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Math Problems
Algebra 1
Solve linear equations with variables on both sides
Solve for
m
m
m
.
\newline
−
7
+
4
m
+
10
=
15
−
2
m
m
=
□
\begin{array}{l} -7+4 m+10=15-2 m \\ m=\square \end{array}
−
7
+
4
m
+
10
=
15
−
2
m
m
=
□
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Solve for
t
t
t
.
\newline
16
−
2
t
=
t
+
9
+
4
t
t
=
□
\begin{array}{l} 16-2 t=t+9+4 t \\ t=\square \end{array}
16
−
2
t
=
t
+
9
+
4
t
t
=
□
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Solve for
d
d
d
.
\newline
2
d
+
4
=
10
+
5
d
d
=
□
\begin{array}{l} 2 d+4=10+5 d \\ d=\square \end{array}
2
d
+
4
=
10
+
5
d
d
=
□
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Solve for
c
c
c
.
\newline
12
c
−
4
=
14
c
−
10
12 c-4=14 c-10
12
c
−
4
=
14
c
−
10
\newline
c
=
c=
c
=
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Solve for
a
a
a
.
\newline
5
+
14
a
=
9
a
−
5
a
=
□
\begin{array}{l} 5+14 a=9 a-5 \\ a=\square \end{array}
5
+
14
a
=
9
a
−
5
a
=
□
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Solve for
b
b
b
.
\newline
4
b
+
5
=
1
+
5
b
b
=
□
\begin{array}{l} 4 b+5=1+5 b \\ b=\square \end{array}
4
b
+
5
=
1
+
5
b
b
=
□
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Solve for
d
d
d
.
\newline
4
d
−
4
=
5
d
−
8
4 d-4=5 d-8
4
d
−
4
=
5
d
−
8
\newline
d
=
d=
d
=
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Solve for
n
n
n
.
\newline
2
−
2
n
=
3
n
+
17
n
=
□
\begin{array}{l} 2-2 n=3 n+17 \\ n=\square \end{array}
2
−
2
n
=
3
n
+
17
n
=
□
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Solve for
a
a
a
.
\newline
a
−
15
=
4
a
−
3
a
=
□
\begin{array}{l} a-15=4 a-3 \\ a=\square \end{array}
a
−
15
=
4
a
−
3
a
=
□
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Solve for
f
f
f
.
\newline
2
+
1.25
f
=
10
−
2.75
f
f
=
□
\begin{array}{l} 2+1.25 f=10-2.75 f \\ f=\square \end{array}
2
+
1.25
f
=
10
−
2.75
f
f
=
□
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Solve for
y
y
y
.
\newline
9
4
y
−
12
=
1
4
y
−
4
y
=
□
\begin{array}{l} \frac{9}{4} y-12=\frac{1}{4} y-4 \\ y=\square \end{array}
4
9
y
−
12
=
4
1
y
−
4
y
=
□
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Solve for
k
k
k
.
\newline
4.5
+
1.5
k
=
18
−
3
k
k
=
□
\begin{array}{l} 4.5+1.5 k=18-3 k \\ k=\square \end{array}
4.5
+
1.5
k
=
18
−
3
k
k
=
□
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Solve for
c
c
c
.
\newline
5
c
+
16.5
=
13.5
+
10
c
c
=
□
\begin{array}{l} 5 c+16.5=13.5+10 c \\ c=\square \end{array}
5
c
+
16.5
=
13.5
+
10
c
c
=
□
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Solve for
g
g
g
.
\newline
9
+
3.5
g
=
11
−
0.5
g
g
=
□
\begin{array}{l} 9+3.5 g=11-0.5 g \\ g=\square \end{array}
9
+
3.5
g
=
11
−
0.5
g
g
=
□
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Solve for
n
n
n
.
\newline
2
−
1
2
n
=
3
n
+
16
n
=
□
\begin{array}{l} 2-\frac{1}{2} n=3 n+16 \\ n=\square \end{array}
2
−
2
1
n
=
3
n
+
16
n
=
□
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Solve for
m
m
m
.
\newline
12.6
+
4
m
=
9.6
+
8
m
m
=
□
\begin{array}{l} 12.6+4 m=9.6+8 m \\ m=\square \end{array}
12.6
+
4
m
=
9.6
+
8
m
m
=
□
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Solve for
d
d
d
.
\newline
2
d
+
4
=
10
+
2.5
d
d
=
□
\begin{array}{l} 2 d+4=10+2.5 d \\ d=\square \end{array}
2
d
+
4
=
10
+
2.5
d
d
=
□
Get tutor help
Solve for
s
s
s
.
\newline
0.5
s
+
1
=
7
+
4.5
s
s
=
□
\begin{array}{l} 0.5 s+1=7+4.5 s \\ s=\square \end{array}
0.5
s
+
1
=
7
+
4.5
s
s
=
□
Get tutor help
Solve for
r
r
r
.
