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Math Problems
Grade 6
Multiply using the distributive property
Reduce to simplest form.
\newline
12
11
−
(
−
2
3
)
=
\frac{12}{11}-\left(-\frac{2}{3}\right)=
11
12
−
(
−
3
2
)
=
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Reduce to simplest form.
\newline
−
3
4
+
7
2
=
-\frac{3}{4}+\frac{7}{2}=
−
4
3
+
2
7
=
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Reduce to simplest form.
\newline
−
9
6
+
(
−
3
5
)
=
-\frac{9}{6}+\left(-\frac{3}{5}\right)=
−
6
9
+
(
−
5
3
)
=
Get tutor help
Reduce to simplest form.
\newline
−
3
7
+
5
2
=
-\frac{3}{7}+\frac{5}{2}=
−
7
3
+
2
5
=
Get tutor help
Reduce to simplest form.
\newline
−
1
3
+
(
−
7
4
)
=
-\frac{1}{3}+\left(-\frac{7}{4}\right)=
−
3
1
+
(
−
4
7
)
=
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Reduce to simplest form.
\newline
−
3
5
+
1
3
=
-\frac{3}{5}+\frac{1}{3}=
−
5
3
+
3
1
=
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Reduce to simplest form.
\newline
5
3
+
(
−
7
6
)
=
\frac{5}{3}+\left(-\frac{7}{6}\right)=
3
5
+
(
−
6
7
)
=
Get tutor help
Reduce to simplest form.
\newline
6
3
+
(
−
1
6
)
=
\frac{6}{3}+\left(-\frac{1}{6}\right)=
3
6
+
(
−
6
1
)
=
Get tutor help
Reduce to simplest form.
\newline
−
5
9
+
(
−
7
12
)
=
-\frac{5}{9}+\left(-\frac{7}{12}\right)=
−
9
5
+
(
−
12
7
)
=
Get tutor help
Reduce to simplest form.
\newline
−
3
5
+
(
−
8
2
)
=
-\frac{3}{5}+\left(-\frac{8}{2}\right)=
−
5
3
+
(
−
2
8
)
=
Get tutor help
Reduce to simplest form.
\newline
−
7
8
+
(
−
1
2
)
=
-\frac{7}{8}+\left(-\frac{1}{2}\right)=
−
8
7
+
(
−
2
1
)
=
Get tutor help
Reduce to simplest form.
\newline
−
3
4
−
(
−
1
6
)
=
-\frac{3}{4}-\left(-\frac{1}{6}\right)=
−
4
3
−
(
−
6
1
)
=
Get tutor help
Reduce to simplest form.
\newline
−
3
7
+
(
−
3
4
)
=
-\frac{3}{7}+\left(-\frac{3}{4}\right)=
−
7
3
+
(
−
4
3
)
=
Get tutor help
Reduce to simplest form.
\newline
8
6
+
(
−
9
5
)
=
\frac{8}{6}+\left(-\frac{9}{5}\right)=
6
8
+
(
−
5
9
)
=
Get tutor help
Reduce to simplest form.
\newline
3
2
+
(
−
6
5
)
=
\frac{3}{2}+\left(-\frac{6}{5}\right)=
2
3
+
(
−
5
6
)
=
Get tutor help
Solve for
k
k
k
.
\newline
k
4
=
3
8
k
=
□
\begin{array}{l} \frac{k}{4}=\frac{3}{8} \\ k=\square \end{array}
4
k
=
8
3
k
=
□
Get tutor help
Solve for
k
k
k
.
\newline
3
k
=
4
5
k
=
□
\begin{array}{l} \frac{3}{k}=\frac{4}{5} \\ k=\square \end{array}
k
3
=
5
4
k
=
□
Get tutor help
Solve for
p
p
p
.
\newline
8
9
=
12
p
p
=
\begin{array}{l} \frac{8}{9}=\frac{12}{p} \\ p= \end{array}
9
8
=
p
12
p
=
Get tutor help
Solve for
t
t
t
.
