Question

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Your question is
4

−
1
⋅
5
3
⋅

−
1
=
4
7
4x-1 \cdot \frac{5}{3} \cdot x-1=\frac{4}{7}
4x−1⋅35​⋅x−1=74​
Solve
1
Multiply the numbers
4

−
1
⋅
5
3

−
1
=
4
7
4x{\color{#c92786}{-1}} \cdot {\color{#c92786}{\frac{5}{3}}}x-1=\frac{4}{7}
4x−1⋅35​x−1=74​
4

−
5
3

−
1
=
4
7
4x{\color{#c92786}{-\frac{5}{3}}}x-1=\frac{4}{7}
4x−35​x−1=74​
2
Combine multiplied terms into a single fraction
4

−
5
3

−
1
=
4
7
4x-\frac{5}{3}x-1=\frac{4}{7}
4x−35​x−1=74​
4

+
−
5

3
−
1
=
4
7
4x+\frac{-5x}{3}-1=\frac{4}{7}
4x+3−5x​−1=74​
3
Add
1
1
1
to both sides of the equation
4

+
−
5

3
−
1
=
4
7
4x+\frac{-5x}{3}-1=\frac{4}{7}
4x+3−5x​−1=74​
4

+
−
5

3
−
1
+
1
=
4
7
+
1
4x+\frac{-5x}{3}-1+{\color{#c92786}{1}}=\frac{4}{7}+{\color{#c92786}{1}}
4x+3−5x​−1+1=74​+1
4
Simplify
Add the numbers
Add the numbers
4

+
−
5

3
=
1
1
7
4x+\frac{-5x}{3}=\frac{11}{7}
4x+3−5x​=711​
5
Multiply all terms by the same value to eliminate fraction denominators
4

+
−
5

3
=
1
1
7
4x+\frac{-5x}{3}=\frac{11}{7}
4x+3−5x​=711​
3
(
4

+
−
5

3
)
=
3
⋅
1
1
7
3(4x+\frac{-5x}{3})=3 \cdot \frac{11}{7}
3(4x+3−5x​)=3⋅711​
6
Simplify
Distribute
Cancel multiplied terms that are in the denominator
Combine like terms
Multiply the numbers
7

=
3
3
7
7x=\frac{33}{7}
7x=733​
7
Divide both sides of the equation by the same term
7

=
3
3
7
7x=\frac{33}{7}
7x=733​
7

7
=
3
3
7
7
\frac{7x}{{\color{#c92786}{7}}}=\frac{\frac{33}{7}}{{\color{#c92786}{7}}}
77x​=7733​​
8
Simplify
Cancel terms that are in both the numerator and denominator
Divide the numbers

=
3
3
4
9
x=\frac{33}{49}
x=4933​
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Solution

=
3
3
4
9​

Answer 1

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