Answer
$(3x+1)^o=40^o
,\\
(7x+49)^o=140^o$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Equate the given supplementary angles to $180$, and use the properties of equality to isolate $x.$ Then substitute the value of $x$ in the following given supplementary angles:
\begin{array}{l}\require{cancel}
(3x+1)^o
,\\
(7x+49)^o
.\end{array}
$\bf{\text{Solution Details:}}$
Since the sum of supplementary angles is $180^o,$ then
\begin{array}{l}\require{cancel}
(3x+1)+(7x+49)=180
.\end{array}
Using the properties of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
3x+7x=180-1-49
\\\\
10x=130
\\\\
x=\dfrac{130}{10}
\\\\
x=13
.\end{array}
Substituting $x=
13
$ in the angle $
(3x+1)^o
$ results to
\begin{array}{l}\require{cancel}
(3\cdot13+1)^o
\\\\=
(39+1)^o
\\\\=
40^o
.\end{array}
Substituting $x=
13
$ in the angle $
(7x+49)^o
$ results to
\begin{array}{l}\require{cancel}
(7\cdot13+49)^o
\\\\=
(91+49)^o
\\\\=
140^o
.\end{array}
Hence, the measures of the supplementary angles are
\begin{array}{l}\require{cancel}
(3x+1)^o=40^o
,\\
(7x+49)^o=140^o
.\end{array}
