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\frac{16}{9}y^{2}-\frac{160}{9}y+\frac{400}{9}+y^{2}=16
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{4}{3}y+\frac{20}{3}\right)^{2}.
\frac{25}{9}y^{2}-\frac{160}{9}y+\frac{400}{9}=16
Combine \frac{16}{9}y^{2} and y^{2} to get \frac{25}{9}y^{2}.
\frac{25}{9}y^{2}-\frac{160}{9}y+\frac{400}{9}-16=0
Subtract 16 from both sides.
\frac{25}{9}y^{2}-\frac{160}{9}y+\frac{256}{9}=0
Subtract 16 from \frac{400}{9} to get \frac{256}{9}.
y=\frac{-\left(-\frac{160}{9}\right)±\sqrt{\left(-\frac{160}{9}\right)^{2}-4\times \frac{25}{9}\times \frac{256}{9}}}{2\times \frac{25}{9}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{9} for a, -\frac{160}{9} for b, and \frac{256}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-\frac{160}{9}\right)±\sqrt{\frac{25600}{81}-4\times \frac{25}{9}\times \frac{256}{9}}}{2\times \frac{25}{9}}
Square -\frac{160}{9} by squaring both the numerator and the denominator of the fraction.
y=\frac{-\left(-\frac{160}{9}\right)±\sqrt{\frac{25600}{81}-\frac{100}{9}\times \frac{256}{9}}}{2\times \frac{25}{9}}
Multiply -4 times \frac{25}{9}.
y=\frac{-\left(-\frac{160}{9}\right)±\sqrt{\frac{25600-25600}{81}}}{2\times \frac{25}{9}}
Multiply -\frac{100}{9} times \frac{256}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
y=\frac{-\left(-\frac{160}{9}\right)±\sqrt{0}}{2\times \frac{25}{9}}
Add \frac{25600}{81} to -\frac{25600}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
y=-\frac{-\frac{160}{9}}{2\times \frac{25}{9}}
Take the square root of 0.
y=\frac{\frac{160}{9}}{2\times \frac{25}{9}}
The opposite of -\frac{160}{9} is \frac{160}{9}.
y=\frac{\frac{160}{9}}{\frac{50}{9}}
Multiply 2 times \frac{25}{9}.
y=\frac{16}{5}
Divide \frac{160}{9} by \frac{50}{9} by multiplying \frac{160}{9} by the reciprocal of \frac{50}{9}.
\frac{16}{9}y^{2}-\frac{160}{9}y+\frac{400}{9}+y^{2}=16
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{4}{3}y+\frac{20}{3}\right)^{2}.
\frac{25}{9}y^{2}-\frac{160}{9}y+\frac{400}{9}=16
Combine \frac{16}{9}y^{2} and y^{2} to get \frac{25}{9}y^{2}.
\frac{25}{9}y^{2}-\frac{160}{9}y=16-\frac{400}{9}
Subtract \frac{400}{9} from both sides.
\frac{25}{9}y^{2}-\frac{160}{9}y=-\frac{256}{9}
Subtract \frac{400}{9} from 16 to get -\frac{256}{9}.
\frac{\frac{25}{9}y^{2}-\frac{160}{9}y}{\frac{25}{9}}=-\frac{\frac{256}{9}}{\frac{25}{9}}
Divide both sides of the equation by \frac{25}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.
y^{2}+\left(-\frac{\frac{160}{9}}{\frac{25}{9}}\right)y=-\frac{\frac{256}{9}}{\frac{25}{9}}
Dividing by \frac{25}{9} undoes the multiplication by \frac{25}{9}.
y^{2}-\frac{32}{5}y=-\frac{\frac{256}{9}}{\frac{25}{9}}
Divide -\frac{160}{9} by \frac{25}{9} by multiplying -\frac{160}{9} by the reciprocal of \frac{25}{9}.
y^{2}-\frac{32}{5}y=-\frac{256}{25}
Divide -\frac{256}{9} by \frac{25}{9} by multiplying -\frac{256}{9} by the reciprocal of \frac{25}{9}.
y^{2}-\frac{32}{5}y+\left(-\frac{16}{5}\right)^{2}=-\frac{256}{25}+\left(-\frac{16}{5}\right)^{2}
Divide -\frac{32}{5}, the coefficient of the x term, by 2 to get -\frac{16}{5}. Then add the square of -\frac{16}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-\frac{32}{5}y+\frac{256}{25}=\frac{-256+256}{25}
Square -\frac{16}{5} by squaring both the numerator and the denominator of the fraction.
y^{2}-\frac{32}{5}y+\frac{256}{25}=0
Add -\frac{256}{25} to \frac{256}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y-\frac{16}{5}\right)^{2}=0
Factor y^{2}-\frac{32}{5}y+\frac{256}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(y-\frac{16}{5}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
y-\frac{16}{5}=0 y-\frac{16}{5}=0
Simplify.
y=\frac{16}{5} y=\frac{16}{5}
Add \frac{16}{5} to both sides of the equation.
y=\frac{16}{5}
The equation is now solved. Solutions are the same.