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Math Problems
Grade 8
Write an equation word problem
{
f
(
1
)
=
15
f
(
n
)
=
f
(
n
−
1
)
⋅
n
f
(
2
)
=
□
\begin{array}{l}\left\{\begin{array}{l}f(1)=15 \\ f(n)=f(n-1) \cdot n\end{array}\right. \\ f(2)=\square\end{array}
{
f
(
1
)
=
15
f
(
n
)
=
f
(
n
−
1
)
⋅
n
f
(
2
)
=
□
Get tutor help
{
h
(
1
)
=
9
h
(
2
)
=
3
h
(
n
)
=
h
(
n
−
2
)
⋅
h
(
n
−
1
)
h
(
3
)
=
□
\begin{array}{l}\left\{\begin{array}{l}h(1)=9 \\ h(2)=3 \\ h(n)=h(n-2) \cdot h(n-1)\end{array}\right. \\ h(3)=\square\end{array}
⎩
⎨
⎧
h
(
1
)
=
9
h
(
2
)
=
3
h
(
n
)
=
h
(
n
−
2
)
⋅
h
(
n
−
1
)
h
(
3
)
=
□
Get tutor help
Rearrange the equation so
x
x
x
is the independent variable.
\newline
y
+
6
=
5
(
x
−
4
)
y
=
□
\begin{array}{l} y+6=5(x-4) \\ y=\square \end{array}
y
+
6
=
5
(
x
−
4
)
y
=
□
\newline
□
\square
□
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Determine the intercepts of the line. Do not round your answers.
−
4
x
+
7
y
=
3
-4x+7y=3
−
4
x
+
7
y
=
3
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\newline
2
3
x
−
4
3
y
=
8
A
(
3
x
−
1
)
=
3
y
\begin{array}{l} \frac{2}{3} x-\frac{4}{3} y=8 \\ A(3 x-1)=3 y \end{array}
3
2
x
−
3
4
y
=
8
A
(
3
x
−
1
)
=
3
y
\newline
In the system of equations,
A
A
A
is a constant. For which value of
A
A
A
is there exactly one solution
(
x
,
y
)
(x, y)
(
x
,
y
)
where
y
=
0
y=0
y
=
0
?
\newline
A)
2
2
2
\newline
B)
3
3
3
\newline
C)
4
4
4
\newline
D)
5
5
5
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−
6
x
+
4
y
=
2
3
x
−
2
y
=
−
1
\begin{aligned} -6 x+4 y & =2 \\ 3 x-2 y & =-1 \end{aligned}
−
6
x
+
4
y
3
x
−
2
y
=
2
=
−
1
\newline
Consider the system of equations. How many
(
x
,
y
)
(x, y)
(
x
,
y
)
solutions does this system have?
\newline
Choose
1
1
1
answer:
\newline
(A) No solutions
\newline
(B) Exactly one solution
\newline
(c) Exactly two solutions
\newline
(D) Infinitely many solutions
\newline
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Solve the system of equations by substitution or elimination.
\newline
4
x
−
y
=
10
4x-y=10
4
x
−
y
=
10
\newline
3
x
+
5
y
=
19
3x+5y=19
3
x
+
5
y
=
19
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Solve system of equations:
\newline
2
x
+
5
y
=
−
11
2x+5y=-11
2
x
+
5
y
=
−
11
\newline
−
8
x
−
5
y
=
−
1
-8x-5y=-1
−
8
x
−
5
y
=
−
1
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4
x
2
+
x
−
29
=
y
4x^2 + x - 29 = y
4
x
2
+
x
−
29
=
y
\newline
x
−
4
=
y
x - 4 = y
x
−
4
=
y
\newline
If
(
x
,
y
)
(x,y)
(
x
,
y
)
is a solution to the system of equations shown and
x
>
0
x > 0
x
>
0
, what is the value of
x
x
x
?
