Solve Matrix

$$\left( \begin{array} { l l } { 2 } & { x } \\ { y } & { - 3 } \end{array} \right) \cdot \left( \begin{array} { c } { 4 } \\ { - 5 } \end{array} \right) = \left( \begin{array} { c } { 1 } \\ { - 3 } \end{array} \right)$$

Solve for x, y

$$x = \frac{7}{5} = 1\frac{2}{5} = 1.4$$

$$y = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

Solution Steps

Consider the first equation. Subtract 8 from both sides.

$$-5x=1-8$$

Subtract 8 from 1 to get -7.

$$-5x=-7$$

Divide both sides by -5.

$$x=\frac{-7}{-5}$$

Fraction \frac{-7}{-5} can be simplified to \frac{7}{5} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{7}{5}$$

Consider the second equation. Subtract 15 from both sides.

$$4y=-3-15$$

Subtract 15 from -3 to get -18.

$$4y=-18$$

Divide both sides by 4.

$$y=\frac{-18}{4}$$

Reduce the fraction \frac{-18}{4} to lowest terms by extracting and canceling out 2.

$$y=-\frac{9}{2}$$

The system is now solved.

$$x=\frac{7}{5} y=-\frac{9}{2}$$