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Math Problems
Grade 8
Write an equation word problem
Solve for
h
h
h
.\begin{cases}h+3h+5h+3h-12=48\h=\square\end{cases}Submit
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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
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{:\begin{align*}x_{
1
1
1
}+x_{
2
2
2
}+
2
2
2
x_{
3
3
3
}&=
−
1
-1
−
1
\x_{
1
1
1
}
−
2
-2
−
2
x_{
2
2
2
}+x_{
3
3
3
}&=
−
5
-5
−
5
(3\)x_{
1
1
1
}+x_{
2
2
2
}+x_{
3
3
3
}&=
3
3
3
\end{align*}:} a) find all solutions by using the Gaussian elimination & Gauss-Jordan Reduction
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\begin{cases}x_{1}+x_{2}+2x_{3}=-1\x_{1}-2x_{2}+x_{3}=-5\3x_{1}+x_{2}+x_{3}=3\end{cases} a) 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
=
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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
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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
?
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Graph the solution of the following system.
\newline
y
<
5
x
>
−
1
\begin{array}{l} y<5 \\ x>-1 \end{array}
y
<
5
x
>
−
1
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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
)
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{:
k
×
34
=
6.29
k\times 34=6.29
k
×
34
=
6.29
,
\newline
k
=
□
k=\square
k
=
□
:}
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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
=
□
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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
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Choose the ordered pair(s) that are solutions to the syste (There may be one or many correct options).
\newline
{
−
4
x
+
3
y
≤
0
8
x
+
2
y
≤
−
2
\left\{\begin{array}{l} -4 x+3 y \leq 0 \\ 8 x+2 y \leq-2 \end{array}\right.
{
−
4
x
+
3
y
≤
0
8
x
+
2
y
≤
−
2
\newline
Show your work here
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Graph the solution to the following system of inequalities.
\newline
{
y
<
−
3
x
−
2
y
≥
2
x
−
5
\begin{cases} y < -3x-2 \ y \geq 2x-5 \end{cases}
{
y
<
−
3
x
−
2
y
≥
2
x
−
5
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−
3
a
+
5
b
=
11
6
a
+
2
b
=
26
\begin{array}{r}-3 a+5 b=11 \\ 6 a+2 b=26\end{array}
−
3
a
+
5
b
=
11
6
a
+
2
b
=
26
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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
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Solve for
s
s
s
.\begin{align*}6s&=99.42\s&=\square\end{align*}Submit
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Complete the equations.
\newline
6
×
□
=
600
6\times\square=600
6
×
□
=
600
,
6
×
□
=
6
,
000
6\times\square=6,000
6
×
□
=
6
,
000
.
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{:\begin{align*}[
3
3
3
a+
5
5
5
u&=
17
17
17
],[
2
2
2
a+u&=
9
9
9
]:\end{align*}
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Learn with an example
\newline
Solve the right triangle.
\newline
Write your answers as integers or as decimals round
\newline
\begin{array}{l}\(\newlineVW=,(\newline\)WX=,(\newline\)m/\_W=.
\newline
\end{array}\)
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{
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
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\begin{cases}x_{1}+x_{2}+2x_{3}=-1\x_{1}-2x_{2}+x_{3}=-5\3x_{1}+x_{2}+x_{3}=3\end{cases} a) Find all solutions by using the Gaussian elimination & 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 by wising the Gaussian elimination \& Gaus-Jordan Reduction
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x
1
+
x
1
+
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_{1}+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
1
+
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 using the Guassian
\newline
elimination and Gauss-Jordan Reduction
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Find all solutions by using the Gaussian elimination & Gauss-Jordan method:
{
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{cases} x_1+x_2+2x_3=-1 \ x_1-2x_2+x_3=-5 \ 3x_1+x_2+x_3=3 \end{cases}
{
x
1
+
x
2
+
2
x
3
=
−
1
x
1
−
2
x
2
+
x
3
=
−
5
3
x
1
+
x
2
+
x
3
=
3
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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
Find all solutions by waing the Gausionan
\newline
elimination& Gaus-Jordan
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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
Find all solutions by nsing the Gaussian elimination \& Gauss-Jordan Reduction
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question:
{
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{cases} x_{1}+x_{2}+2x_{3}=-1 \ x_{1}-2x_{2}+x_{3}=-5 \ 3x_{1}+x_{2}+x_{3}=3 \end{cases}
{
x
1
+
x
2
+
2
x
3
=
−
1
x
1
−
2
x
2
+
x
3
=
−
5
3
x
1
+
x
2
+
x
3
=
3
a) Find all solutions by using the Gaussian elimination & Gaus-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 by uning the Gaussian elimination \& 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 by using the Gaussian elimination \& Gauss-Jordan Reduction
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L
=
[
4
1
3
−
2
]
m
=
[
3
0
0
1
]
N
=
[
2
3
4
1
]
Evaluate:
2
N
−
L
(
N
M
)
L
\begin{array}{l}L=\left[\begin{array}{cc}4 & 1 \\ 3 & -2\end{array}\right] \quad m=\left[\begin{array}{ll}3 & 0 \\ 0 & 1\end{array}\right] \quad N=\left[\begin{array}{ll}2 & 3 \\ 4 & 1\end{array}\right] \\ \text { Evaluate: } 2 N-L \\ (N M) L\end{array}
L
=
[
4
3
1
−
2
]
m
=
[
3
0
0
1
]
N
=
[
2
4
3
1
]
Evaluate:
2
N
−
L
(
NM
)
L
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NAME
\newline
5
5
5
.
