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Solving Literal Equations
Choose Course
Algebra 1
One-Variable Equations
Solving Literal Equations
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Solving Literal Equations 1.11 - Solution
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Return to Solving Literal Equations
a
Solving the formula for
m
m
m
with
inverse operations
.
d
=
m
V
d=\dfrac{m}{V}
d
=
V
m
MultEqn
LHS
⋅
V
=
RHS
⋅
V
\text{LHS} \cdot V=\text{RHS}\cdot V
LHS
⋅
V
=
RHS
⋅
V
d
V
=
m
dV=m
d
V
=
m
RearrangeEqn
Rearrange equation
m
=
d
V
m=dV
m
=
d
V
b
The density
d
d
d
is
5.23
5.23
5
.
2
3
g
/
cm
3
\text{g}/\text{cm}^3
g
/
cm
3
and the volume is
V
=
1.1
V= 1.1
V
=
1
.
1
cm
3
\text{cm}^3
cm
3
. In order to find the mass,
m
,
m,
m
,
we substitute these values into our rewritten equation.
m
=
d
V
m=dV
m
=
d
V
SubstituteII
d
=
5.23
d={\color{#0000FF}{5.23}}
d
=
5
.
2
3
,
V
=
1.1
V={\color{#009600}{1.1}}
V
=
1
.
1
m
=
5.23
⋅
1.1
m={\color{#0000FF}{5.23}} \cdot {\color{#009600}{1.1}}
m
=
5
.
2
3
⋅
1
.
1
Multiply
Multiply
m
=
5.753
m=5.753
m
=
5
.
7
5
3
RoundDec
Round to
2
decimals
{\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}
m
≈
5.75
m\approx5.75
m
≈
5
.
7
5
Therefore, the mass is approximately
5.75
5.75
5
.
7
5
g.