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Math Problems
Grade 8
Solve multi-step equations with fractional coefficients
Solve. Show your work.
\newline
1
4
(
12
−
8
x
)
=
4
(
−
2
x
−
6
)
\frac{1}{4}(12-8 x)=4(-2 x-6)
4
1
(
12
−
8
x
)
=
4
(
−
2
x
−
6
)
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x
(
1
x
x
3
)
−
x
x
−
2
x
=
0
I
x(\frac{1}{x\sqrt{x^3}})-x^x-\frac{2}{x}=0\ I
x
(
x
x
3
1
)
−
x
x
−
x
2
=
0
I
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Solve for
b
b
b
.
\newline
b
+
1
=
1
4
(
3
b
+
2
)
b + 1 = \frac{1}{4}(3b + 2)
b
+
1
=
4
1
(
3
b
+
2
)
\newline
b
=
b =
b
=
__
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Solve for
a
a
a
.
\newline
3
(
2
3
a
−
1
3
)
=
5
a
+
3
3\left(\frac{2}{3}a - \frac{1}{3}\right) = 5a + 3
3
(
3
2
a
−
3
1
)
=
5
a
+
3
\newline
a = ____
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Solve for
j
j
j
.
\newline
5
(
4
5
j
+
2
)
=
3
−
6
j
5\left(\frac{4}{5}j + 2\right) = 3 - 6j
5
(
5
4
j
+
2
)
=
3
−
6
j
\newline
j
=
j =
j
=
__
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Solve for
x
x
x
.
\newline
2
−
x
=
2
(
3
−
1
4
x
)
2 - x = 2(3 - \frac{1}{4}x)
2
−
x
=
2
(
3
−
4
1
x
)
\newline
x
=
x =
x
=
__
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Solve for
c
c
c
.
\newline
1
2
c
=
2
(
1
2
c
−
3
)
\frac{1}{2}c = 2\left(\frac{1}{2}c - 3\right)
2
1
c
=
2
(
2
1
c
−
3
)
\newline
c = ____
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Solve for
p
p
p
.
\newline
5
p
=
4
(
1
2
p
−
1
)
5p = 4\left(\frac{1}{2}p - 1\right)
5
p
=
4
(
2
1
p
−
1
)
\newline
p
=
p =
p
=
__
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Solve for
m
m
m
.
\newline
4
m
+
2
=
4
(
1
2
m
−
1
4
)
4m + 2 = 4\left(\frac{1}{2}m - \frac{1}{4}\right)
4
m
+
2
=
4
(
2
1
m
−
4
1
)
\newline
m
=
m =
m
=
__
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Solve for
t
t
t
.
\newline
3
(
1
3
t
+
2
3
)
=
4
t
−
2
3\left(\frac{1}{3}t + \frac{2}{3}\right) = 4t - 2
3
(
3
1
t
+
3
2
)
=
4
t
−
2
\newline
t
=
t =
t
=
__
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Solve for
q
q
q
.
\newline
3
(
1
4
q
+
1
)
=
q
3\left(\frac{1}{4}q + 1\right) = q
3
(
4
1
q
+
1
)
=
q
\newline
q = ____
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Solve for
k
k
k
.
\newline
8
(
3
4
k
+
3
)
=
8
−
2
k
8\left(\frac{3}{4}k + 3\right) = 8 - 2k
8
(
4
3
k
+
3
)
=
8
−
2
k
\newline
k
=
k =
k
=
__
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Solve for
m
m
m
.
\newline
1
4
(
8
m
+
24
)
=
21
−
13
m
\frac{1}{4}(8m + 24) = 21 - 13m
4
1
(
8
m
+
24
)
=
21
−
13
m
\newline
m
=
m =
m
=
__
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Solve for
s
s
s
.
\newline
−
s
=
4
(
1
2
s
+
4
)
-s = 4(\frac{1}{2}s + 4)
−
s
=
4
(
2
1
s
+
4
)
\newline
s
=
s =
s
=
__
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Solve for
b
b
b
.
\newline
b
+
3
=
1
2
(
b
−
4
)
b + 3 = \frac{1}{2}(b - 4)
b
+
3
=
2
1
(
b
−
4
)
\newline
b = ____
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Solve for
z
z
z
.
\newline
−
3
+
z
=
4
(
1
5
z
−
3
)
-3 + z = 4(\frac{1}{5}z - 3)
−
3
+
z
=
4
(
5
1
z
−
3
)
\newline
z
=
z =
z
=
__
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Solve for
j
j
j
.
