z/2-z/7+31=7z
solve
Answers
How to solve your problem
2
−
1
⋅
7
+
3
1
=
7
\frac{z}{2}-1 \cdot \frac{z}{7}+31=7z
2z−1⋅7z+31=7z
Solve
1
Find common denominator
2
+
−
7
+
3
1
=
7
\frac{z}{2}+\frac{-z}{7}+31=7z
2z+7−z+31=7z
7
1
4
+
2
(
−
1
)
1
4
+
1
4
⋅
3
1
1
4
=
7
\frac{7z}{14}+\frac{2\left(-1\right)z}{14}+\frac{14 \cdot 31}{14}=7z
147z+142(−1)z+1414⋅31=7z
2
Combine fractions with common denominator
3
Multiply the numbers
4
Multiply the numbers
5
Combine like terms
6
Subtract
7
7z
7z
from both sides of the equation
7
Simplify
8
Multiply all terms by the same value to eliminate fraction denominators
9
Cancel multiplied terms that are in the denominator
10
Subtract
4
3
4
434
434
from both sides of the equation
11
Simplify
12
Divide both sides of the equation by the same term
13
Simplify
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Solution
=
1
4
3
Answer:
Step-by-step explanation:
2
−
1
⋅
7
+
3
1
=
7
2
+
−
7
+
3
1
=
7
\frac{z}{2}+\frac{-z}{7}+31=7z
2z+7−z+31=7z
7
1
4
+
2
(
−
1
)
1
4
+
1
4
⋅
3
1
1
4
=
7
7
1
4
+
2
(
−
1
)
1
4
+
1
4
⋅
3
1
1
4
=
7
\frac{7z}{14}+\frac{2\left(-1\right)z}{14}+\frac{14 \cdot 31}{14}=7z
147z+142(−1)z+1414⋅31=7z
7
+
2
(
−
1
)
+
1
4
⋅
3
1
1
4
=
7
7
+
2
(
−
1
)
+
1
4
⋅
3
1
1
4
=
7
\frac{7z+{\color{#c92786}{2}}\left({\color{#c92786}{-1}}\right)z+14 \cdot 31}{14}=7z
147z+2(−1)z+14⋅31=7z
7
−
2
+
1
4
⋅
3
1
1
4
=
7
=
1
4
3