = 8
If you want to eliminate the x-terms, leave the top equation alone and multiply the bottom equation by -3:
3x + 2y = -6 ---> 3x + 2y = -6
x + 4y = 8 ---> multiply by -3 ---> -3x - 12y = -24
Add down the columns: -10y = -30
y = 3
Substitue this value back into an original equation: x + 4y = 8 ---> x + 4(3) = 8 ---> x + 12 = 8
---> x = -4
+23254 First, I would get rid of fractions:
--- multiply the first equation by 6: 6(x/2) + 6(y/3) = 6(-1) ---> 3x + 2y = -6
--- multiply the second equation by 2: 2(x/2) + 2(2y) = 2(4) ---> x + 4y = 8
Now, combine these: 3x + 2y = 6
x + 4y = 8
If you want to eliminate the x-terms, leave the top equation alone and multiply the bottom equation by -3:
3x + 2y = -6 ---> 3x + 2y = -6
x + 4y = 8 ---> multiply by -3 ---> -3x - 12y = -24
Add down the columns: -10y = -30
y = 3
Substitue this value back into an original equation: x + 4y = 8 ---> x + 4(3) = 8 ---> x + 12 = 8
---> x = -4
