To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
$$x+y=12,y^{2}+x^{2}=72$$
Solve $x+y=12$ for $x$ by isolating $x$ on the left hand side of the equal sign.
$$x+y=12$$
Subtract $y$ from both sides of the equation.
$$x=-y+12$$
Substitute $-y+12$ for $x$ in the other equation, $y^{2}+x^{2}=72$.
$$y^{2}+\left(-y+12\right)^{2}=72$$
Square $-y+12$.
$$y^{2}+y^{2}-24y+144=72$$
Add $y^{2}$ to $y^{2}$.
$$2y^{2}-24y+144=72$$
Subtract $72$ from both sides of the equation.
$$2y^{2}-24y+72=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1+1\left(-1\right)^{2}$ for $a$, $1\times 12\left(-1\right)\times 2$ for $b$, and $72$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
The opposite of $1\times 12\left(-1\right)\times 2$ is $24$.
$$y=\frac{24}{2\times 2}$$
Multiply $2$ times $1+1\left(-1\right)^{2}$.
$$y=\frac{24}{4}$$
Divide $24$ by $4$.
$$y=6$$
There are two solutions for $y$: $6$ and $6$. Substitute $6$ for $y$ in the equation $x=-y+12$ to find the corresponding solution for $x$ that satisfies both equations.
