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Please help 1/7x-3 5/7 > 4/7x-5 6/7

Sagot :

[tex]\frac { 1 }{ 7 } x-\left( 3+\frac { 5 }{ 7 } \right) >\frac { 4 }{ 7 } x-\left( 5+\frac { 6 }{ 7 } \right)[/tex]

[tex]\\ \\ -\left( 3+\frac { 5 }{ 7 } \right) +\left( 5+\frac { 6 }{ 7 } \right) >\frac { 4 }{ 7 } x-\frac { 1 }{ 7 } x[/tex]

[tex]\\ \\ -3-\frac { 5 }{ 7 } +5+\frac { 6 }{ 7 } >\frac { 3 }{ 7 } x\\ \\ 2+\frac { 1 }{ 7 } >\frac { 3 }{ 7 } x[/tex]

[tex]\\ \\ \frac { 14 }{ 7 } +\frac { 1 }{ 7 } >\frac { 3 }{ 7 } x\\ \\ \frac { 15 }{ 7 } >\frac { 3 }{ 7 } x[/tex]

[tex]\\ \\ \frac { 7 }{ 3 } \cdot \frac { 15 }{ 7 } >\frac { 3 }{ 7 } x\cdot \frac { 7 }{ 3 } \\ \\ \frac { 15 }{ 3 } >x\\ \\ \therefore \quad x<5[/tex]
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