Since one equation has a negative y and the other has a positive y, I'm going to use those since they cancel each other out. Before that, the two y's need to be equal to each other.
x+2y=6
x-y=3
Multiply the bottom equation by two so then you have:
x+2y=6
2x-2y=6
The y's now cancel out:
x=6
2x=6
Add them together
3x=12
Divide
x=4.
To find y, plug x into either equation (*don't have to do both, but I will)
(4)+2y=6
(4)-y=3
Subtract four
2y=2
-y=-1
Divide each
2y/2 = 2/2
y=1
-y/-1 = -1/-1
y=1
The answer is:
x=4
y=1
I hope that helps!
You could rewrite
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as

and be tempted to cancel out the factors of

. But this cancellation is only valid when

.
When

, you end up with the indeterminate form
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, which is why

is not a zero.
Answer:
0.5<
Step-by-step explanation:
5/10<5+65/10+100<65/100
=5/10<5+65/10+100<65/100
=1/2<1+13/2+20<13/20
=1/2<14/22<
I perfer the the gx are square root of the output