Solve {l}{2k+1=3Q}{k^2+1=2Q^2}

$$\left\{ \begin{array} { l } { 2 k + 1 = 3 Q } \\ { k ^ { 2 } + 1 = 2 Q ^ { 2 } } \end{array} \right.$$

Solve for k, Q

$$k=7\text{, }Q=5$$

$$k=1\text{, }Q=1$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3Q from both sides.

$$2k+1-3Q=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$2k-3Q=-1$$

Consider the second equation. Subtract 2Q^{2} from both sides.

$$k^{2}+1-2Q^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$k^{2}-2Q^{2}=-1$$

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.

$$2k-3Q=-1,-2Q^{2}+k^{2}=-1$$

Solve 2k-3Q=-1 for k by isolating k on the left hand side of the equal sign.

$$2k-3Q=-1$$

Subtract -3Q from both sides of the equation.

$$2k=3Q-1$$

Divide both sides by 2.

$$k=\frac{3}{2}Q-\frac{1}{2}$$

Substitute \frac{3}{2}Q-\frac{1}{2} for k in the other equation, -2Q^{2}+k^{2}=-1.

$$-2Q^{2}+\left(\frac{3}{2}Q-\frac{1}{2}\right)^{2}=-1$$

Square \frac{3}{2}Q-\frac{1}{2}.

$$-2Q^{2}+\frac{9}{4}Q^{2}-\frac{3}{2}Q+\frac{1}{4}=-1$$

Add -2Q^{2} to \frac{9}{4}Q^{2}.

$$\frac{1}{4}Q^{2}-\frac{3}{2}Q+\frac{1}{4}=-1$$

Add 1 to both sides of the equation.

$$\frac{1}{4}Q^{2}-\frac{3}{2}Q+\frac{5}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2+1\times \left(\frac{3}{2}\right)^{2} for a, 1\left(-\frac{1}{2}\right)\times \frac{3}{2}\times 2 for b, and \frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$Q=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\left(-\frac{3}{2}\right)^{2}-4\times \frac{1}{4}\times \frac{5}{4}}}{2\times \frac{1}{4}}$$

Square 1\left(-\frac{1}{2}\right)\times \frac{3}{2}\times 2.

$$Q=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4\times \frac{1}{4}\times \frac{5}{4}}}{2\times \frac{1}{4}}$$

Multiply -4 times -2+1\times \left(\frac{3}{2}\right)^{2}.

$$Q=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-\frac{5}{4}}}{2\times \frac{1}{4}}$$

Multiply -1 times \frac{5}{4}.

$$Q=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9-5}{4}}}{2\times \frac{1}{4}}$$

Add \frac{9}{4} to -\frac{5}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$Q=\frac{-\left(-\frac{3}{2}\right)±\sqrt{1}}{2\times \frac{1}{4}}$$

Take the square root of 1.

$$Q=\frac{-\left(-\frac{3}{2}\right)±1}{2\times \frac{1}{4}}$$

The opposite of 1\left(-\frac{1}{2}\right)\times \frac{3}{2}\times 2 is \frac{3}{2}.

$$Q=\frac{\frac{3}{2}±1}{2\times \frac{1}{4}}$$

Multiply 2 times -2+1\times \left(\frac{3}{2}\right)^{2}.

$$Q=\frac{\frac{3}{2}±1}{\frac{1}{2}}$$

Now solve the equation Q=\frac{\frac{3}{2}±1}{\frac{1}{2}} when ± is plus. Add \frac{3}{2} to 1.

$$Q=\frac{\frac{5}{2}}{\frac{1}{2}}$$

Divide \frac{5}{2} by \frac{1}{2} by multiplying \frac{5}{2} by the reciprocal of \frac{1}{2}.

$$Q=5$$

Now solve the equation Q=\frac{\frac{3}{2}±1}{\frac{1}{2}} when ± is minus. Subtract 1 from \frac{3}{2}.

$$Q=\frac{\frac{1}{2}}{\frac{1}{2}}$$

Divide \frac{1}{2} by \frac{1}{2} by multiplying \frac{1}{2} by the reciprocal of \frac{1}{2}.

$$Q=1$$

There are two solutions for Q: 5 and 1. Substitute 5 for Q in the equation k=\frac{3}{2}Q-\frac{1}{2} to find the corresponding solution for k that satisfies both equations.

$$k=\frac{3}{2}\times 5-\frac{1}{2}$$

Multiply \frac{3}{2} times 5.

$$k=\frac{15-1}{2}$$

Add \frac{3}{2}\times 5 to -\frac{1}{2}.

$$k=7$$

Now substitute 1 for Q in the equation k=\frac{3}{2}Q-\frac{1}{2} and solve to find the corresponding solution for k that satisfies both equations.

$$k=\frac{3-1}{2}$$

Add 1\times \frac{3}{2} to -\frac{1}{2}.

$$k=1$$

The system is now solved.

$$k=7,Q=5\text{ or }k=1,Q=1$$