\newline
12
−
1
5
r
=
2
r
+
1
r
=
□
\begin{array}{l} 12-\frac{1}{5} r=2 r+1 \\ r=\square \end{array}
12
−
5
1
r
=
2
r
+
1
r
=
□
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Solve for
k
k
k
.
\newline
20
−
6
+
4
k
=
2
−
2
k
k
=
□
\begin{array}{l} 20-6+4 k=2-2 k \\ k=\square \end{array}
20
−
6
+
4
k
=
2
−
2
k
k
=
□
Get tutor help
Solve for
b
b
b
.
\newline
2
3
b
+
5
=
20
−
b
b
=
□
\begin{array}{l} \frac{2}{3} b+5=20-b \\ b=\square \end{array}
3
2
b
+
5
=
20
−
b
b
=
□
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Solve for
c
c
c
.
\newline
6
c
+
14
=
−
5
c
+
4
+
9
c
c
=
\begin{array}{l} 6 c+14=-5 c+4+9 c \\ c= \end{array}
6
c
+
14
=
−
5
c
+
4
+
9
c
c
=
Get tutor help
Solve for
p
p
p
.
\newline
16
−
3
p
=
2
3
p
+
5
p
=
\begin{array}{l} 16-3 p=\frac{2}{3} p+5 \\ p= \end{array}
16
−
3
p
=
3
2
p
+
5
p
=
Get tutor help
Solve for
t
t
t
.
\newline
16
−
2
t
=
3
2
t
+
9
t
=
□
\begin{array}{l} 16-2 t=\frac{3}{2} t+9 \\ t=\square \end{array}
16
−
2
t
=
2
3
t
+
9
t
=
□
Get tutor help
Solve for
a
a
a
.
\newline
4
a
+
5
=
2
+
3.25
a
a
=
□
\begin{array}{l} 4 a+5=2+3.25 a \\ a=\square \end{array}
4
a
+
5
=
2
+
3.25
a
a
=
□
Get tutor help
Solve for
c
c
c
.
\newline
8
3
c
−
2
=
2
3
c
−
12
c
=
□
\begin{array}{l} \frac{8}{3} c-2=\frac{2}{3} c-12 \\ c=\square \end{array}
3
8
c
−
2
=
3
2
c
−
12
c
=
□
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Solve for
a
a
a
.
\newline
−
1
4
a
−
4
=
7
4
a
−
3
a
=
□
\begin{array}{l} -\frac{1}{4} a-4=\frac{7}{4} a-3 \\ a=\square \end{array}
−
4
1
a
−
4
=
4
7
a
−
3
a
=
□
Get tutor help
Solve for
d
d
d
.
\newline
6
d
−
11
2
=
2
d
−
13
2
d
=
\begin{array}{l} 6 d-\frac{11}{2}=2 d-\frac{13}{2} \\ d= \end{array}
6
d
−
2
11
=
2
d
−
2
13
d
=
Get tutor help
Solve for
e
e
e
.
\newline
9
e
−
7
=
7
e
−
11
e
=
□
\begin{array}{l} 9 e-7=7 e-11 \\ e=\square \end{array}
9
e
−
7
=
7
e
−
11
e
=
□
Get tutor help
Solve for
x
x
x
.
\newline
4
x
+
8
=
7.2
+
5
x
x
=
□
\begin{array}{l} 4 x+8=7.2+5 x \\ x=\square \end{array}
4
x
+
8
=
7.2
+
5
x
x
=
□
Get tutor help
Solve for
f
f
f
.
\newline
−
f
+
2
+
4
f
=
8
−
3
f
f
=
□
\begin{array}{l} -f+2+4 f=8-3 f \\ f=\square \end{array}
−
f
+
2
+
4
f
=
8
−
3
f
f
=
□
Get tutor help
Solve for
s
s
s
.
\newline
8
−
4
s
=
s
+
13
s
=
□
\begin{array}{l} 8-4 s=s+13 \\ s=\square \end{array}
8
−
4
s
=
s
+
13
s
=
□
Get tutor help
Solve for
b
b
b
.
\newline
7
b
−
15
=
5
b
−
3
b
=
\begin{array}{l} 7 b-15=5 b-3 \\ b= \end{array}
7
b
−
15
=
5
b
−
3
b
=
Get tutor help
Solve for
h
h
h
.
\newline
15.3
+
1
h
=
1.3
−
1
h
h
=
□
\begin{array}{l} 15.3+1 h=1.3-1 h \\ h=\square \end{array}
15.3
+
1
h
=
1.3
−
1
h
h
=
□
Get tutor help
Solve for
b
b
b
.
\newline
7
b
−
2
5
=
6
b
−
7
5
b
=
□
\begin{array}{l} 7 b-\frac{2}{5}=6 b-\frac{7}{5} \\ b=\square \end{array}
7
b
−
5
2
=
6
b
−
5
7
b
=
□
Get tutor help
Solve for
h
h
h
.
