((a+7)/(a+4))-1=((a+10)/(2a+8))

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Solution for ((a+7)/(a+4))-1=((a+10)/(2a+8)) equation:


D( a )

2*a+8 = 0

a+4 = 0

2*a+8 = 0

2*a+8 = 0

2*a+8 = 0 // - 8

2*a = -8 // : 2

a = -8/2

a = -4

a+4 = 0

a+4 = 0

a+4 = 0 // - 4

a = -4

a in (-oo:-4) U (-4:+oo)

(a+7)/(a+4)-1 = (a+10)/(2*a+8) // - (a+10)/(2*a+8)

(a+7)/(a+4)-((a+10)/(2*a+8))-1 = 0

(a+7)/(a+4)+(-1*(a+10))/(2*a+8)-1 = 0

((a+7)*(2*a+8))/((a+4)*(2*a+8))+(-1*(a+10)*(a+4))/((a+4)*(2*a+8))+(-1*(a+4)*(2*a+8))/((a+4)*(2*a+8)) = 0

(a+7)*(2*a+8)-1*(a+10)*(a+4)-1*(a+4)*(2*a+8) = 0

a^2-2*a^2+8*a-16*a-32+16 = 0

-a^2-8*a-16 = 0

-a^2-8*a-16 = 0

-1*(a^2+8*a+16) = 0

a^2+8*a+16 = 0

DELTA = 8^2-(1*4*16)

DELTA = 0

a = -8/(1*2)

a = -4 or a = -4

-1*(a+4)^2 = 0

(-1*(a+4)^2)/((a+4)*(2*a+8)) = 0

(-1*(a+4)^2)/((a+4)*(2*a+8)) = 0 // * (a+4)*(2*a+8)

-1*(a+4)^2 = 0

a+4 = 0 // - 4

a = -4

a in { -4}

a belongs to the empty set

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