Resources
Testimonials
Plans
Sign in
Sign up
Resources
Testimonials
Plans
Home
Math Problems
Grade 8
Write an equation word problem
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
=
??
Get tutor help
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
]
⇒
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
Get tutor help
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
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
De. XLI 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
>
≠
>\neq
>
=
T.
9
9
9
Solve two-step equations QEB
\newline
Solve for
u
u
u
.
\newline
u
−
63
6
=
5
u
=
□
\begin{array}{l} \frac{u-63}{6}=5 \\ u=\square \end{array}
6
u
−
63
=
5
u
=
□
\newline
Submit
Get tutor help
T.
9
9
9
Solve two-step equations QEB
\newline
Solve for
j
j
j
.
\newline
j
4
−
2
=
2
j
=
□
\begin{array}{l} \frac{j}{4}-2=2 \\ j=\square \end{array}
4
j
−
2
=
2
j
=
□
\newline
Submit
Get tutor help
16
16
16
.
\newline
y
−
4
=
3
(
x
+
2
)
2
x
+
6
y
=
10
\begin{array}{l} y-4=3(x+2) \\ 2 x+6 y=10 \end{array}
y
−
4
=
3
(
x
+
2
)
2
x
+
6
y
=
10
\newline
Write an equation in slope-ir given point and is perpendic
\newline
19
19
19
.
(
0
,
0
)
;
y
=
−
3
x
+
2
(0,0) ; y=-3 x+2
(
0
,
0
)
;
y
=
−
3
x
+
2
\newline
22
22
22
.
(
−
3
,
2
)
;
x
−
2
y
=
7
(-3,2) ; x-2 y=7
(
−
3
,
2
)
;
x
−
2
y
=
7
\newline
25
25
25
. Urban Planning A path connect the park entran path should be perpend
Get tutor help
−
3
x
−
6
y
=
45
−
5
y
=
x
\begin{aligned}-3 x-6 y & =45 \\ -5 y & =x\end{aligned}
−
3
x
−
6
y
−
5
y
=
45
=
x
Get tutor help
{
x
+
2
y
=
−
1
x
+
y
=
2
\left\{\begin{array}{l}x+2 y=-1 \\ x+y=2\end{array}\right.
{
x
+
2
y
=
−
1
x
+
y
=
2
Get tutor help
n
−
18
=
3
21
15
22
54
\begin{array}{l}n-18=3 \\ 21 \\ 15 \\ 22 \\ 54 \\\end{array}
n
−
18
=
3
21
15
22
54
Get tutor help
y
=
3
x
+
1
y
=
−
1
2
x
+
4
\begin{array}{l} y=3 x+1 \\ y=-\frac{1}{2} x+4 \end{array}
y
=
3
x
+
1
y
=
−
2
1
x
+
4
\newline
Select Choice
Get tutor help
Solve each of the following systems of linear equations by the given method.
\newline
Write your answers CLEARLY as an ordered pair (X,Y). For example:
(
1
,
2
)
(1,2)
(
1
,
2
)
.
\newline
(
4
4
4
points each)
\newline
1
1
1
) Solve with Substitution:
\newline
2
2
2
) Solve with Elimination:
\newline
y
=
2
x
−
7
y
=
−
x
+
8
\begin{array}{l} y=2 x-7 \\ y=-x+8 \end{array}
y
=
2
x
−
7
y
=
−
x
+
8
\newline
3
x
+
3
y
=
6
−
3
x
+
7
y
=
−
26
\begin{array}{c} 3 x+3 y=6 \\ -3 x+7 y=-26 \end{array}
3
x
+
3
y
=
6
−
3
x
+
7
y
=
−
26
\newline
3
3
3
) Solve with Substitution:
\newline
4
4
4
) Solve with Elimination:
\newline
4.5
x
−
6
y
=
30
y
=
1.5
x
−
20
\begin{array}{c} 4.5 x-6 y=30 \\ y=1.5 x-20 \end{array}
4.5
x
−
6
y
=
30
y
=
1.5
x
−
20
\newline
8
x
+
14
y
=
4
−
6
x
−
7
y
=
−
10
\begin{array}{c} 8 x+14 y=4 \\ -6 x-7 y=-10 \end{array}
8
x
+
14
y
=
4
−
6
x
−
7
y
=
−
10
Get tutor help
Solve for
k
k
k
.
\newline
5
k
−
2
3
k
=
7
k
=
□
\begin{array}{l} \frac{5 k-2}{3 k}=7 \\ k=\square \end{array}
3
k
5
k
−
2
=
7
k
=
□
Get tutor help
−
2
v
+
12
=
2
v
=
?
\begin{array}{l}-2 v+12=2 \\ v=?\end{array}
−
2
v
+
12
=
2
v
=
?
Get tutor help
a
3
+
9
=
22
a
=
?
?
\begin{array}{l}\frac{a}{3}+9=22 \\ a=? ?\end{array}
3
a
+
9
=
22
a
=
??
