Solve {l}{ba=-4}{c+2ab+8b=1}{3-2b=5}

$$\left. \begin{array} { l } { b a = - 4 } \\ { c + 2 a b + 8 b = 1 } \\ { 3 - 2 b = 5 } \end{array} \right.$$

Solve for b, a, c

$$b=-1$$

$$a=4$$

$$c=9-2t_{1744830476}$$

Solution Steps

Consider the third equation. Subtract 3 from both sides.

$$-2b=5-3$$

Subtract 3 from 5 to get 2.

$$-2b=2$$

Divide both sides by -2.

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

Divide 2 by -2 to get -1.

$$b=-1$$

Consider the first equation. Insert the known values of variables into the equation.

$$-a=-4$$

Divide both sides by -1.

$$a=\frac{-4}{-1}$$

Fraction \frac{-4}{-1} can be simplified to 4 by removing the negative sign from both the numerator and the denominator.

$$a=4$$

Consider the second equation. Insert the known values of variables into the equation.

$$c+2\times 4\left(-1\right)+8\left(-1\right)=1$$

Do the multiplications.

$$c+8\left(-1\right)-8=1$$

Multiply 8 and -1 to get -8.

$$c-8-8=1$$

Subtract 8 from -8 to get -16.

$$c-16=1$$

Add 16 to both sides.

$$c=1+16$$

Add 1 and 16 to get 17.

$$c=17$$

The system is now solved.

$$b=-1 a=4 c=17$$