Solve {l}{0=4a-2b}{-sqrt{3}=a+b}

$$\left\{ \begin{array} { l } { 0 = 4 a - 2 b } \\ { - \sqrt { 3 } = a + b } \end{array} \right.$$

Solve for a, b

$$a=-\frac{\sqrt{3}}{3}\approx -0.577350269$$

$$b = -\frac{2 \sqrt{3}}{3} \approx -1.154700538$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$4a-2b=0$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$a+b=-\sqrt{3}$$

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.

$$4a-2b=0,a+b=-\sqrt{3}$$

Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.

$$4a-2b=0$$

Add 2b to both sides of the equation.

$$4a=2b$$

Divide both sides by 4.

$$a=\frac{1}{4}\times 2b$$

Multiply \frac{1}{4} times 2b.

$$a=\frac{1}{2}b$$

Substitute \frac{b}{2} for a in the other equation, a+b=-\sqrt{3}.

$$\frac{1}{2}b+b=-\sqrt{3}$$

Add \frac{b}{2} to b.

$$\frac{3}{2}b=-\sqrt{3}$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=-\frac{2\sqrt{3}}{3}$$

Substitute -\frac{2\sqrt{3}}{3} for b in a=\frac{1}{2}b. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{1}{2}\left(-\frac{2\sqrt{3}}{3}\right)$$

Multiply \frac{1}{2} times -\frac{2\sqrt{3}}{3}.

$$a=-\frac{\sqrt{3}}{3}$$

The system is now solved.

$$a=-\frac{\sqrt{3}}{3},b=-\frac{2\sqrt{3}}{3}$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$4a-2b=0$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$a+b=-\sqrt{3}$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$4a-2b=0,a+b=-\sqrt{3}$$

To make 4a and a equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 4.

$$4a-2b=0,4a+4b=4\left(-\sqrt{3}\right)$$

Simplify.

$$4a-2b=0,4a+4b=-4\sqrt{3}$$

Subtract 4a+4b=-4\sqrt{3} from 4a-2b=0 by subtracting like terms on each side of the equal sign.

$$4a-4a-2b-4b=4\sqrt{3}$$

Add 4a to -4a. Terms 4a and -4a cancel out, leaving an equation with only one variable that can be solved.

$$-2b-4b=4\sqrt{3}$$

Add -2b to -4b.

$$-6b=4\sqrt{3}$$

Divide both sides by -6.

$$b=-\frac{2\sqrt{3}}{3}$$

Substitute -\frac{2\sqrt{3}}{3} for b in a+b=-\sqrt{3}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a-\frac{2\sqrt{3}}{3}=-\sqrt{3}$$

Add \frac{2\sqrt{3}}{3} to both sides of the equation.

$$a=-\frac{\sqrt{3}}{3}$$

The system is now solved.

$$a=-\frac{\sqrt{3}}{3},b=-\frac{2\sqrt{3}}{3}$$