Solve {l}{4x-3y=9}{6y=1.8}{z=20x-1}{text{Solvefor}atext{where}}{a=z}

$$\left. \begin{array} { l } { 4 x - 3 y = 9 }\\ { 6 y = 1.8 }\\ { z = 20 x - 1 }\\ { \text{Solve for } a \text{ where} } \\ { a = z } \end{array} \right.$$

Solve for x, y, z, a

$$a=48.5$$

Solution Steps

Consider the second equation. Divide both sides by 6.

$$y=\frac{1.8}{6}$$

Expand \frac{1.8}{6} by multiplying both numerator and the denominator by 10.

$$y=\frac{18}{60}$$

Reduce the fraction \frac{18}{60} to lowest terms by extracting and canceling out 6.

$$y=\frac{3}{10}$$

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

$$4x-3\times \frac{3}{10}=9$$

Multiply -3 and \frac{3}{10} to get -\frac{9}{10}.

$$4x-\frac{9}{10}=9$$

Add \frac{9}{10} to both sides.

$$4x=9+\frac{9}{10}$$

Add 9 and \frac{9}{10} to get \frac{99}{10}.

$$4x=\frac{99}{10}$$

Divide both sides by 4.

$$x=\frac{\frac{99}{10}}{4}$$

Express \frac{\frac{99}{10}}{4} as a single fraction.

$$x=\frac{99}{10\times 4}$$

Multiply 10 and 4 to get 40.

$$x=\frac{99}{40}$$

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

$$z=20\times \frac{99}{40}-1$$

Multiply 20 and \frac{99}{40} to get \frac{99}{2}.

$$z=\frac{99}{2}-1$$

Subtract 1 from \frac{99}{2} to get \frac{97}{2}.

$$z=\frac{97}{2}$$

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

$$a=\frac{97}{2}$$

The system is now solved.

$$x=\frac{99}{40} y=\frac{3}{10} z=\frac{97}{2} a=\frac{97}{2}$$