\newline
4
3
=
t
7
t
=
□
\begin{array}{l} \frac{4}{3}=\frac{t}{7} \\ t=\square \end{array}
3
4
=
7
t
t
=
□
Get tutor help
Solve for
k
k
k
.
\newline
9
10
=
10
k
k
=
\begin{array}{l} \frac{9}{10}=\frac{10}{k} \\ k= \end{array}
10
9
=
k
10
k
=
Get tutor help
Solve for
q
q
q
.
\newline
10
9
=
3
q
q
=
□
\begin{array}{l} \frac{10}{9}=\frac{3}{q} \\ q=\square \end{array}
9
10
=
q
3
q
=
□
Get tutor help
Solve for
n
n
n
.
\newline
8
3
=
12
n
n
=
\begin{array}{l} \frac{8}{3}=\frac{12}{n} \\ n= \end{array}
3
8
=
n
12
n
=
Get tutor help
Solve for
y
y
y
.
\newline
3
11
=
6
y
y
=
\begin{array}{l} \frac{3}{11}=\frac{6}{y} \\ y= \end{array}
11
3
=
y
6
y
=
Get tutor help
Solve for
p
p
p
.
\newline
p
8
=
7
9
p
=
\begin{array}{l} \frac{p}{8}=\frac{7}{9} \\ p= \end{array}
8
p
=
9
7
p
=
Get tutor help
Solve for
k
k
k
.
\newline
4
k
=
6
11
k
=
\begin{array}{l} \frac{4}{k}=\frac{6}{11} \\ k= \end{array}
k
4
=
11
6
k
=
Get tutor help
Solve for
y
y
y
.
\newline
6
11
=
y
3
y
=
\begin{array}{l} \frac{6}{11}=\frac{y}{3} \\ y= \end{array}
11
6
=
3
y
y
=
Get tutor help
Solve for
n
n
n
.
\newline
12
5
=
n
8
n
=
\begin{array}{l} \frac{12}{5}=\frac{n}{8} \\ n= \end{array}
5
12
=
8
n
n
=
Get tutor help
Solve for
n
n
n
.
\newline
11
n
=
8
5
n
=
\begin{array}{l} \frac{11}{n}=\frac{8}{5} \\ n= \end{array}
n
11
=
5
8
n
=
Get tutor help
Solve for
k
k
k
.
\newline
10
3
=
k
9
k
=
\begin{array}{l} \frac{10}{3}=\frac{k}{9} \\ k= \end{array}
3
10
=
9
k
k
=
Get tutor help
Solve for
t
t
t
.
\newline
7
3
=
4
t
t
=
\begin{array}{l} \frac{7}{3}=\frac{4}{t} \\ t= \end{array}
3
7
=
t
4
t
=
Get tutor help
Solve for
k
k
k
.
\newline
8
7
=
12
k
k
=
\begin{array}{l} \frac{8}{7}=\frac{12}{k} \\ k= \end{array}
7
8
=
k
12
k
=
Get tutor help
Solve for
q
q
q
.
\newline
3
5
=
q
10
q
=
\begin{array}{l} \frac{3}{5}=\frac{q}{10} \\ q= \end{array}
5
3
=
10
q
q
=
Get tutor help
Solve for
p
p
p
.
\newline
7
p
=
8
9
p
=
\begin{array}{l} \frac{7}{p}=\frac{8}{9} \\ p= \end{array}
p
7
=
9
8
p
=
Get tutor help
Solve for
k
k
k
.
\newline
12
7
=
k
8
k
=
\begin{array}{l} \frac{12}{7}=\frac{k}{8} \\ k= \end{array}
7
12
=
8
k
k
=
Get tutor help
Solve for
k
k
k
.
