Solve sqrt{1+(y+3)^2+(y-1)^2}=sqrt{25+(y-7)^2+(y+5)^2}

$$\sqrt { 1 + ( y + 3 ) ^ { 2 } + ( y - 1 ) ^ { 2 } } = \sqrt { 25 + ( y - 7 ) ^ { 2 } + ( y + 5 ) ^ { 2 } }$$

Solve for y

$$y=11$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{1+\left(y+3\right)^{2}+\left(y-1\right)^{2}}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+3\right)^{2}.

$$\left(\sqrt{1+y^{2}+6y+9+\left(y-1\right)^{2}}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Add 1 and 9 to get 10.

$$\left(\sqrt{10+y^{2}+6y+\left(y-1\right)^{2}}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-1\right)^{2}.

$$\left(\sqrt{10+y^{2}+6y+y^{2}-2y+1}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$\left(\sqrt{10+2y^{2}+6y-2y+1}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Combine 6y and -2y to get 4y.

$$\left(\sqrt{10+2y^{2}+4y+1}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Add 10 and 1 to get 11.

$$\left(\sqrt{11+2y^{2}+4y}\right)^{2}=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Calculate \sqrt{11+2y^{2}+4y} to the power of 2 and get 11+2y^{2}+4y.

$$11+2y^{2}+4y=\left(\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-7\right)^{2}.

$$11+2y^{2}+4y=\left(\sqrt{25+y^{2}-14y+49+\left(y+5\right)^{2}}\right)^{2}$$

Add 25 and 49 to get 74.

$$11+2y^{2}+4y=\left(\sqrt{74+y^{2}-14y+\left(y+5\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+5\right)^{2}.

$$11+2y^{2}+4y=\left(\sqrt{74+y^{2}-14y+y^{2}+10y+25}\right)^{2}$$

Combine y^{2} and y^{2} to get 2y^{2}.

$$11+2y^{2}+4y=\left(\sqrt{74+2y^{2}-14y+10y+25}\right)^{2}$$

Combine -14y and 10y to get -4y.

$$11+2y^{2}+4y=\left(\sqrt{74+2y^{2}-4y+25}\right)^{2}$$

Add 74 and 25 to get 99.

$$11+2y^{2}+4y=\left(\sqrt{99+2y^{2}-4y}\right)^{2}$$

Calculate \sqrt{99+2y^{2}-4y} to the power of 2 and get 99+2y^{2}-4y.

$$11+2y^{2}+4y=99+2y^{2}-4y$$

Subtract 2y^{2} from both sides.

$$11+2y^{2}+4y-2y^{2}=99-4y$$

Combine 2y^{2} and -2y^{2} to get 0.

$$11+4y=99-4y$$

Add 4y to both sides.

$$11+4y+4y=99$$

Combine 4y and 4y to get 8y.

$$11+8y=99$$

Subtract 11 from both sides.

$$8y=99-11$$

Subtract 11 from 99 to get 88.

$$8y=88$$

Divide both sides by 8.

$$y=\frac{88}{8}$$

Divide 88 by 8 to get 11.

$$y=11$$

Substitute 11 for y in the equation \sqrt{1+\left(y+3\right)^{2}+\left(y-1\right)^{2}}=\sqrt{25+\left(y-7\right)^{2}+\left(y+5\right)^{2}}.

$$\sqrt{1+\left(11+3\right)^{2}+\left(11-1\right)^{2}}=\sqrt{25+\left(11-7\right)^{2}+\left(11+5\right)^{2}}$$

Simplify. The value y=11 satisfies the equation.

$$3\times 33^{\frac{1}{2}}=3\times 33^{\frac{1}{2}}$$

Equation \sqrt{\left(y+3\right)^{2}+\left(y-1\right)^{2}+1}=\sqrt{\left(y+5\right)^{2}+\left(y-7\right)^{2}+25} has a unique solution.

$$y=11$$