Algebra Examples

Solve for c f=9/5c+32
f=95c+32
Step 1
Rewrite the equation as 95c+32=f.
95c+32=f
Step 2
Combine 95 and c.
9c5+32=f
Step 3
Subtract 32 from both sides of the equation.
9c5=f-32
Step 4
Multiply both sides of the equation by 59.
599c5=59(f-32)
Step 5
Simplify both sides of the equation.
Tap for more steps...
Step 5.1
Simplify the left side.
Tap for more steps...
Step 5.1.1
Simplify 599c5.
Tap for more steps...
Step 5.1.1.1
Cancel the common factor of 5.
Tap for more steps...
Step 5.1.1.1.1
Cancel the common factor.
599c5=59(f-32)
Step 5.1.1.1.2
Rewrite the expression.
19(9c)=59(f-32)
19(9c)=59(f-32)
Step 5.1.1.2
Cancel the common factor of 9.
Tap for more steps...
Step 5.1.1.2.1
Factor 9 out of 9c.
19(9(c))=59(f-32)
Step 5.1.1.2.2
Cancel the common factor.
19(9c)=59(f-32)
Step 5.1.1.2.3
Rewrite the expression.
c=59(f-32)
c=59(f-32)
c=59(f-32)
c=59(f-32)
Step 5.2
Simplify the right side.
Tap for more steps...
Step 5.2.1
Simplify 59(f-32).
Tap for more steps...
Step 5.2.1.1
Apply the distributive property.
c=59f+59-32
Step 5.2.1.2
Combine 59 and f.
c=5f9+59-32
Step 5.2.1.3
Multiply 59-32.
Tap for more steps...
Step 5.2.1.3.1
Combine 59 and -32.
c=5f9+5-329
Step 5.2.1.3.2
Multiply 5 by -32.
c=5f9+-1609
c=5f9+-1609
Step 5.2.1.4
Move the negative in front of the fraction.
c=5f9-1609
c=5f9-1609
c=5f9-1609
c=5f9-1609
f=95c+32
(
(
)
)
|
|
[
[
]
]
7
7
8
8
9
9
4
4
5
5
6
6
/
/
^
^
×
×
>
>
1
1
2
2
3
3
-
-
+
+
÷
÷
<
<
π
π
,
,
0
0
.
.
%
%
=
=
 [x2  12  π  xdx ]