Math, asked by mukesh4133, 2 months ago

Solve the equation 4x+5/4 +3x+ 5/6 >= 2X-7/2​

Answers

Answered by Anonymous
1

Given equation:-

\implies\sf 4x+\dfrac 5 4+3x+\dfrac5 6 =2x-\dfrac 7 2

\rule{200}{1}

Solution:-

\implies\sf 4x+\dfrac 5 4+3x+\dfrac5 6 =2x-\dfrac 7 2

Adding similar terms::

\implies\sf\Big (4x+3x\Big)+\Bigg(\dfrac 5 4+\dfrac5 6\Bigg) =2x-\dfrac 7 2

\implies\sf\Big (7x\Big)+\Bigg(\dfrac 5 4+\dfrac5 6\Bigg) =2x-\dfrac 7 2

Taking LCM::

\implies\sf 7x+\Bigg(\dfrac{15+10}{12}\Bigg) =2x-\dfrac 7 2

\implies\sf 7x+\Bigg(\dfrac{25}{12}\Bigg) =2x-\dfrac 7 2

Now transporting LHS and RHS::

\implies\sf\Big( 7x-2x\Big)+\Bigg(\dfrac{25}{12}+\dfrac 7 2\Bigg) =0

Solving by taking LCM::

\implies\sf\Big( 5x\Big)+\Bigg(\dfrac{25+42}{12}\Bigg) =0

\implies\sf\Big( 5x\Big)+\Bigg(\dfrac{67}{12}\Bigg) =0

\implies\sf5x=-\dfrac{67}{12}

\implies\sf x=-\dfrac{67}{12\times 5}

\underline{\boxed{\implies\bf{\red{ x=-\dfrac{67}{60}}}}}

» This is the required answer.

Answered by Mbappe007
1

Answer:

Given equation:-

\implies\sf 4x+\dfrac 5 4+3x+\dfrac5 6 =2x-\dfrac 7 2⟹4x+

4

5

+3x+

6

5

=2x−

2

7

\rule{200}{1}

Solution:-

\implies\sf 4x+\dfrac 5 4+3x+\dfrac5 6 =2x-\dfrac 7 2⟹4x+

4

5

+3x+

6

5

=2x−

2

7

Adding similar terms::

\implies\sf\Big (4x+3x\Big)+\Bigg(\dfrac 5 4+\dfrac5 6\Bigg) =2x-\dfrac 7 2⟹(4x+3x)+(

4

5

+

6

5

)=2x−

2

7

\implies\sf\Big (7x\Big)+\Bigg(\dfrac 5 4+\dfrac5 6\Bigg) =2x-\dfrac 7 2⟹(7x)+(

4

5

+

6

5

)=2x−

2

7

Taking LCM::

\implies\sf 7x+\Bigg(\dfrac{15+10}{12}\Bigg) =2x-\dfrac 7 2⟹7x+(

12

15+10

)=2x−

2

7

\implies\sf 7x+\Bigg(\dfrac{25}{12}\Bigg) =2x-\dfrac 7 2⟹7x+(

12

25

)=2x−

2

7

Now transporting LHS and RHS::

\implies\sf\Big( 7x-2x\Big)+\Bigg(\dfrac{25}{12}+\dfrac 7 2\Bigg) =0⟹(7x−2x)+(

12

25

+

2

7

)=0

Solving by taking LCM::

\implies\sf\Big( 5x\Big)+\Bigg(\dfrac{25+42}{12}\Bigg) =0⟹(5x)+(

12

25+42

)=0

\implies\sf\Big( 5x\Big)+\Bigg(\dfrac{67}{12}\Bigg) =0⟹(5x)+(

12

67

)=0

\implies\sf5x=-\dfrac{67}{12}⟹5x=−

12

67

\implies\sf x=-\dfrac{67}{12\times 5}⟹x=−

12×5

67

\underline{\boxed{\implies\bf{\red{ x=-\dfrac{67}{60}}}}}

⟹x=−

60

67

» This is the required answer.

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