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What is
(
q
∗
c
)
(
x
)
?
(q*c)(x) ?
(
q
∗
c
)
(
x
)?
\newline
q
(
x
)
=
2
(
x
)
q(x) = 2(x)
q
(
x
)
=
2
(
x
)
\newline
c
(
x
)
=
−
2
x
2
+
9
x
c(x) = -2x^2+ 9x
c
(
x
)
=
−
2
x
2
+
9
x
\newline
Write your answer as a polynomial or a rational function.
\newline
□
\square
□
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Solve the following system of equations.
\newline
8
x
−
5
y
=
25
8x-5y=25
8
x
−
5
y
=
25
\newline
−
2
x
+
7
y
=
11
-2x+7y=11
−
2
x
+
7
y
=
11
\newline
x
=
x =
x
=
□
\square
□
\newline
y
=
y =
y
=
□
\square
□
Get tutor help
Solve for
h
h
h
.
\newline
h
+
3
h
+
5
h
+
3
h
−
12
=
48
h+3h+5h+3h-12=48
h
+
3
h
+
5
h
+
3
h
−
12
=
48
\newline
h
=
□
h=\square
h
=
□
Get tutor help
Determine whether
(
2
,
1
,
8
)
(2,1,8)
(
2
,
1
,
8
)
is a solution to the system.
\newline
4
x
−
4
y
−
5
z
=
−
36
2
x
−
5
y
+
5
z
=
39
−
14
x
−
3
y
+
10
z
=
49
\begin{array}{rr} 4 x-4 y-5 z= & -36 \\ 2 x-5 y+5 z= & 39 \\ -14 x-3 y+10 z= & 49 \end{array}
4
x
−
4
y
−
5
z
=
2
x
−
5
y
+
5
z
=
−
14
x
−
3
y
+
10
z
=
−
36
39
49
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Find the solution to the system by the substitution method. Check your answers.
\newline
5
x
+
8
y
=
10
x
+
2
y
=
−
2
\begin{aligned} 5 x+8 y & =10 \\ x+2 y & =-2 \end{aligned}
5
x
+
8
y
x
+
2
y
=
10
=
−
2
Get tutor help
x
1
+
x
2
+
2
x
3
=
−
1
x_{1}+x_{2}+2x_{3}=-1
x
1
+
x
2
+
2
x
3
=
−
1
\newline
x
1
−
2
x
2
+
x
3
=
−
5
x_{1}-2x_{2}+x_{3}=-5
x
1
−
2
x
2
+
x
3
=
−
5
\newline
3
x
1
+
x
2
+
x
3
=
3
3x_{1}+x_{2}+x_{3}=3
3
x
1
+
x
2
+
x
3
=
3
\newline
find all solutions by using the Gaussian elimination or Gauss-Jordan Reduction.
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x
1
+
x
2
+
2
x
3
=
−
1
x
1
−
2
x
2
+
x
3
=
−
5
3
x
1
+
x
2
+
x
3
=
3
\begin{array}{l} x_{1}+x_{2}+2 x_{3}=-1 \\ x_{1}-2 x_{2}+x_{3}=-5 \\ 3 x_{1}+x_{2}+x_{3}=3 \end{array}
x
1
+
x
2
+
2
x
3
=
−
1
x
1
−
2
x
2
+
x
3
=
−
5
3
x
1
+
x
2
+
x
3
=
3
\newline
a)Find all solutions using Gaussian elimination or Gauss-Jordan reduction.
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Solve the system of equations.
\newline
6
x
−
5
y
=
15
x
=
y
+
3
x
=
y
=
\begin{array}{l} 6 x-5 y=15 \\ x=y+3 \\ x= \\ y= \end{array}
6
x
−
5
y
=
15
x
=
y
+
3
x
=
y
=
Get tutor help
Determine whether
(
8
,
9
,
9
)
(8,9,9)
(
8
,
9
,
9
)
is a solution to the system.
