Solve (-7y+8/4)^2+(y-2)^2=72/65

$$( \frac { - 7 y + 8 } { 4 } ) ^ { 2 } + ( y - 2 ) ^ { 2 } = \frac { 72 } { 65 }$$

Solve for y

$$y=\frac{64}{65}\approx 0.984615385$$

$$y = \frac{112}{65} = 1\frac{47}{65} \approx 1.723076923$$

Steps Using the Quadratic Formula

To raise \frac{-7y+8}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-7y+8\right)^{2}}{4^{2}}+\left(y-2\right)^{2}=\frac{72}{65}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-2\right)^{2}.

$$\frac{\left(-7y+8\right)^{2}}{4^{2}}+y^{2}-4y+4=\frac{72}{65}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y^{2}-4y+4 times \frac{4^{2}}{4^{2}}.

$$\frac{\left(-7y+8\right)^{2}}{4^{2}}+\frac{\left(y^{2}-4y+4\right)\times 4^{2}}{4^{2}}=\frac{72}{65}$$

Since \frac{\left(-7y+8\right)^{2}}{4^{2}} and \frac{\left(y^{2}-4y+4\right)\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(-7y+8\right)^{2}+\left(y^{2}-4y+4\right)\times 4^{2}}{4^{2}}=\frac{72}{65}$$

Do the multiplications in \left(-7y+8\right)^{2}+\left(y^{2}-4y+4\right)\times 4^{2}.

$$\frac{49y^{2}-112y+64+16y^{2}-64y+64}{4^{2}}=\frac{72}{65}$$

Combine like terms in 49y^{2}-112y+64+16y^{2}-64y+64.

$$\frac{65y^{2}-176y+128}{4^{2}}=\frac{72}{65}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{65y^{2}-176y+128}{16}=\frac{72}{65}$$

Divide each term of 65y^{2}-176y+128 by 16 to get \frac{65}{16}y^{2}-11y+8.

$$\frac{65}{16}y^{2}-11y+8=\frac{72}{65}$$

Subtract \frac{72}{65} from both sides.

$$\frac{65}{16}y^{2}-11y+8-\frac{72}{65}=0$$

Subtract \frac{72}{65} from 8 to get \frac{448}{65}.

$$\frac{65}{16}y^{2}-11y+\frac{448}{65}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{65}{16} for a, -11 for b, and \frac{448}{65} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times \frac{65}{16}\times \frac{448}{65}}}{2\times \frac{65}{16}}$$

Square -11.

$$y=\frac{-\left(-11\right)±\sqrt{121-4\times \frac{65}{16}\times \frac{448}{65}}}{2\times \frac{65}{16}}$$

Multiply -4 times \frac{65}{16}.

$$y=\frac{-\left(-11\right)±\sqrt{121-\frac{65}{4}\times \frac{448}{65}}}{2\times \frac{65}{16}}$$

Multiply -\frac{65}{4} times \frac{448}{65} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-11\right)±\sqrt{121-112}}{2\times \frac{65}{16}}$$

Add 121 to -112.

$$y=\frac{-\left(-11\right)±\sqrt{9}}{2\times \frac{65}{16}}$$

Take the square root of 9.

$$y=\frac{-\left(-11\right)±3}{2\times \frac{65}{16}}$$

The opposite of -11 is 11.

$$y=\frac{11±3}{2\times \frac{65}{16}}$$

Multiply 2 times \frac{65}{16}.

$$y=\frac{11±3}{\frac{65}{8}}$$

Now solve the equation y=\frac{11±3}{\frac{65}{8}} when ± is plus. Add 11 to 3.

$$y=\frac{14}{\frac{65}{8}}$$

Divide 14 by \frac{65}{8} by multiplying 14 by the reciprocal of \frac{65}{8}.

$$y=\frac{112}{65}$$

Now solve the equation y=\frac{11±3}{\frac{65}{8}} when ± is minus. Subtract 3 from 11.

$$y=\frac{8}{\frac{65}{8}}$$

Divide 8 by \frac{65}{8} by multiplying 8 by the reciprocal of \frac{65}{8}.

