What is the answer to a^2+r=(a+r)/(2a+1) ?

The answer to a^2+r=(a+r)/(2a+1) is r=(1-2a^2-a)/2 ;\quad a\ne-1/2 ,a\ne 0

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a^2+r=(a+r)/(2a+1)

a^{2} + r = \frac{a + r}{2 a + 1}

r = \frac{1 - 2 a ^{2} - a}{2}; a \neq = - \frac{1}{2}, a \neq = 0

Solution steps

a^{2} + r = \frac{a + r}{2 a + 1}

Multiply both sides by

2 a + 1

a^{2} (2 a + 1) + r (2 a + 1) = a + r; a \neq = - \frac{1}{2}

Expand

a^{2} (2 a + 1) : 2 a^{3} + a^{2}

2 a^{3} + a^{2} + r (2 a + 1) = a + r

Expand

r (2 a + 1) : 2 a r + r

2 a^{3} + a^{2} + 2 a r + r = a + r; a \neq = - \frac{1}{2}

Move

2 a^{3}

to the right side

a^{2} + 2 a r + r = a + r - 2 a^{3}; a \neq = - \frac{1}{2}

Move

a^{2}

to the right side

2 a r + r = a + r - 2 a^{3} - a^{2}; a \neq = - \frac{1}{2}

Move

r

to the left side

2 a r = a - 2 a^{3} - a^{2}; a \neq = - \frac{1}{2}

Divide both sides by

2 a; a \neq = - \frac{1}{2}, a \neq = 0

r = \frac{1 - 2 a ^{2} - a}{2}; a \neq = - \frac{1}{2}, a \neq = 0

2142 - 0.08 y = y - 0.18

\frac{3 ^{2} x + 3}{9 ^{3}} = 9^{14}

x, \frac{111}{11} x - \frac{301}{13} y = 0

x, 0.2 x + 0.8 y = 9.1

x, x + 13 = 12 y

What is the answer to a^2+r=(a+r)/(2a+1) ?
The answer to a^2+r=(a+r)/(2a+1) is r=(1-2a^2-a)/2 ;\quad a\ne-1/2 ,a\ne 0