Solve {l}{x^2+p^2=100}{p-x=-2}

$$\left. \begin{array} { l } { x ^ { 2 } + p ^ { 2 } = 100 } \\ { p - x = - 2 } \end{array} \right.$$

Solve for x, p

$$x=8\text{, }p=6$$

$$x=-6\text{, }p=-8$$

Steps Using Substitution and Quadratic Formula

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.

$$p-x=-2,x^{2}+p^{2}=100$$

Solve p-x=-2 for p by isolating p on the left hand side of the equal sign.

$$p-x=-2$$

Subtract -x from both sides of the equation.

$$p=x-2$$

Substitute x-2 for p in the other equation, x^{2}+p^{2}=100.

$$x^{2}+\left(x-2\right)^{2}=100$$

Square x-2.

$$x^{2}+x^{2}-4x+4=100$$

Add x^{2} to x^{2}.

$$2x^{2}-4x+4=100$$

Subtract 100 from both sides of the equation.

$$2x^{2}-4x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\left(-2\right)\times 1\times 2 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-96\right)}}{2\times 2}$$

Square 1\left(-2\right)\times 1\times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-96\right)}}{2\times 2}$$

Multiply -4 times 1+1\times 1^{2}.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-96\right)}}{2\times 2}$$

Multiply -8 times -96.

$$x=\frac{-\left(-4\right)±\sqrt{16+768}}{2\times 2}$$

Add 16 to 768.

$$x=\frac{-\left(-4\right)±\sqrt{784}}{2\times 2}$$

Take the square root of 784.

$$x=\frac{-\left(-4\right)±28}{2\times 2}$$

The opposite of 1\left(-2\right)\times 1\times 2 is 4.

$$x=\frac{4±28}{2\times 2}$$

Multiply 2 times 1+1\times 1^{2}.

$$x=\frac{4±28}{4}$$

Now solve the equation x=\frac{4±28}{4} when ± is plus. Add 4 to 28.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{4±28}{4} when ± is minus. Subtract 28 from 4.

$$x=-\frac{24}{4}$$

Divide -24 by 4.

$$x=-6$$

There are two solutions for x: 8 and -6. Substitute 8 for x in the equation p=x-2 to find the corresponding solution for p that satisfies both equations.

$$p=8-2$$

Add 1\times 8 to -2.

$$p=6$$

Now substitute -6 for x in the equation p=x-2 and solve to find the corresponding solution for p that satisfies both equations.

$$p=-6-2$$

Add -6 to -2.

$$p=-8$$

The system is now solved.

$$p=6,x=8\text{ or }p=-8,x=-6$$