$$y=\frac{64}{65}$$

The equation is now solved.

$$y=\frac{112}{65} y=\frac{64}{65}$$

Steps for Completing the Square

To raise \frac{-7y+8}{4} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(-7y+8\right)^{2}}{4^{2}}+\left(y-2\right)^{2}=\frac{72}{65}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-2\right)^{2}.

$$\frac{\left(-7y+8\right)^{2}}{4^{2}}+y^{2}-4y+4=\frac{72}{65}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply y^{2}-4y+4 times \frac{4^{2}}{4^{2}}.

$$\frac{\left(-7y+8\right)^{2}}{4^{2}}+\frac{\left(y^{2}-4y+4\right)\times 4^{2}}{4^{2}}=\frac{72}{65}$$

Since \frac{\left(-7y+8\right)^{2}}{4^{2}} and \frac{\left(y^{2}-4y+4\right)\times 4^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(-7y+8\right)^{2}+\left(y^{2}-4y+4\right)\times 4^{2}}{4^{2}}=\frac{72}{65}$$

Do the multiplications in \left(-7y+8\right)^{2}+\left(y^{2}-4y+4\right)\times 4^{2}.

$$\frac{49y^{2}-112y+64+16y^{2}-64y+64}{4^{2}}=\frac{72}{65}$$

Combine like terms in 49y^{2}-112y+64+16y^{2}-64y+64.

$$\frac{65y^{2}-176y+128}{4^{2}}=\frac{72}{65}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{65y^{2}-176y+128}{16}=\frac{72}{65}$$

Divide each term of 65y^{2}-176y+128 by 16 to get \frac{65}{16}y^{2}-11y+8.

$$\frac{65}{16}y^{2}-11y+8=\frac{72}{65}$$

Subtract 8 from both sides.

$$\frac{65}{16}y^{2}-11y=\frac{72}{65}-8$$

Subtract 8 from \frac{72}{65} to get -\frac{448}{65}.

$$\frac{65}{16}y^{2}-11y=-\frac{448}{65}$$

Divide both sides of the equation by \frac{65}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{65}{16}y^{2}-11y}{\frac{65}{16}}=-\frac{\frac{448}{65}}{\frac{65}{16}}$$

Dividing by \frac{65}{16} undoes the multiplication by \frac{65}{16}.

$$y^{2}+\left(-\frac{11}{\frac{65}{16}}\right)y=-\frac{\frac{448}{65}}{\frac{65}{16}}$$

Divide -11 by \frac{65}{16} by multiplying -11 by the reciprocal of \frac{65}{16}.

$$y^{2}-\frac{176}{65}y=-\frac{\frac{448}{65}}{\frac{65}{16}}$$

Divide -\frac{448}{65} by \frac{65}{16} by multiplying -\frac{448}{65} by the reciprocal of \frac{65}{16}.

$$y^{2}-\frac{176}{65}y=-\frac{7168}{4225}$$

Divide -\frac{176}{65}, the coefficient of the x term, by 2 to get -\frac{88}{65}. Then add the square of -\frac{88}{65} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{176}{65}y+\left(-\frac{88}{65}\right)^{2}=-\frac{7168}{4225}+\left(-\frac{88}{65}\right)^{2}$$

Square -\frac{88}{65} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{176}{65}y+\frac{7744}{4225}=\frac{-7168+7744}{4225}$$

Add -\frac{7168}{4225} to \frac{7744}{4225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{176}{65}y+\frac{7744}{4225}=\frac{576}{4225}$$

Factor y^{2}-\frac{176}{65}y+\frac{7744}{4225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{88}{65}\right)^{2}=\frac{576}{4225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{88}{65}\right)^{2}}=\sqrt{\frac{576}{4225}}$$

Simplify.

$$y-\frac{88}{65}=\frac{24}{65} y-\frac{88}{65}=-\frac{24}{65}$$

Add \frac{88}{65} to both sides of the equation.

$$y=\frac{112}{65} y=\frac{64}{65}$$