\newline
−
2
x
+
3
y
+
4
z
=
47
−
5
x
+
4
y
−
4
z
=
−
40
11
x
+
2
y
−
8
z
=
34
\begin{array}{rr} -2 x+3 y+4 z= & 47 \\ -5 x+4 y-4 z= & -40 \\ 11 x+2 y-8 z= & 34 \end{array}
−
2
x
+
3
y
+
4
z
=
−
5
x
+
4
y
−
4
z
=
11
x
+
2
y
−
8
z
=
47
−
40
34
Get tutor help
Find the solution to the system by the substitution method. Check your answers.
\newline
3
x
+
4
y
=
10
x
+
2
y
=
−
2
\begin{array}{r} 3 x+4 y=10 \\ x+2 y=-2 \end{array}
3
x
+
4
y
=
10
x
+
2
y
=
−
2
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3
x
−
4
y
=
10
2
x
−
4
y
=
6
\begin{array}{l} 3 x-4 y=10 \\ 2 x-4 y=6 \end{array}
3
x
−
4
y
=
10
2
x
−
4
y
=
6
\newline
If
(
x
,
y
)
(x, y)
(
x
,
y
)
satisfies the given system of equations, what is the value of
y
y
y
?
Get tutor help
3
x
+
2
y
=
8
2
x
−
y
=
3
\begin{array}{c} 3 x+2 y=8 \\ 2 x-y=3 \end{array}
3
x
+
2
y
=
8
2
x
−
y
=
3
\newline
What is the solution
(
x
,
y
)
(x, y)
(
x
,
y
)
to the given system of equations?
\newline
A)
(
3
,
2
)
(3,2)
(
3
,
2
)
\newline
B)
(
2
,
3
)
(2,3)
(
2
,
3
)
\newline
C)
(
1
,
2
)
(1,2)
(
1
,
2
)
\newline
D)
(
2
,
1
)
(2,1)
(
2
,
1
)
Get tutor help
k
×
34
=
6.29
k\times 34=6.29
k
×
34
=
6.29
\newline
k
=
□
k=\square
k
=
□
Get tutor help
Solve for
m
\mathrm{m}
m
and write your result in the empty box provided below.
\newline
−
20
+
14
m
=
10
m
+
16
m
=
□
\begin{array}{l} -20+14 \mathrm{~m}=10 \mathrm{~m}+16 \\ \mathrm{~m}=\square \end{array}
−
20
+
14
m
=
10
m
+
16
m
=
□
Get tutor help
Consider the following system of equations,
\newline
−
2
x
−
9
y
=
12
−
2
x
−
7
y
=
8
\begin{aligned} -2 x-9 y & =12 \\ -2 x-7 y & =8 \end{aligned}
−
2
x
−
9
y
−
2
x
−
7
y
=
12
=
8
Get tutor help
Solve for
s
s
s
.
\newline
6
s
=
99.42
s
=
\begin{array}{l} 6 s=99.42 \\ s= \end{array}
6
s
=
99.42
s
=
\newline
Get tutor help
Solve for
s
s
s
.
\newline
6
s
=
99.42
6s=99.42
6
s
=
99.42
.
\newline
s
=
s =
s
=
□
\square
□
Get tutor help
Complete the equations.
\newline
6
×
□
=
600
6\times\square=600
6
×
□
=
600
,
\newline
6
×
□
=
6
,
000
6\times\square=6,000
6
×
□
=
6
,
000
.
