Question

m-3/m+4=m+1/m-2( slove the equation)​

Answer 1

m-3÷m+4=m+1÷m-2
= (m-3)(m-2)=(m+4)(m+1)
= m2-2m-3m+6=m2+m+4m+4
= -5m+6=5m+4
= 2=10m
= m=1/5

Answer 2

The value of m =

$\frac{2}{3}$

GIVEN

Algebraic expression -

$m - \frac{3}{m} + 4 = m + \frac{1}{m} - 2$

TO FIND

The value of m

SOLUTION

We can simply solve the above problem as follows -

We are given an algebraic equation-

$m - \frac{3}{m} + 4 = m + \frac{1}{m} - 2$

We can rewrite it as :

$\frac{m}{1} - \frac{3}{m} + \frac{4}{1} = \frac{m}{1} + \frac{1}{m} - \frac{2}{1}$

$\frac{ {m}^{2} - 3 + 4m }{m} = \frac{ {m}^{2} + 1 - 2m }{m}$

Rearranging so that the Constant is on left-

$\frac{ {m}^{2}+ 4m - 3}{m} = \frac{ {m}^{2}- 2m + 1}{m}$

Now, multiplying the terms with, 'm' to eliminate the fraction denominator

$m. \frac{ {m}^{2} + 4m - 3 }{m} = m. \frac{ {m}^{2} - 2m + 1 }{m}$

In the equation after canceling the denominator

${m}^{2} + 4m - 3 = {m}^{2} - 2m + 1$

Moving terms to the left side,

${m}^{2} + 4m - 3 - ( {m}^{2} - 2m + 1 ) = 0$

${m}^{2} + 4m - 3 - {m}^{2} + 2m - 1 = 0$

Subtracting the digits :

${m}^{2} + 4m - 4 - {m}^{2}+2m = 0$

Adding the like terms :

$6m - 4 = 0$

Moving the digit 4 to RHS, the negative sign becomes positive.

$6m = 4$

$m = \frac{4}{6} = \frac{2}{3}$

Hence, the value of m is $\frac{2}{3}$

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