Solve {l}{k+q=6}{q^2/28+k^2/frac{28{3}}=1}

$$\left\{ \begin{array} { l } { k + q = 6 } \\ { \frac { q ^ { 2 } } { 28 } + \frac { k ^ { 2 } } { \frac { 28 } { 3 } } = 1 } \end{array} \right.$$

Solve for k, q

$$k=2\text{, }q=4$$

$$k=1\text{, }q=5$$

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.

$$k+q=6,\frac{1}{28}q^{2}+\frac{3}{28}k^{2}=1$$

Solve k+q=6 for k by isolating k on the left hand side of the equal sign.

$$k+q=6$$

Subtract q from both sides of the equation.

$$k=-q+6$$

Substitute -q+6 for k in the other equation, \frac{1}{28}q^{2}+\frac{3}{28}k^{2}=1.

$$\frac{1}{28}q^{2}+\frac{3}{28}\left(-q+6\right)^{2}=1$$

Square -q+6.

$$\frac{1}{28}q^{2}+\frac{3}{28}\left(q^{2}-12q+36\right)=1$$

Multiply \frac{3}{28} times q^{2}-12q+36.

$$\frac{1}{28}q^{2}+\frac{3}{28}q^{2}-\frac{9}{7}q+\frac{27}{7}=1$$

Add \frac{1}{28}q^{2} to \frac{3}{28}q^{2}.

$$\frac{1}{7}q^{2}-\frac{9}{7}q+\frac{27}{7}=1$$

Subtract 1 from both sides of the equation.

$$\frac{1}{7}q^{2}-\frac{9}{7}q+\frac{20}{7}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{28}+\frac{3}{28}\left(-1\right)^{2} for a, \frac{3}{28}\times 6\left(-1\right)\times 2 for b, and \frac{20}{7} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-\frac{9}{7}\right)±\sqrt{\left(-\frac{9}{7}\right)^{2}-4\times \frac{1}{7}\times \frac{20}{7}}}{2\times \frac{1}{7}}$$

Square \frac{3}{28}\times 6\left(-1\right)\times 2.

$$q=\frac{-\left(-\frac{9}{7}\right)±\sqrt{\frac{81}{49}-4\times \frac{1}{7}\times \frac{20}{7}}}{2\times \frac{1}{7}}$$

Multiply -4 times \frac{1}{28}+\frac{3}{28}\left(-1\right)^{2}.

$$q=\frac{-\left(-\frac{9}{7}\right)±\sqrt{\frac{81}{49}-\frac{4}{7}\times \frac{20}{7}}}{2\times \frac{1}{7}}$$

Multiply -\frac{4}{7} times \frac{20}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$q=\frac{-\left(-\frac{9}{7}\right)±\sqrt{\frac{81-80}{49}}}{2\times \frac{1}{7}}$$

Add \frac{81}{49} to -\frac{80}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$q=\frac{-\left(-\frac{9}{7}\right)±\sqrt{\frac{1}{49}}}{2\times \frac{1}{7}}$$

Take the square root of \frac{1}{49}.

$$q=\frac{-\left(-\frac{9}{7}\right)±\frac{1}{7}}{2\times \frac{1}{7}}$$

The opposite of \frac{3}{28}\times 6\left(-1\right)\times 2 is \frac{9}{7}.

$$q=\frac{\frac{9}{7}±\frac{1}{7}}{2\times \frac{1}{7}}$$

Multiply 2 times \frac{1}{28}+\frac{3}{28}\left(-1\right)^{2}.

$$q=\frac{\frac{9}{7}±\frac{1}{7}}{\frac{2}{7}}$$

Now solve the equation q=\frac{\frac{9}{7}±\frac{1}{7}}{\frac{2}{7}} when ± is plus. Add \frac{9}{7} to \frac{1}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$q=\frac{\frac{10}{7}}{\frac{2}{7}}$$

Divide \frac{10}{7} by \frac{2}{7} by multiplying \frac{10}{7} by the reciprocal of \frac{2}{7}.

$$q=5$$

Now solve the equation q=\frac{\frac{9}{7}±\frac{1}{7}}{\frac{2}{7}} when ± is minus. Subtract \frac{1}{7} from \frac{9}{7} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$q=\frac{\frac{8}{7}}{\frac{2}{7}}$$

Divide \frac{8}{7} by \frac{2}{7} by multiplying \frac{8}{7} by the reciprocal of \frac{2}{7}.

$$q=4$$

There are two solutions for q: 5 and 4. Substitute 5 for q in the equation k=-q+6 to find the corresponding solution for k that satisfies both equations.

$$k=-5+6$$

Add -5 to 6.

$$k=1$$

Now substitute 4 for q in the equation k=-q+6 and solve to find the corresponding solution for k that satisfies both equations.

$$k=-4+6$$

Add -4 to 6.

$$k=2$$

The system is now solved.

$$k=1,q=5\text{ or }k=2,q=4$$