Solve {l}{4b_{3}=3*-5/4}{text{Solvefor}atext{where}}{a=-7/4-4}

$$\left. \begin{array} { l } { 4 b_{3} = 3 \cdot -\frac{5}{4} }\\ { \text{Solve for } a \text{ where} } \\ { a = -\frac{7}{4} - 4 } \end{array} \right.$$

Solve for b_3, a

$$a = -\frac{23}{4} = -5\frac{3}{4} = -5.75$$

Solution Steps

Consider the first equation. Multiply 3 and -\frac{5}{4} to get -\frac{15}{4}.

$$4b_{3}=-\frac{15}{4}$$

Divide both sides by 4.

$$b_{3}=\frac{-\frac{15}{4}}{4}$$

Express \frac{-\frac{15}{4}}{4} as a single fraction.

$$b_{3}=\frac{-15}{4\times 4}$$

Multiply 4 and 4 to get 16.

$$b_{3}=\frac{-15}{16}$$

Fraction \frac{-15}{16} can be rewritten as -\frac{15}{16} by extracting the negative sign.

$$b_{3}=-\frac{15}{16}$$

Consider the second equation. Subtract 4 from -\frac{7}{4} to get -\frac{23}{4}.

$$a=-\frac{23}{4}$$

The system is now solved.

$$b_{3}=-\frac{15}{16} a=-\frac{23}{4}$$