\newline
4
j
−
2
=
3
(
1
3
j
+
2
)
4j - 2 = 3\left(\frac{1}{3}j + 2\right)
4
j
−
2
=
3
(
3
1
j
+
2
)
\newline
j
=
j =
j
=
__
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Solve for
h
h
h
.
\newline
7
−
h
=
3
(
2
3
h
+
4
)
7 - h = 3\left(\frac{2}{3}h + 4\right)
7
−
h
=
3
(
3
2
h
+
4
)
\newline
h
=
h =
h
=
__
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Solve for
v
v
v
.
\newline
v
−
5
=
1
2
(
v
+
10
)
v - 5 = \frac{1}{2}(v + 10)
v
−
5
=
2
1
(
v
+
10
)
\newline
v
=
v =
v
=
__
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Solve for
s
s
s
.
\newline
9
(
2
3
s
−
1
)
=
−
2
s
9\left(\frac{2}{3}s - 1\right) = -2s
9
(
3
2
s
−
1
)
=
−
2
s
\newline
s
=
s =
s
=
__
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Solve for
r
r
r
.
\newline
r
=
4
(
1
5
r
+
2
)
r = 4\left(\frac{1}{5}r + 2\right)
r
=
4
(
5
1
r
+
2
)
\newline
r = ____
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Solve for
v
v
v
.
\newline
2
(
3
−
1
4
v
)
=
2
−
v
2(3 - \frac{1}{4}v) = 2 - v
2
(
3
−
4
1
v
)
=
2
−
v
\newline
v = ____
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Solve for
d
d
d
.
\newline
1
2
(
d
+
4
)
=
d
−
1
\frac{1}{2}(d + 4) = d - 1
2
1
(
d
+
4
)
=
d
−
1
\newline
d = ____
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Solve for
y
y
y
.
\newline
−
2
+
y
=
2
(
1
3
y
−
3
)
-2 + y = 2(\frac{1}{3}y - 3)
−
2
+
y
=
2
(
3
1
y
−
3
)
\newline
y
=
y =
y
=
__
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Solve for
v
v
v
.
\newline
5
v
−
1
2
=
4
+
3
2
v
5v - \frac{1}{2} = 4 + \frac{3}{2}v
5
v
−
2
1
=
4
+
2
3
v
\newline
v
=
v =
v
=
__
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Solve for
v
v
v
.
\newline
1
2
v
+
8
=
3
−
1
3
v
\frac{1}{2}v + 8 = 3 - \frac{1}{3}v
2
1
v
+
8
=
3
−
3
1
v
\newline
v
=
v =
v
=
__
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(
y
−
4
5
)
2
+
(
x
+
7
10
)
2
=
30
(y-\frac{4}{5})^2+(x+\frac{7}{10})^2=30
(
y
−
5
4
)
2
+
(
x
+
10
7
)
2
=
30
find the radius
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d.)
4
x
−
3
2
x
+
1
=
5
7
\frac{4 x-3}{2 x+1}=\frac{5}{7}
2
x
+
1
4
x
−
3
=
7
5
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(
6
6
6
)
(
5
x
+
2
y
)
(
5
x
−
2
y
)
=
(5 x+2 y)(5 x-2 y)=
(
5
x
+
2
y
)
(
5
x
−
2
y
)
=
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(
3
+
2
2
)
(
3
−
2
)
=
(\sqrt{3}+2 \sqrt{2})(\sqrt{3}-2)=
(
3
+
2
2
)
(
3
−
2
)
=
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prove
(
sin
x
+
cos
x
)
2
=
1
+
sin
2
x
(\sin x+\cos x)^2=1+\sin 2x
(
sin
x
+
cos
x
)
2
=
1
+
sin
2
x
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(
3
2
+
4
)
(
3
2
−
4
)
(3\sqrt{2}+4)(3\sqrt{2}-4)
(
3
2
+
4
)
(
3
2
−
4
)
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36
36
36
.
