Solve {l}{21x+124=84}{7x+4y=28}

$$\left. \begin{array} { l } { 21 x + 124 = 84 } \\ { 7 x + 4 y = 28 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{40}{21} = -1\frac{19}{21} \approx -1.904761905$$

$$y = \frac{31}{3} = 10\frac{1}{3} \approx 10.333333333$$

Solution Steps

Consider the first equation. Subtract 124 from both sides.

$$21x=84-124$$

Subtract 124 from 84 to get -40.

$$21x=-40$$

Divide both sides by 21.

$$x=-\frac{40}{21}$$

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

$$7\left(-\frac{40}{21}\right)+4y=28$$

Multiply 7 and -\frac{40}{21} to get -\frac{40}{3}.

$$-\frac{40}{3}+4y=28$$

Add \frac{40}{3} to both sides.

$$4y=28+\frac{40}{3}$$

Add 28 and \frac{40}{3} to get \frac{124}{3}.

$$4y=\frac{124}{3}$$

Divide both sides by 4.

$$y=\frac{\frac{124}{3}}{4}$$

Express \frac{\frac{124}{3}}{4} as a single fraction.

$$y=\frac{124}{3\times 4}$$

Multiply 3 and 4 to get 12.

$$y=\frac{124}{12}$$

Reduce the fraction \frac{124}{12} to lowest terms by extracting and canceling out 4.

$$y=\frac{31}{3}$$

The system is now solved.

$$x=-\frac{40}{21} y=\frac{31}{3}$$