Sagot :
[tex]──────────────────────┈━ \\ \\
\huge\tt{ \green{ q} \orange{u} \red{e} \blue{s} \pink{t} \purple{i} \green{o} \red{ n} \red{} \blue{} \pink{} \purple{} \green{} \orange{} \purple{} }[/tex]
. What is the equation of the line in standard form given slope = 3 and b = 2?
[tex]\huge\mathbb{✏Answer:}[/tex]
[tex]The standard form is
2
x
−
3
y
=
−
12
Explanation:
Start by finding the slope-intercept form of the equation, then converting that to the standard form.
The slope-intercept form is
y
=
m
x
+
b
The slope
m
is given as
2
3
, so the equation up to that point is
y
=
2
3
x
+
b
To find
b
, sub in the values for
x
and
y
from the ordered pair given in the problem.
6
=
2
3
(
3
1
)
+
b
Solve for
b
1) Clear the parentheses by multiplying the fractions
6
=
2
+
b
2) Subtract
2
from both sides to isolate
b
4
=
b
So the slope-intercept form of the equation is
y
=
(
2
3
)
x
+
4
←
slope-intercept form
Change the slope-intercept form into standard form.
Standard form is
a
x
+
b
y
=
c
where
a
is a positive whole digit
1) Clear the fraction by multiplying all the terms on both sides by
3
and letting the denominator cancel
3
y
=
2
x
+
12
2) Subtract
2
x
from both sides to get the
x
and
y
terms on the same side
−
2
x
+
3
y
=
12
3) Multiply through by
−
1
to clear the minus sign
2
x
−
3
y
=
−
12
←
standard form[/tex]
Answer:
The standard form is
2x−3y=−12
Step-by-step explanation:
Start by finding the slope-intercept form of the equation, then converting that to the standard form.
The slope-intercept form is
y
=
m
x
+
b
The slope
m
is given as
2
3
, so the equation up to that point is
y
=
2
3
x
+
b
To find
b
, sub in the values for
x
and
y
from the ordered pair given in the problem.
6
=
2
3
(
3
1
)
+
b
Solve for
b
1) Clear the parentheses by multiplying the fractions
6
=
2
+
b
2) Subtract
2
from both sides to isolate
b
4
=
b
So the slope-intercept form of the equation is
y
=
(
2
3
)
x
+
4
←
slope-intercept form
Change the slope-intercept form into standard form.
Standard form is
ax+by=c where a is a positive whole digit
1) Clear the fraction by multiplying all the terms on both sides by
3
and letting the denominator cancel
3
y
=
2
x
+
12
2) Subtract
2
x
from both sides to get the
x
and
y
terms on the same side
−
2
x
+
3
y
=
12
3) Multiply through by
−
1
to clear the minus sign
2
x
−
3
y
=
−
12
←
standard form