Get tutor help
Simplify the expression:
\newline
3
a
+
5
u
=
17
3a+5u = 17
3
a
+
5
u
=
17
,
\newline
2
a
+
u
=
9
2a+u=9
2
a
+
u
=
9
Get tutor help
{
f
(
1
)
=
−
2
,
f
(
2
)
=
5
,
f
(
n
)
=
f
(
n
−
2
)
⋅
f
(
n
−
1
)
\begin{cases} f(1) = -2, \ f(2) = 5, \ f(n) = f(n-2) \cdot f(n-1) \end{cases}
{
f
(
1
)
=
−
2
,
f
(
2
)
=
5
,
f
(
n
)
=
f
(
n
−
2
)
⋅
f
(
n
−
1
)
,
f
(
3
)
=
0
f(3) = \boxed{\phantom{0}}
f
(
3
)
=
0
Get tutor help
x
1
+
x
2
+
2
x
3
=
−
1
x_{1}+x_{2}+2x_{3}=-1
x
1
+
x
2
+
2
x
3
=
−
1
\newline
x
1
−
2
x
2
+
x
3
=
−
5
x_{1}-2x_{2}+x_{3}=-5
x
1
−
2
x
2
+
x
3
=
−
5
\newline
3
x
1
+
x
2
+
x
3
=
3
3x_{1}+x_{2}+x_{3}=3
3
x
1
+
x
2
+
x
3
=
3
\newline
Find all solutions by using the Gaussian elimination & Gauss-Jordan Reduction.
Get tutor help
x
1
+
x
2
+
2
x
3
=
−
1
x
1
−
2
x
2
+
x
3
=
−
5
3
x
1
+
x
2
+
x
3
=
3
\begin{array}{l} x_{1}+x_{2}+2 x_{3}=-1 \\ x_{1}-2 x_{2}+x_{3}=-5 \\ 3 x_{1}+x_{2}+x_{3}=3 \end{array}
x
1
+
x
2
+
2
x
3
=
−
1
x
1
−
2
x
2
+
x
3
=
−
5
3
x
1
+
x
2
+
x
3
=
3
\newline
Find all solutions by waing the Gausionan
\newline
elimination& Gaus-Jordan
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Use substitution to solve equations:
\newline
y
=
5
x
+
1
y=5x+1
y
=
5
x
+
1
\newline
4
x
+
y
=
10
4x+y=10
4
x
+
y
=
10
Get tutor help
9
x
+
4
y
<
8
−
3
x
−
7
y
≥
5
\begin{array}{l} 9 x+4 y<8 \\ -3 x-7 y \geq 5 \end{array}
9
x
+
4
y
<
8
−
3
x
−
7
y
≥
5
\newline
Is
(
1
,
−
2
)
(1,-2)
(
1
,
−
2
)
a solution of the system?
Get tutor help
Solve for
c
c
c
.
\newline
c
−
−
488
=
742
c--488=742
c
−
−
488
=
742
\newline
c
=
c=
c
=
□
\square
□
Get tutor help
A systern of equations is shown
\newline
2
x
+
2
y
=
17
2x+2y=17
2
x
+
2
y
=
17
\newline
4
x
−
y
=
25
4x-y=25
4
x
−
y
=
25
\newline
Enter your answer in the box.
\newline
□
\square
□
,
□
\square
□
Get tutor help
Solve for
u
u
u
.
\newline
(
u
−
63
6
)
=
5
\left(\frac{u-63}{6}\right)=5
(
6
u
−
63
)
=
5
,
\newline
u
=
□
u=\square
u
=
□
\newline
Get tutor help
8
x
−
y
=
12
8x-y=12
8
x
−
y
=
12
\newline
2
x
−
6
y
=
3
2x-6y=3
2
x
−
6
y
=
3
\newline
Consider the system of equations. If
(
x
,
y
)
(x,y)
(
x
,
y
)
is the solution to the system, then what is the value of
x
+
y
x+y
x
+
y
?
\newline
□
\square
□
Get tutor help
7
x
+
10
y
=
60
7x+10y=60
7
x
+
10
y
=
60
\newline
7
x
−
2
y
=
240
7x-2y=240
7
x
−
2
y
=
240
\newline
If
(
x
,
y
)
(x,y)
(
x
,
y
)
satisfies the given system of equations, what is the value of
x
x
x
?
