2^-2 would become 1/2^2 which becomes 1/4.
Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8
The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.
After doing this, we realize that only -1 works, so the answer is x=-1
Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)
Best wishes:)
Answer:
Step-by-step explanation:
2x = 3y + 1/2
standard : Ax + By = C
2x = 3y + 1/2....multiply everything by 2 to get rid of the fractions
4x = 6y + 1 ....subtract 6y from both sides
4x - 6y = 1 <==
slope intercept : y = mx + b
2x = 3y + 1/2....subtract 1/2 from both sides
2x - 1/2 = 3y....divide everything by 3
2/3x - 1/6 = y...rearrange
y = 2/3x - 1/6 <===
