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1.
Here we only have
x
{\displaystyle x}
x
+
x
=
2
x
{\displaystyle x+x=2x}
2.
We have a similar situation to problem 1, but this time with
x
2
{\displaystyle x^{2}}
x
2
+
3
x
2
=
(
1
+
3
)
x
2
=
4
x
2
{\displaystyle x^{2}+3x^{2}=(1+3)x^{2}=4x^{2}}
3.
Here we have three different terms in the mix (
x
{\displaystyle x}
x
2
{\displaystyle x^{2}}
x
3
{\displaystyle x^{3}}
x
{\displaystyle x}
x
2
{\displaystyle x^{2}}
x
3
{\displaystyle x^{3}}
(
3
x
+
2
x
)
+
(
2
x
2
−
2
x
2
)
+
3
x
3
{\displaystyle (3x+2x)+(2x^{2}-2x^{2})+3x^{3}}
Now combine the coefficients of the like terms:
(
3
+
2
)
x
+
(
2
−
2
)
x
2
+
3
x
3
=
5
x
+
3
x
3
{\displaystyle (3+2)x+(2-2)x^{2}+3x^{3}=5x+3x^{3}}
4.
This is another case with three distinct terms (
z
y
{\displaystyle zy}
z
{\displaystyle z}
y
{\displaystyle y}
z
y
+
2
z
y
+
2
z
+
2
y
=
(
1
+
2
)
z
y
+
2
z
+
2
y
=
3
z
y
+
2
z
+
2
y
{\displaystyle {\begin{aligned}zy+2zy+2z+2y&=(1+2)zy+2z+2y\\&=3zy+2z+2y\end{aligned}}}
−
x
2
+
4
x
2
y
−
4
x
y
2
+
7
x
y
{\displaystyle -x^{2}+4x^{2}y-4xy^{2}+7xy}
4
x
2
{\displaystyle 4x^{2}}
18
x
2
y
2
{\displaystyle 18x^{2}y^{2}}
36
a
b
2
x
z
{\displaystyle 36ab^{2}xz}
60
x
5
y
4
z
2
{\displaystyle 60x^{5}y^{4}z^{2}}
x
5
{\displaystyle {\sqrt {x^{5}}}}
x
{\displaystyle x}
7
x
12
{\displaystyle {\frac {7x}{12}}}
16
x
y
15
{\displaystyle {\frac {16xy}{15}}}
2
x
−
y
+
z
{\displaystyle 2x-y+z}
x
2
+
y
2
x
y
{\displaystyle {\frac {x^{2}+y^{2}}{xy}}}
Solve for x.
x
=
y
2
{\displaystyle x={\frac {y}{2}}}
Solve for z.
z
=
x
−
8
3
{\displaystyle z={\frac {x-8}{3}}}
Solve for y.
y
=
b
2
{\displaystyle y=b^{2}}
Solve for x.
x
=
y
+
9
{\displaystyle x={\sqrt {y+9}}}
Solve for b.
b
=
6
y
+
7
z
6
{\displaystyle b={\frac {6y+7z}{6}}}
Find the Roots of:
x
=
−
2
o
r
3
{\displaystyle x=-2\ or\ 3}
x
=
7
o
r
3
2
{\displaystyle x=7\ or\ {\frac {3}{2}}}
x
=
2
o
r
3
{\displaystyle x=2\ or\ 3}
x
=
−
1
o
r
0
{\displaystyle x=-1\ or\ 0}
x
=
−
3
o
r
4
{\displaystyle x=-3\ or\ 4}
a
=
4
,
s
=
2
{\displaystyle a=4,s=2}
c
=
2
5
,
d
=
1
{\displaystyle c={\frac {2}{5}},d=1}
b
=
7
,
g
=
12
{\displaystyle b=7,g=12}
