Solve {c}{A^2+3B^2=12}{6A-6B=12}

$$\left. \begin{array} { c } { A ^ { 2 } + 3 B ^ { 2 } = 12 } \\ { 6 A - 6 B = 12 } \end{array} \right.$$

Solve for A, B

$$A=0\text{, }B=-2$$

$$A=3\text{, }B=1$$

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.

$$6A-6B=12,3B^{2}+A^{2}=12$$

Solve 6A-6B=12 for A by isolating A on the left hand side of the equal sign.

$$6A-6B=12$$

Subtract -6B from both sides of the equation.

$$6A=6B+12$$

Divide both sides by 6.

$$A=B+2$$

Substitute B+2 for A in the other equation, 3B^{2}+A^{2}=12.

$$3B^{2}+\left(B+2\right)^{2}=12$$

Square B+2.

$$3B^{2}+B^{2}+4B+4=12$$

Add 3B^{2} to B^{2}.

$$4B^{2}+4B+4=12$$

Subtract 12 from both sides of the equation.

$$4B^{2}+4B-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+1\times 1^{2} for a, 1\times 2\times 1\times 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

Multiply -4 times 3+1\times 1^{2}.

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

Multiply -16 times -8.

$$B=\frac{-4±\sqrt{16+128}}{2\times 4}$$

Add 16 to 128.

$$B=\frac{-4±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$B=\frac{-4±12}{2\times 4}$$

Multiply 2 times 3+1\times 1^{2}.

$$B=\frac{-4±12}{8}$$

Now solve the equation B=\frac{-4±12}{8} when ± is plus. Add -4 to 12.

$$B=\frac{8}{8}$$

Divide 8 by 8.

$$B=1$$

Now solve the equation B=\frac{-4±12}{8} when ± is minus. Subtract 12 from -4.

$$B=-\frac{16}{8}$$

Divide -16 by 8.

$$B=-2$$

There are two solutions for B: 1 and -2. Substitute 1 for B in the equation A=B+2 to find the corresponding solution for A that satisfies both equations.

$$A=1+2$$

Add 1\times 1 to 2.

$$A=3$$

Now substitute -2 for B in the equation A=B+2 and solve to find the corresponding solution for A that satisfies both equations.

$$A=-2+2$$

Add -2 to 2.

$$A=0$$

The system is now solved.

$$A=3,B=1\text{ or }A=0,B=-2$$