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Math Problems
Grade 8
Write an equation word problem
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?
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{:\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
Get tutor help
1
1
1
.
\newline
x
=
y
x
+
2
y
=
3
\begin{array}{l} x=y \\ x+2 y=3 \end{array}
x
=
y
x
+
2
y
=
3
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=
=
=
Step-by-Step Solutions begin
\newline
sy graphing.
\newline
2
2
2
.
\newline
y
=
x
2
−
4
y
=
−
3
\begin{array}{l} y=x^{2}-4 \\ y=-3 \end{array}
y
=
x
2
−
4
y
=
−
3
\newline
4
4
4
.
\newline
y
=
−
2
x
2
+
5
y
=
2
x
+
1
\begin{array}{l} y=-2 x^{2}+5 \\ y=2 x+1 \end{array}
y
=
−
2
x
2
+
5
y
=
2
x
+
1
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What is the value of
x
x
x
in the solution to this system of equations?
\newline
2
x
+
4
y
=
−
17
y
=
−
3
x
+
2
\begin{array}{c} 2 x+4 y=-17 \\ y=-3 x+2 \end{array}
2
x
+
4
y
=
−
17
y
=
−
3
x
+
2
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10
10
10
. Solve the following simultaneous equations.
\newline
x
+
2
y
=
9
y
=
3
x
+
1
\begin{array}{l} x+2 y=9 \\ y=3 x+1 \end{array}
x
+
2
y
=
9
y
=
3
x
+
1
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Rearrange the equation so
q
q
q
is the independent variable.
\newline
9
q
−
4
=
3
r
−
6
r
=
□
\begin{array}{l} 9 q-4=3 r-6 \\ r=\square \end{array}
9
q
−
4
=
3
r
−
6
r
=
□
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c
=
21
a
a
+
c
=
330
\begin{array}{l}c=21 a \\ a+c=330\end{array}
c
=
21
a
a
+
c
=
330
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x
,
y
,
z
∈
Z
+
7
x
+
4
y
+
5
z
=
20
⇒
y
=
?
\begin{array}{l}x, y, z \in Z^{+} \\ 7 x+4 y+5 z=20 \\ \Rightarrow y=?\end{array}
x
,
y
,
z
∈
Z
+
7
x
+
4
y
+
5
z
=
20
⇒
y
=
?
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2
+
7
=
27
4
+
4
=
24
5
+
9
=
42
6
+
0
=
?
?
\begin{array}{l}2+7=27 \\ 4+4=24 \\ 5+9=42 \\ 6+0=? ?\end{array}
2
+
7
=
27
4
+
4
=
24
5
+
9
=
42
6
+
0
=
??
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Solve for
X
X
X
in the equation, where
\newline
A
=
[
−
2
1
3
−
4
2
4
]
and
B
=
[
0
4
−
5
5
3
4
]
.
X
=
2
A
+
2
B
\begin{array}{l} A=\left[\begin{array}{rrr} -2 & 1 & 3 \\ -4 & 2 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 0 & 4 & -5 \\ 5 & 3 & 4 \end{array}\right] . \\ X=2 A+2 B \end{array}
A
=
[
−
2
−
4
1
2
3
4
]
and
B
=
[
0
5
4
3
−
5
4
]
.