2
2
2
Solving Systems Practice
\newline
Use substitution to solve each s
\newline
1
1
1
.
\newline
{
y
=
5
x
+
1
4
x
+
y
=
10
\begin{cases} y=5x+1 \ 4x+y=10 \end{cases}
{
y
=
5
x
+
1
4
x
+
y
=
10
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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
{:\begin{align*}[
5
5
5
y&=
8
8
8
],[y&=
0
0
0
],[
−
4
-4
−
4
x+
5
5
5
y&=
33
33
33
],[
8
8
8
x+y&=
33
33
33
]:\end{align*}}
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Solve for
c
c
c
.
\newline
c
−
−
488
=
742
c--488=742
c
−
−
488
=
742
\newline
c
=
c=
c
=
□
\square
□
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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}^{\prime}+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 Gausian elimination
2
2
2
Gauss-Jordan
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\begin{cases}x_{1}-2x_{2}+x_{3}=-5\3x_{1}+x_{2}+x_{3}=3\end{cases} a) Find all solutions by using the Gaussian elimination? Gauss-Jordan Reduction
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x
1
+
2
x
2
+
3
x
3
x
1
+
(
2
)
(
2
)
+
3
(
3
)
x
1
+
5
=
6
\begin{array}{r} x_{1}+2 x_{2}+3 x_{3} \\ x_{1}+(2)(2)+3(3) \\ x_{1}+5=6 \end{array}
x
1
+
2
x
2
+
3
x
3
x
1
+
(
2
)
(
2
)
+
3
(
3
)
x
1
+
5
=
6
\newline
An
m
×
n
m \times n
m
×
n
matnix
A
A
A
is
x
1
+
5
=
6
\quad x_{1}+5=6
x
1
+
5
=
6
\newline
In recuad row ecchelon form if it satisfies
x
1
=
6
\quad x_{1}=6
x
1
=
6
the follouinn annorties:
\newline
x
1
+
x
2
1
+
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}^{1}+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
1
+
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 by unang the Gausion elimination? Gause-Jordan Reduction
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Question
5
5
5
(
1
1
1
point)
\newline
This graph represents a linear function.
\newline
a
y
=
3
x
+
2
\quad y=3 x+2
y
=
3
x
+
2
\newline
b
y
=
1
3
x
+
2
y=\frac{1}{3} x+2
y
=
3
1
x
+
2
\newline
c
y
=
3
x
−
2
y=3 x-2
y
=
3
x
−
2
\newline
d
y
=
1
3
x
−
2
y=\frac{1}{3} x-2
y
=
3
1
x
−
2
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Solve the system with the addition method:
\newline
{
−
12
x
−
8
y
=
−
6
6
x
+
4
y
=
3
\left\{\begin{array}{ll} -12 x-8 y & =-6 \\ 6 x+4 y & =3 \end{array}\right.
{
−
12
x
−
8
y
6
x
+
4
y
=
−
6
=
3
\newline
Answer:
□
\square
□
,
□
\square
□
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ara
1
1
1
-Soskion
2
2
2
Cilsulator A systern of equations is shom \left\{\begin{array}{l}2x+2y=17\4x-y=25\end{array}\right. Enter your answer in the box.
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A system of equations is shom.
\newline
{
2
x
+
2
y
=
17
4
x
−
y
=
25
\left\{\begin{array}{l} 2 x+2 y=17 \\ 4 x-y=25 \end{array}\right.
{
2
x
+
2
y
=
17
4
x
−
y
=
25
\newline
Enter your answer in the bex.
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Agebra
1
1
1
-Sonwion
2
2
2
: Crtivatator
\newline
18
18
18
. A syatem of equations is ahown.
\newline
{
2
x
+
2
y
=
17
4
x
−
y
=
25
\left\{\begin{array}{l} 2 x+2 y=17 \\ 4 x-y=25 \end{array}\right.
{
2
x
+
2
y
=
17
4
x
−
y
=
25
\newline
Enter your answer in the box.
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−
7
−
(
−
2
)
=
39
−
46
=
\begin{array}{r}-7-(-2)= \\ 39-46=\end{array}
−
7
−
(
−
2
)
=
39
−
46
=
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4
+
1
=
□
0
+
1
=
8
+
1
=
□
6
+
1
=
3
+
1
=
□
8
+
1
=
\begin{array}{l}4+1=\square \quad 0+1= \\ 8+1=\square \quad 6+1= \\ 3+1=\square \quad 8+1= \\\end{array}
4
+
1
=
□
0
+
1
=
8
+
1
=
□
6
+
1
=
3
+
1
=
□
8
+
1
=
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b
0.69
=
2.7
b
=
[
?
]
\begin{array}{c}\frac{b}{0.69}=2.7 \\ b=[?]\end{array}
0.69
b
=
2.7
b
=
[
?]
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3
x
+
3
y
=
42
x
−
3
y
=
33
\begin{array}{r}3 x+3 y=42 \\ x-3 y=33\end{array}
3
x
+
3
y
=
42
x
−
3
y
=
33
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1
2
3
...
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