\newline
◻
Get tutor help
w
x
+
2
y
=
3
(
1
+
y
)
+
1
\ w x + 2 y= 3 (1 + y) + 1
w
x
+
2
y
=
3
(
1
+
y
)
+
1
\newline
8
−
y
=
2
(
1
−
y
)
+
3
x
\ 8 - y = 2 (1 - y) + 3 x
8
−
y
=
2
(
1
−
y
)
+
3
x
\newline
In the system of equations,
w
w
w
is a constant. For what value of
w
w
w
will the system of equations have exactly one solution
(
x
,
y
)
(x,y)
(
x
,
y
)
with
x
=
1
x=1
x
=
1
?
\newline
□
\square
□
Get tutor help
Solve the equation
\newline
(
2
sin
x
−
1
)
(
2
cos
x
+
1
)
=
0
(2\sin x-1)(2\cos x+1)=0
(
2
sin
x
−
1
)
(
2
cos
x
+
1
)
=
0
within the interval
−
180
≤
x
≤
180
-180 \leq x \leq 180
−
180
≤
x
≤
180
,
Get tutor help
Find the solution of the system of equations.
\newline
−
3
x
−
8
y
=
−
25
6
x
+
8
y
=
10
\begin{aligned} -3 x-8 y & =-25 \\ 6 x+8 y & =10 \end{aligned}
−
3
x
−
8
y
6
x
+
8
y
=
−
25
=
10
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
−
3
x
+
4
y
=
−
7
3
x
+
6
y
=
27
\begin{aligned} -3 x+4 y & =-7 \\ 3 x+6 y & =27 \end{aligned}
−
3
x
+
4
y
3
x
+
6
y
=
−
7
=
27
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
2
x
+
3
y
=
19
2
x
−
6
y
=
−
44
\begin{array}{l} 2 x+3 y=19 \\ 2 x-6 y=-44 \end{array}
2
x
+
3
y
=
19
2
x
−
6
y
=
−
44
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
9
x
+
4
y
=
47
−
9
x
−
y
=
−
32
\begin{array}{l} 9 x+4 y=47 \\ -9 x-y=-32 \end{array}
9
x
+
4
y
=
47
−
9
x
−
y
=
−
32
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
9
x
+
y
=
−
12
7
x
+
y
=
−
10
\begin{array}{l} 9 x+y=-12 \\ 7 x+y=-10 \end{array}
9
x
+
y
=
−
12
7
x
+
y
=
−
10
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
3
x
−
6
y
=
−
45
x
−
6
y
=
−
47
\begin{array}{r} 3 x-6 y=-45 \\ x-6 y=-47 \end{array}
3
x
−
6
y
=
−
45
x
−
6
y
=
−
47
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
x
+
3
y
=
13
−
x
+
2
y
=
12
\begin{aligned} x+3 y & =13 \\ -x+2 y & =12 \end{aligned}
x
+
3
y
−
x
+
2
y
=
13
=
12
\newline
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
−
4
x
−
2
y
=
−
30
−
4
x
−
7
y
=
−
25
\begin{array}{l} -4 x-2 y=-30 \\ -4 x-7 y=-25 \end{array}
−
4
x
−
2
y
=
−
30
−
4
x
−
7
y
=
−
25
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
x
−
8
y
=
16
−
x
−
10
y
=
2
\begin{aligned} x-8 y & =16 \\ -x-10 y & =2 \end{aligned}
x
−
8
y
−
x
−
10
y
=
16
=
2
\newline
(_________,________)
Get tutor help
Find the solution of the system of equations.
\newline
8
x
−
5
y
=
−
14
−
8
x
+
y
=
22
\begin{array}{l} 8 x-5 y=-14 \\ -8 x+y=22 \end{array}
8
x
−
5
y
=
−
14
−
8
x
+
y
=
22
\newline
(_________,________)
Get tutor help
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