\newline
k
6
=
4
3
k
=
□
\begin{array}{l} \frac{k}{6}=\frac{4}{3} \\ k=\square \end{array}
6
k
=
3
4
k
=
□
Get tutor help
Solve for
k
k
k
.
\newline
11
8
=
k
4
k
=
\begin{array}{l} \frac{11}{8}=\frac{k}{4} \\ k= \end{array}
8
11
=
4
k
k
=
Get tutor help
Solve for
p
p
p
.
\newline
12
11
=
p
12
p
=
\begin{array}{l} \frac{12}{11}=\frac{p}{12} \\ p= \end{array}
11
12
=
12
p
p
=
Get tutor help
Solve for
x
x
x
.
\newline
x
5
=
8
9
x
=
□
\begin{array}{l} \frac{x}{5}=\frac{8}{9} \\ x=\square \end{array}
5
x
=
9
8
x
=
□
Get tutor help
Solve for
x
x
x
.
\newline
8
x
=
5
9
x
=
□
\begin{array}{l} \frac{8}{x}=\frac{5}{9} \\ x=\square \end{array}
x
8
=
9
5
x
=
□
Get tutor help
Solve for
k
k
k
.
\newline
4
3
=
11
k
k
=
\begin{array}{l} \frac{4}{3}=\frac{11}{k} \\ k= \end{array}
3
4
=
k
11
k
=
Get tutor help
Solve for
y
y
y
.
\newline
2
y
−
4
y
+
7
=
1
3
y
=
□
\begin{array}{l} \frac{2 y-4}{y+7}=\frac{1}{3} \\ y=\square \end{array}
y
+
7
2
y
−
4
=
3
1
y
=
□
Get tutor help
Solve for
k
k
k
.
\newline
6
2
k
−
6
=
1
3
k
=
□
\begin{array}{l} \frac{6}{2 k-6}=\frac{1}{3} \\ k=\square \end{array}
2
k
−
6
6
=
3
1
k
=
□
Get tutor help
Solve for
r
r
r
.
\newline
−
8
r
+
6
=
4
r
=
\begin{array}{l} \frac{-8}{r+6}=4 \\ r= \end{array}
r
+
6
−
8
=
4
r
=
Get tutor help
Solve for
p
p
p
.
\newline
2
p
+
1
5
p
+
6
=
1
7
p
=
□
\begin{array}{l} \frac{2 p+1}{5 p+6}=\frac{1}{7} \\ p=\square \end{array}
5
p
+
6
2
p
+
1
=
7
1
p
=
□
Get tutor help
Solve for
q
q
q
.
\newline
2
2
q
+
3
=
1
8
q
=
□
\begin{array}{l} \frac{2}{2 q+3}=\frac{1}{8} \\ q=\square \end{array}
2
q
+
3
2
=
8
1
q
=
□
Get tutor help
Solve for
q
q
q
.
\newline
10
2
q
+
8
=
1
10
q
=
□
\begin{array}{l} \frac{10}{2 q+8}=\frac{1}{10} \\ q=\square \end{array}
2
q
+
8
10
=
10
1
q
=
□
Get tutor help
Evaluate.
\newline
3
1
5
−
96
5
=
\frac{3^{\frac{1}{5}}}{\sqrt[5]{-96}}=
5
−
96
3
5
1
=
Get tutor help
Evaluate.
\newline
−
54
3
⋅
1
2
3
=
\sqrt[3]{-54} \cdot \sqrt[3]{\frac{1}{2}}=
3
−
54
⋅
3
2
1
=
Get tutor help
Evaluate.
\newline
−
16
5
⋅
2
5
=
\sqrt[5]{-16} \cdot \sqrt[5]{2}=
5
−
16
⋅
5
2
=
Get tutor help
Rewrite the expression in the form
b
n
b^{n}
b
n
.
\newline
b
1
5
b
=
□
\frac{b^{\frac{1}{5}}}{b}=\square
b
b
5
1
=
□
Get tutor help
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