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yaroslaw [1]
3 years ago
12

Please help me I’m lost

Mathematics
1 answer:
netineya [11]3 years ago
3 0

We are given with an equation and need to find it's solution set also we are given with a condition that <em>m ≠ {0, 1}</em> . So , let's start ;

So we are given with :

{:\implies \quad \sf \dfrac{6}{m}+2=\dfrac{m+3}{m-1}}

First , we will simplify the given Equation and then we will solve it . So , now <em>taking LCM </em>in <em>LHS</em> ;

{:\implies \quad \sf \dfrac{6+2m}{m}=\dfrac{m+3}{m-1}}

Now , <em>Cross Multiplying</em> ;

{:\implies \quad \sf (m-1)(6+2m)=m(m+3)}

{:\implies \quad \sf m(6+2m)-1(6+2m)=m^{2}+3m}

{:\implies \quad \sf 6m+2m^{2}-6-2m=m^{2}+3m}

Now , transposing whole <em>RHS</em> to <em>LHS</em> and <em>collecting like terms</em> ;

{:\implies \quad \sf (2m^{2}-m^{2})+(6m-2m-3m)-6=0}

{:\implies \quad \sf m^{2}+m-6=0}

Now , we will <em>solve this by splitting</em> the <em>Middle term</em> ;

{:\implies \quad \sf m^{2}+3m-2m-6=0}

{:\implies \quad \sf m(m+3)-2(m+3)=0}

{:\implies \quad \sf (m+3)(m-2)=0}

So , now either <em>, m+3 = 0</em> or <em>m-2 = 0</em><em> </em>. On simplification we will get , <em>m = 2 , -3.</em> So <em>m = {-3, 2</em><em>}</em>

<em>Hence Option 1) {-3, 2} is correct</em>

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