X
=
2
A
+
2
B
\newline
x
=
[
−
4
−
4
10
2
10
]
⇒
x=\left[\begin{array}{lll} \boxed{-4} & \boxed{-4} & \\ & \frac{10}{2} & 10 \end{array}\right] \Rightarrow
x
=
[
−
4
−
4
2
10
10
]
⇒
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y
=
−
5
x
+
7
y
=
−
x
−
5
\begin{array}{l}y=-5 x+7 \\ y=-x-5\end{array}
y
=
−
5
x
+
7
y
=
−
x
−
5
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Solve the System by Graphing:
\newline
y
=
−
3
2
x
+
1
y
=
1
2
x
−
3
\begin{array}{l} y=-\frac{3}{2} x+1 \\ y=\frac{1}{2} x-3 \end{array}
y
=
−
2
3
x
+
1
y
=
2
1
x
−
3
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x
+
4
y
=
8
y
=
−
1
4
x
+
2
\begin{array}{l}x+4 y=8 \\ y=-\frac{1}{4} x+2\end{array}
x
+
4
y
=
8
y
=
−
4
1
x
+
2
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Focus by adjusting the corners
\newline
08
08
08
Practice Quiz Final [CA]
\newline
Question
12
12
12
of
\newline
list
\newline
Multiply
\newline
(
8
x
+
9
)
(
3
x
2
+
4
x
+
9
)
(8 x+9)\left(3 x^{2}+4 x+9\right)
(
8
x
+
9
)
(
3
x
2
+
4
x
+
9
)
\newline
The answer is
\newline
y your a
\newline
SOLVING STEPS
\newline
Simplify the expression
\newline
(
8
x
+
9
)
×
(
3
x
2
+
4
x
+
9
(8 x+9) \times\left(3 x^{2}+4 x+9\right.
(
8
x
+
9
)
×
(
3
x
2
+
4
x
+
9
\newline
Simplify using distributive property
\newline
24
x
3
+
59
x
2
+
108
x
+
81
24 x^{3}+59 x^{2}+108 x+81
24
x
3
+
59
x
2
+
108
x
+
81
\newline
Show Solving Steps
→
\rightarrow
→
\newline
ANIMATED TUTORIAL
\newline
PLUS
\newline
Multiply binomials and trinomials using the vertical method
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b)
−
2
x
+
3
≤
7
−
5
x
≤
7
/
5
1
x
≥
7
x
=
21
\begin{array}{c}-2 x+3 \leq 7 \\ -5 x \leq 7 / 5 \\ 1 x \geq 7 \\ x=21\end{array}
−
2
x
+
3
≤
7
−
5
x
≤
7/5
1
x
≥
7
x
=
21
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If
A
=
[
1
1
1
2
]
,
B
=
[
−
2
6
4
7
]
then find
X
when
X
+
4
A
=
7
B
∫
18
38
X
+
4
A
=
7
B
\begin{array}{l}\text { If } A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{cc}-2 & 6 \\ 4 & 7\end{array}\right] \text { then find } X \text { when } \\ X+4 A=7 B \\ \int \text { 18 } 38 \\ X+4 A=7 B\end{array}
If
A
=
[
1
1
1
2
]
,
B
=
[
−
2
4
6
7
]
then find
X
when
X
+
4
A
=
7
B
∫
18
38
X
+
4
A
=
7
B
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x
+
3
y
=
0
2
x
−
y
=
−
7
\begin{array}{l}x+3 y=0 \\ 2 x-y=-7\end{array}
x
+
3
y
=
0
2
x
−
y
=
−
7
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y
=
3
x
+
16
5
x
+
7
y
=
8
\begin{array}{l}y=3 x+16 \\ 5 x+7 y=8\end{array}
y
=
3
x
+
16
5
x
+
7
y
=
8
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3
x
−
5
y
=
9
y
=
3
x
−
9
\begin{array}{l}3 x-5 y=9 \\ y=3 x-9\end{array}
3
x
−
5
y
=
9
y
=
3
x
−
9
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y
=
7
x
−
11
y
=
−
4
x
\begin{array}{l}y=7 x-11 \\ y=-4 x\end{array}
y
=
7
x
−
11
y
=
−
4
x
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XL| Solve two-step equations
\newline
ixl.com/math/grade
−
7
-7
−
7
/solve-two-step-equations
\newline
24
24
24
\newline
My IXL
\newline
Learning
\newline
Assessmen
\newline
Seventh grade
\newline
T.
9
9
9
Solve two-step equations
\newline
QEB
\newline
Solve for
\newline
u
u
u
.
\newline
(
u
−
63
6
)
=
5
\left(\frac{u-63}{6}\right)=5
(
6
u
−
63
)
=
5
,
u
=
□
u=\square
u
=
□
\newline
Submit
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