Solve {c}{1.4-2B=7}{3A+2B=13}

$$\left. \begin{array} { c } { 1.4 - 2 B = 7 } \\ { 3 A + 2 B = 13 } \end{array} \right.$$

Solve for B, A

$$B=-2.8$$

$$A=6.2$$

Solution Steps

Consider the first equation. Subtract 1.4 from both sides.

$$-2B=7-1.4$$

Subtract 1.4 from 7 to get 5.6.

$$-2B=5.6$$

Divide both sides by -2.

$$B=\frac{5.6}{-2}$$

Expand \frac{5.6}{-2} by multiplying both numerator and the denominator by 10.

$$B=\frac{56}{-20}$$

Reduce the fraction \frac{56}{-20} to lowest terms by extracting and canceling out 4.

$$B=-\frac{14}{5}$$

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

$$3A+2\left(-\frac{14}{5}\right)=13$$

Multiply 2 and -\frac{14}{5} to get -\frac{28}{5}.

$$3A-\frac{28}{5}=13$$

Add \frac{28}{5} to both sides.

$$3A=13+\frac{28}{5}$$

Add 13 and \frac{28}{5} to get \frac{93}{5}.

$$3A=\frac{93}{5}$$

Divide both sides by 3.

$$A=\frac{\frac{93}{5}}{3}$$

Express \frac{\frac{93}{5}}{3} as a single fraction.

$$A=\frac{93}{5\times 3}$$

Multiply 5 and 3 to get 15.

$$A=\frac{93}{15}$$

Reduce the fraction \frac{93}{15} to lowest terms by extracting and canceling out 3.

$$A=\frac{31}{5}$$

The system is now solved.

$$B=-\frac{14}{5} A=\frac{31}{5}$$