Solve a=98/-321quad;quadb=-(-(439/-12))

$$a = \frac { 98 } { - 321 } \quad ; \quad b = - ( - ( \frac { 439 } { - 12 } ) )$$

Solve for a, b

$$a=-\frac{98}{321}\approx -0.30529595$$

$$b = -\frac{439}{12} = -36\frac{7}{12} \approx -36.583333333$$

Solution Steps

Consider the first equation. Fraction \frac{98}{-321} can be rewritten as -\frac{98}{321} by extracting the negative sign.

$$a=-\frac{98}{321}$$

Consider the second equation. Fraction \frac{439}{-12} can be rewritten as -\frac{439}{12} by extracting the negative sign.

$$b=-\left(-\left(-\frac{439}{12}\right)\right)$$

The opposite of -\frac{439}{12} is \frac{439}{12}.

$$b=-\frac{439}{12}$$

The system is now solved.

$$a=-\frac{98}{321} b=-\frac{439}{12}$$