d
d
+
2
−
2
2
−
d
=
d
+
6
d
2
−
4
\frac{d}{d+2}-\frac{2}{2-d}=\frac{d+6}{d^{2}-4}
d
+
2
d
−
2
−
d
2
=
d
2
−
4
d
+
6
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9
x
−
2
(
x
+
8
)
=
5
x
−
11
(
2
−
x
9x-2(x+8)=5x-11(2-x
9
x
−
2
(
x
+
8
)
=
5
x
−
11
(
2
−
x
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10
9
y
(
y
+
3
)
−
y
y
+
3
=
10 \frac{9}{y(y+3)}-\frac{y}{y+3}=
10
y
(
y
+
3
)
9
−
y
+
3
y
=
\newline
A
9
+
y
2
y
(
y
+
3
)
\frac{9+y^{2}}{y(y+3)}
y
(
y
+
3
)
9
+
y
2
\newline
C
3
−
y
y
\frac{3-y}{y}
y
3
−
y
\newline
B
9
−
y
y
+
3
\frac{9-y}{y+3}
y
+
3
9
−
y
\newline
D
3
+
y
y
\frac{3+y}{y}
y
3
+
y
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2
x
=
4
x
+
5
2
−
3
2x= 4x+\frac{5}{2}-3
2
x
=
4
x
+
2
5
−
3
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2
(
4
x
−
3
)
+
log
2
(
3
x
+
5
)
=
3
2(4 x-3)+\log _{2}(3 x+5)=3
2
(
4
x
−
3
)
+
lo
g
2
(
3
x
+
5
)
=
3
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q
−
3
/
q
−
7
/
q
−
5
=
3
q-3/q - 7/q-5 = 3
q
−
3/
q
−
7/
q
−
5
=
3
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Identify the properties used.
\newline
2
y
=
2 y=
2
y
=
\newline
y
=
y=
y
=
\newline
\qquad
\newline
\qquad
4
4
4
\newline
4
4
4
.
n
4
−
2
=
10
\frac{n}{4}-2=10
4
n
−
2
=
10
\newline
n
4
=
\frac{n}{4}=
4
n
=
\newline
\qquad
\newline
\qquad
\newline
n
=
n=
n
=
\newline
\qquad
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Solve.
\newline
2
x
+
8
5
=
3
x
+
2
6
\frac{2 x+8}{5}=\frac{3 x+2}{6}
5
2
x
+
8
=
6
3
x
+
2
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(
3
x
+
5
4
)
+
4
3
x
−
5
(3x+\frac{5}{4})+\frac{4}{3}x-5
(
3
x
+
4
5
)
+
3
4
x
−
5
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TOWERS A lookout tower sits on a network of struts and postś. Leslie measured three angles on the tower. If
m
∠
1
=
(
7
x
−
7
)
∘
m \angle 1=(7 x-7)^{\circ}
m
∠1
=
(
7
x
−
7
)
∘
,
m
∠
2
=
(
4
x
+
2
)
∘
m \angle 2=(4 x+2)^{\circ}
m
∠2
=
(
4
x
+
2
)
∘
, and
m
∠
3
=
(
2
x
+
6
)
∗
m \angle 3=(2 x+6)^{*}
m
∠3
=
(
2
x
+
6
)
∗
, what is
m
∠
1
?
m \angle 1 ?
m
∠1
?
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12
(
5
+
2
y
)
=
4
y
−
(
5
−
9
y
)
12(5+2y)=4y - (5 - 9y)
12
(
5
+
2
y
)
=
4
y
−
(
5
−
9
y
)
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Solve the equation
2
x
+
7
3
=
2
−
x
4
\frac{2 x+7}{3}=\frac{2-x}{4}
3
2
x
+
7
=
4
2
−
x
.
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Find the solution to this equation.
\newline
−
x
+
2
(
x
+
3
)
=
6
+
x
-x+2(x+3)=6+x
−
x
+
2
(
x
+
3
)
=
6
+
x
\newline
(A) No solution
\newline
(B) All real numbers
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3
3
3
.
2
x
(
2
x
−
3
)
≥
4
x
2
−
3
(
2
x
+
1
)
2 x(2 x-3) \geq 4 x^{2}-3(2 x+1)
2
x
(
2
x
−
3
)
≥
4
x
2
−
3
(
2
x
+
1
)
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rac{
7
7
7
}{
3
3
3
}igg( rac{
1
1
1
}{
2
2
2
}x - rac{
10
10
10
}{
3
3
3
}igg) = - rac{
35
35
35
}{
3
3
3
} - rac{
1
1
1
}{
2
2
2
}x
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if
(
3
y
−
5
x
)
/
(
7
x
−
4
y
)
=
3
/
4
(3y -5x)/(7x-4y) = 3/4
(
3
y
−
5
x
)
/
(
7
x
−
4
y
)
=
3/4
, find the value of
x
/
y
x/y
x
/
y
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find all solutions to the equations
15
x
(
x
+
4
)
=
x
+
2
x
+
4
\frac{15}{x(x+4)} = \frac{x+2}{x+4}
x
(
x
+
4
)
15
=
x
+
4
x
+
2
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Solve for
x
x
x
. Express your answer as a proper or improper fraction in simplest terms.
\newline
−
1
4
=
1
5
x
+
1
4
-\frac{1}{4}=\frac{1}{5} x+\frac{1}{4}
−
4
1
=
5
1
x
+
4
1
\newline
Answer:
x
=
x=
x
=
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1
2
3
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