Get tutor help
Investigate:
\newline
Consider the system of equations
\newline
y
=
4
x
2
−
9
x
+
20
y
=
15
x
+
k
\begin{array}{l} y=4 x^{2}-9 x+20 \\ y=15 x+k \end{array}
y
=
4
x
2
−
9
x
+
20
y
=
15
x
+
k
\newline
There are two values of
x
x
x
which are solutions to the system, one of which is three times as large as the other.
\newline
Determine the value of
k
k
k
and hence the solution to the system.
Get tutor help
2
a
+
b
=
6
b
=
21
−
a
2
\begin{array}{l}2 a+b=6 \\ b=21-a^{2}\end{array}
2
a
+
b
=
6
b
=
21
−
a
2
Get tutor help
3
a
+
b
=
7
b
=
25
−
a
2
\begin{array}{l}3 a+b=7 \\ b=25-a^{2}\end{array}
3
a
+
b
=
7
b
=
25
−
a
2
Get tutor help
2
a
+
b
=
8
b
=
43
−
a
2
\begin{array}{l}2 a+b=8 \\ b=43-a^{2}\end{array}
2
a
+
b
=
8
b
=
43
−
a
2
Get tutor help
bcps.schoology.com
\newline
Show What You Know: Systems of Equations
\newline
At what point do these two lines intersect? Find the solution to this system of equations:
\newline
y
=
−
3
x
−
3
y
=
4
x
+
4
\begin{array}{l} y=-3 x-3 \\ y=4 x+4 \end{array}
y
=
−
3
x
−
3
y
=
4
x
+
4
Get tutor help
Solve the system by the elimination method.
\newline
−
8
x
+
2
y
=
−
10
2
x
−
2
y
=
10
\begin{array}{c} -8 x+2 y=-10 \\ 2 x-2 y=10 \end{array}
−
8
x
+
2
y
=
−
10
2
x
−
2
y
=
10
Get tutor help
1
1
1
) Solve the system of linear equations using elimination by multiplication.
\newline
{
2
x
+
4
y
=
−
4
3
x
=
−
5
y
−
3
\left\{\begin{array}{l} 2 x+4 y=-4 \\ 3 x=-5 y-3 \end{array}\right.
{
2
x
+
4
y
=
−
4
3
x
=
−
5
y
−
3
Get tutor help
3
3
3
.
y
=
x
+
2
y
=
−
x
+
2
\begin{array}{l}y=x+2 \\ y=-x+2\end{array}
y
=
x
+
2
y
=
−
x
+
2
Get tutor help
{:\begin{align*}[
3
3
3
x+
2
2
2
y&=
4
4
4
(x-y
−
6
-6
−
6
)],\[\(6\)(y+x)&=\(7\)x\(-24\)]:\end{align*}\(\newline\)Which of the following accurately describes all solutions to the system of equations shown?\(\newline\)Choose \(1\) answer:\(\newline\)(A) \(\newline\)\(x=3\) and \(\newline\)\(y=-1\)\(\newline\)(B) \(\newline\)\(x=12\) and \(\newline\)\(y=-2\)\(\newline\)(C) There are infinite solutions to the system.\(\newline\)(D) There are no solutions to the system.
Get tutor help
(
10
solution)
undefined
solve for
x
.
2
5
(
3
x
−
1
)
−
6
=
−
5
(
2
x
−
2
)
−
5
2
(
3
x
−
1
)
−
30
=
−
2502
\begin{array}{l}\underbrace{(10 \text { solution) }}_{\text {solve for } x \text {. }} \\ \frac{2}{5}(3 x-1)-6=-5(2 x-2) \\ -5 \\ 2(3 x-1)-30=-2502 \\\end{array}
solve for
x
.
(
10
solution)
5
2
(
3
x
−
1
)
−
6
=
−
5
(
2
x
−
2
)
−
5
2
(
3
x
−
1
)
−
30
=
−
2502
Get tutor help
f
(
x
)
=
5
x
−
3
f
(
5
)
=
□
\begin{array}{l}f(x)=5 x-3 \\ f(5)=\square\end{array}
f
(
x
)
=
5
x
−
3
f
(
5
)
=
□
Get tutor help
5
{
y
=
−
11
x
+
20
6
x
+
2
y
=
8
5\left\{\begin{array}{l}y=-11 x+20 \\ 6 x+2 y=8\end{array}\right.
5
{
y
=
−
11
x
+
20
6
x
+
2
y
=
8
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
y
=
3
x
2
y
2
−
x
y
=
15
\begin{array}{l}y=3 x \\ 2 y^{2}-x y=15\end{array}
y
=
3
x
2
y
2
−
x
y
=
15
Get tutor help
3
a
+
2
b
=
4
4
a
−
b
=
−
13
\begin{array}{l}3 a+2 b=4 \\ 4 a-b=-13\end{array}
3
a
+
2
b
=
4
4
a
−
b
=
−
13
Get tutor help
b)Write an equation for the sum of their ages and determine how old each girl is.
