Solve {l}{a^2+b^2=150}{a-b=5}

$$\left. \begin{array} { l } { a ^ { 2 } + b ^ { 2 } = 150 } \\ { a - b = 5 } \end{array} \right.$$

Solve for a, b

$$a=\frac{5-5\sqrt{11}}{2}\approx -5.791561976\text{, }b=\frac{-5\sqrt{11}-5}{2}\approx -10.791561976$$

$$a=\frac{5\sqrt{11}+5}{2}\approx 10.791561976\text{, }b=\frac{5\sqrt{11}-5}{2}\approx 5.791561976$$

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.

$$a-b=5,b^{2}+a^{2}=150$$

Solve a-b=5 for a by isolating a on the left hand side of the equal sign.

$$a-b=5$$

Subtract -b from both sides of the equation.

$$a=b+5$$

Substitute b+5 for a in the other equation, b^{2}+a^{2}=150.

$$b^{2}+\left(b+5\right)^{2}=150$$

Square b+5.

$$b^{2}+b^{2}+10b+25=150$$

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

$$2b^{2}+10b+25=150$$

Subtract 150 from both sides of the equation.

$$2b^{2}+10b-125=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 5\times 1\times 2 for b, and -125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-10±\sqrt{10^{2}-4\times 2\left(-125\right)}}{2\times 2}$$

Square 1\times 5\times 1\times 2.

$$b=\frac{-10±\sqrt{100-4\times 2\left(-125\right)}}{2\times 2}$$

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

$$b=\frac{-10±\sqrt{100-8\left(-125\right)}}{2\times 2}$$

Multiply -8 times -125.

$$b=\frac{-10±\sqrt{100+1000}}{2\times 2}$$

Add 100 to 1000.

$$b=\frac{-10±\sqrt{1100}}{2\times 2}$$

Take the square root of 1100.

$$b=\frac{-10±10\sqrt{11}}{2\times 2}$$

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

$$b=\frac{-10±10\sqrt{11}}{4}$$

Now solve the equation b=\frac{-10±10\sqrt{11}}{4} when ± is plus. Add -10 to 10\sqrt{11}.

$$b=\frac{10\sqrt{11}-10}{4}$$

Divide -10+10\sqrt{11} by 4.

$$b=\frac{5\sqrt{11}-5}{2}$$

Now solve the equation b=\frac{-10±10\sqrt{11}}{4} when ± is minus. Subtract 10\sqrt{11} from -10.

$$b=\frac{-10\sqrt{11}-10}{4}$$

Divide -10-10\sqrt{11} by 4.

$$b=\frac{-5\sqrt{11}-5}{2}$$

There are two solutions for b: \frac{-5+5\sqrt{11}}{2} and \frac{-5-5\sqrt{11}}{2}. Substitute \frac{-5+5\sqrt{11}}{2} for b in the equation a=b+5 to find the corresponding solution for a that satisfies both equations.

$$a=\frac{5\sqrt{11}-5}{2}+5$$

Now substitute \frac{-5-5\sqrt{11}}{2} for b in the equation a=b+5 and solve to find the corresponding solution for a that satisfies both equations.

$$a=\frac{-5\sqrt{11}-5}{2}+5$$

The system is now solved.

$$a=\frac{5\sqrt{11}-5}{2}+5,b=\frac{5\sqrt{11}-5}{2}\text{ or }a=\frac{-5\sqrt{11}-5}{2}+5,b=\frac{-5\sqrt{11}-5}{2}$$