\newline
17
+
4
h
+
2
=
1
−
5
h
h
=
□
\begin{array}{l} 17+4 h+2=1-5 h \\ h=\square \end{array}
17
+
4
h
+
2
=
1
−
5
h
h
=
□
Get tutor help
Solve for
s
s
s
.
\newline
2
−
2
s
=
3
4
s
+
13
s
=
□
\begin{array}{l} 2-2 s=\frac{3}{4} s+13 \\ s=\square \end{array}
2
−
2
s
=
4
3
s
+
13
s
=
□
Get tutor help
Solve for
g
g
g
.
\newline
−
3
+
5
+
6
g
=
11
−
3
g
g
=
□
\begin{array}{l} -3+5+6 g=11-3 g \\ g=\square \end{array}
−
3
+
5
+
6
g
=
11
−
3
g
g
=
□
Get tutor help
Solve for
e
e
e
.
\newline
9
e
+
4
=
−
5
e
+
14
+
13
e
e
=
□
\begin{array}{l} 9 e+4=-5 e+14+13 e \\ e=\square \end{array}
9
e
+
4
=
−
5
e
+
14
+
13
e
e
=
□
Get tutor help
Solve for
p
p
p
.
\newline
17
−
2
p
=
2
p
+
5
+
2
p
p
=
□
\begin{array}{l} 17-2 p=2 p+5+2 p \\ p=\square \end{array}
17
−
2
p
=
2
p
+
5
+
2
p
p
=
□
Get tutor help
Solve for
r
r
r
.
\newline
16
−
2
r
=
−
3
r
+
6
r
+
1
r
=
□
\begin{array}{l} 16-2 r=-3 r+6 r+1 \\ r=\square \end{array}
16
−
2
r
=
−
3
r
+
6
r
+
1
r
=
□
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17
+
27
=
−
7
w
17+27=-7 w
17
+
27
=
−
7
w
\newline
Given the equation, what is the value of
10
(
−
21
w
+
1
)
10(-21 w+1)
10
(
−
21
w
+
1
)
?
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41
=
12
d
−
7
41=12 d-7
41
=
12
d
−
7
\newline
What is the value of
d
d
d
in the equation shown?
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−
5
x
+
1
=
−
6
-5 x+1=-6
−
5
x
+
1
=
−
6
\newline
What is the solution to the equation shown?
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Jaquan passed a checkpoint jogging at a constant speed of
9
km
h
9\frac{\text{km}}{\text{h}}
9
h
km
. Then,
2
2
2
minutes later, Odalis passed the same checkpoint. If Odalis runs at a constant speed of
12
km
h
12\frac{\text{km}}{\text{h}}
12
h
km
, then how far past the checkpoint will Odalis be when they catch up to Jaquan?
\newline
km
\text{km}
km
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Xóchitl, who has already warmed up, immediately starts to pedal at a rate of
18
km
h
18\frac{\text{km}}{\text{h}}
18
h
km
on a stationary bike.
\newline
She sees that her friend Cowessess has already pedaled
12
12
12
minutes at an average rate of
10
km
h
10\frac{\text{km}}{\text{h}}
10
h
km
.
\newline
Assuming Cowessess's rate stays the same, how long would Xóchitl have to pedal to catch up to Cowessess's distance?
\newline
minutes
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Marcel plugged in his work tablet and phone. The phone had a battery charge of
13
%
13\%
13%
and started increasing by
2
2
2
percentage points every
3
3
3
minutes. The tablet had charge of
25
%
25\%
25%
and started increasing by
1
1
1
percentage point every
3
3
3
minutes.
\newline
Let
t
t
t
represent the time, in minutes, since Marcel plugged in the phone and tablet.
\newline
Complete the inequality to represent the times when the phone would have at least as much battery charge as the tablet.
\newline
t
t
t
select inequality symbol
□
\square
□
minutes
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A computer is rendering two scenes. The computer has already rendered
540
540
540
thousand pixels (
k
p
x
kpx
k
p
x
) of a kitchen scene and renders another
90
k
p
x
90kpx
90
k
p
x
each minute. The computer has rendered
960
k
p
x
960kpx
960
k
p
x
of a garden scene and renders
60
k
p
x
60kpx
60
k
p
x
more pixels each minute. After how many more minutes will the scenes have the same amounts rendered? minutes
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Aspen has
306
306
306
green beads and
210
210
210
silver beads. For every
5
cm
5\text{cm}
5
cm
of a belt that Aspen sews, they use
40
40
40
green beads and
24
24
24
silver beads. After how many centimeters of the belt will Aspen have equal numbers of green and silver beads left?
cm
\text{cm}
cm
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Chiamaka is
149
km
149\text{km}
149
km
from the university and drives
95
km
95\text{km}
95
km
closer every hour. Valente is
170
km
170\text{km}
170
km
from the university and drives
110
km
110\text{km}
110
km
closer every hour. Let
t
t
t
represent the time, in hours, since Chiamaka and Valente started driving toward the university. Complete the inequality to represent the times when Valente is closer than Chiamaka to the university.
t
t
t
select inequality symbol hours
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