\newline
8
8
8
. An isosceles triangle and an equilateral triangle have the same perimeter. Find the length of the longest side of the isosceles triangle.
\newline
3
x
+
1
=
x
+
3
x
3
x
−
x
=
3
−
1
2
x
=
2
\begin{aligned} 3 x+1 & =x+3 x \\ & 3 x-x=3-1 \\ & 2 x=2 \end{aligned}
3
x
+
1
=
x
+
3
x
3
x
−
x
=
3
−
1
2
x
=
2
Get tutor help
Solve the following system of equations.
\newline
−
5
x
−
6
y
=
16
7
x
−
5
y
=
−
9
\begin{array}{l} -5 x-6 y=16 \\ 7 x-5 y=-9 \end{array}
−
5
x
−
6
y
=
16
7
x
−
5
y
=
−
9
Get tutor help
1
1
1
:
21
21
21
PM Mon Apr
15
15
15
\newline
Exit
\newline
M.
14
14
14
Solve a right triangle
\newline
0
0
0
\newline
Solve the right triangle.
\newline
Write your answers in simplified, rationalized form. Do not round.
\newline
H
J
=
□
H
I
=
□
m
∠
J
=
□
。
\begin{aligned} H J & =\square \\ H I & =\square \\ m \angle J & =\square 。 \end{aligned}
H
J
H
I
m
∠
J
=
□
=
□
=
□
。
Get tutor help
−
2
y
=
4
x
−
18
y
=
2
x
+
9
\begin{array}{l} -2 y=4 x-18 \\ y=2 x+9 \end{array}
−
2
y
=
4
x
−
18
y
=
2
x
+
9
\newline
Ive each system of equations. Shch
\newline
8
8
8
.
{
x
+
y
=
4
2
x
−
5
y
=
15
\left\{\begin{array}{l}x+y=4 \\ 2 x-5 y=15\end{array}\right.
{
x
+
y
=
4
2
x
−
5
y
=
15
Get tutor help
Solve for
b
b
b
.
\newline
3
+
2
b
=
5
b
=
\begin{array}{l} 3+2 b=5 \\ b= \end{array}
3
+
2
b
=
5
b
=
\newline
Submit
Get tutor help
24
−
6
y
=
2
x
6
(
y
−
2
)
=
3
+
x
\begin{array}{c} 24-6 y=2 x \\ 6(y-2)=3+x \end{array}
24
−
6
y
=
2
x
6
(
y
−
2
)
=
3
+
x
\newline
Consider the system of equations. If
(
x
,
y
)
(x, y)
(
x
,
y
)
is the solution to the system, then
Get tutor help
3
3
3
. Gabby and Chris went on a scavenger hunt. Together they collected a total of
24
24
24
items. Chris collected
3
3
3
less than twice as many items as Gabby collected. Let
g
g
g
be the number of items Gabby collected and
c
\mathrm{c}
c
be the number of items Chris collected. Which system of equations can be used to find the number of items each child collected?
\newline
A.
\newline
A.
=
2
g
−
3
c
+
g
=
24
\begin{array}{l} \text { A. }=2 g-3 \\ c+g=24 \end{array}
A.
=
2
g
−
3
c
+
g
=
24
\newline
c.
\newline
2
c
+
3
g
=
24
c
−
g
=
3
\begin{array}{l} 2 c+3 g=24 \\ c-g=3 \end{array}
2
c
+
3
g
=
24
c
−
g
=
3
\newline
B.
g
=
2
c
+
3
c
+
g
=
24
\text { B. } \begin{aligned} g & =2 c+3 \\ c & +g=24 \end{aligned}
B.
g
c
=
2
c
+
3
+
g
=
24
\newline
D.
g
=
2
c
−
3
c
+
g
=
24
\text { D. } \begin{array}{r} g=2 c-3 \\ c+g=24 \end{array}
D.
g
=
2
c
−
3
c
+
g
=
24
Get tutor help
A
X
=
14
,
C
X
=
9
B
X
=
18
,
X
D
=
?
\begin{array}{l}A X=14, C X=9 \\ B X=18, X D=?\end{array}
A
X
=
14
,
CX
=
9
BX
=
18
,
X
D
=
?
Get tutor help
How many solutions does the system have?
\newline
{
y
=
5
x
+
1
y
=
−
2
x
−
8
\left\{\begin{array}{l} y=5 x+1 \\ y=-2 x-8 \end{array}\right.
{
y
=
5
x
+
1
y
=
−
2
x
−
8
\newline
Choose
1
1
1
answer:
\newline
(A) Exactly one solution
\newline
(B) No solutions
\newline
(C) Infinitely many solutions
Get tutor help
Previous
1
...
2
3
4
...
8
Next