Commenter jdoe said it right: solve for y and leave the rest on the other side.
-x + 3y = 6
3y = 6 + x add x on both sides
3y = x + 6 rearrange to get the x first
y = (x + 6) /3 divide both sides by 3
y = x/3 + 6/3 split the numerator (caution - never split denominators)
y = x/3 + 2 simplify 6/3
Thus the line in slope intercept form of y = mx + b is y = 
Answer:
y
=
2
x
−
1
Explanation:
First, we need to determine the slope of the line. The formula for determining the slope of a line is:
m
=
y
2
−
y
1
x
2
−
x
1
where
m
is the slope and the x and y terms are for the points:
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
For this problem the slope is:
m
=
3
−
−
1
2
−
0
m
=
3
+
1
2
m
=
4
2
m
=
2
Now, selecting one of the points we can use the point slope formula to find the equation.
The point slope formula is:
y
−
y
1
=
m
(
x
−
x
1
)
Substituting one of our points gives:
y
−
−
1
=
2
(
x
−
0
)
y
+
1
=
2
x
Solving for
y
to put this in standard form gives:
y
+
1
−
1
=
2
x
−
1
y
+
0
=
2
x
−
1
y
=
2
x
−
1
Answer linky
=
2
x
−
1
Explanation:
First, we need to determine the slope of the line. The formula for determining the slope of a line is:
m
=
y
2
−
y
1
x
2
−
x
1
where
m
is the slope and the x and y terms are for the points:
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
For this problem the slope is:
m
=
3
−
−
1
2
−
0
m
=
3
+
1
2
m
=
4
2
m
=
2
Now, selecting one of the points we can use the point slope formula to find the equation.
The point slope formula is:
y
−
y
1
=
m
(
x
−
x
1
)
Substituting one of our points gives:
y
−
−
1
=
2
(
x
−
0
)
y
+
1
=
2
x
Solving for
y
to put this in standard form gives:
y
+
1
−
1
=
2
x
−
1
y
+
0
=
2
x
−
1
y
=
2
x
−
1
Answer link
Simplify (we cannot solve if there is no equals)
6s^3(5s^2)(3s^4)
multiply the coefficients, add the exponents since the bases are the same
(6*5*3) s^(3+2+4)
90 s^(9)
Answer:
Express 15 2/3 in simplest radical form 1
Solution :
We have been given a parametric curve :
x = sin t , y = cos t , 0 < t < π
In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.
Therefore,
Taking double derivatives of the above equation:
For the concave up, we have
∴ 
For the concave down, we have
