Solve {r}{a+b=-1}{a^2+b^2=25}

$$\left\{ \begin{array} { r } { a + b = - 1 } \\ { a ^ { 2 } + b ^ { 2 } = 25 } \end{array} \right.$$

Solve for a, b

$$a=3\text{, }b=-4$$

$$a=-4\text{, }b=3$$

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.

$$a+b=-1,b^{2}+a^{2}=25$$

Solve a+b=-1 for a by isolating a on the left hand side of the equal sign.

$$a+b=-1$$

Subtract b from both sides of the equation.

$$a=-b-1$$

Substitute -b-1 for a in the other equation, b^{2}+a^{2}=25.

$$b^{2}+\left(-b-1\right)^{2}=25$$

Square -b-1.

$$b^{2}+b^{2}+2b+1=25$$

Add b^{2} to b^{2}.

$$2b^{2}+2b+1=25$$

Subtract 25 from both sides of the equation.

$$2b^{2}+2b-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\left(-1\right)\left(-1\right)\times 2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-2±\sqrt{2^{2}-4\times 2\left(-24\right)}}{2\times 2}$$

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

$$b=\frac{-2±\sqrt{4-4\times 2\left(-24\right)}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$b=\frac{-2±\sqrt{4-8\left(-24\right)}}{2\times 2}$$

Multiply -8 times -24.

$$b=\frac{-2±\sqrt{4+192}}{2\times 2}$$

Add 4 to 192.

$$b=\frac{-2±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$b=\frac{-2±14}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$b=\frac{-2±14}{4}$$

Now solve the equation b=\frac{-2±14}{4} when ± is plus. Add -2 to 14.

$$b=\frac{12}{4}$$

Divide 12 by 4.

$$b=3$$

Now solve the equation b=\frac{-2±14}{4} when ± is minus. Subtract 14 from -2.

$$b=-\frac{16}{4}$$

Divide -16 by 4.

$$b=-4$$

There are two solutions for b: 3 and -4. Substitute 3 for b in the equation a=-b-1 to find the corresponding solution for a that satisfies both equations.

$$a=-3-1$$

Add -3 to -1.

$$a=-4$$

Now substitute -4 for b in the equation a=-b-1 and solve to find the corresponding solution for a that satisfies both equations.

$$a=-\left(-4\right)-1$$

Multiply -1 times -4.

$$a=4-1$$

Add -4\left(-1\right) to -1.

$$a=3$$

The system is now solved.

$$a=-4,b=3\text{ or }a=3,b=-4$$