Solve {l}{x/3+y/5+(11,5-x-y)/5=21/10}{frac{y}{3}+frac{x}{5}+frac{(11,5-x-y)}{4}=frac{31}{10}}

$$\left. \begin{array} { l } { \frac { x } { 3 } + \frac { y } { 5 } + \frac { ( 11,5 - x - y ) } { 5 } = \frac { 21 } { 10 } } \\ { \frac { y } { 3 } + \frac { x } { 5 } + \frac { ( 11,5 - x - y ) } { 4 } = \frac { 31 } { 10 } } \end{array} \right.$$

Solve for x, y

$$x=-1,5$$

$$y=1,8$$

Solution Steps

Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 3;5;10.

$$10x+6y+6\left(11,5-x-y\right)=63$$

Use the distributive property to multiply 6 by 11,5-x-y.

$$10x+6y+69-6x-6y=63$$

Combine 10x and -6x to get 4x.

$$4x+6y+69-6y=63$$

Combine 6y and -6y to get 0.

$$4x+69=63$$

Subtract 69 from both sides.

$$4x=63-69$$

Subtract 69 from 63 to get -6.

$$4x=-6$$

Divide both sides by 4.

$$x=\frac{-6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{2}$$

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

$$\frac{y}{3}+\frac{-\frac{3}{2}}{5}+\frac{11,5-\left(-\frac{3}{2}\right)-y}{4}=\frac{31}{10}$$

Multiply both sides of the equation by 60, the least common multiple of 3;5;4;10.

$$20y+12\left(-\frac{3}{2}\right)+15\left(11,5-\left(-\frac{3}{2}\right)-y\right)=186$$

Multiply 12 and -\frac{3}{2} to get -18.

$$20y-18+15\left(11,5-\left(-\frac{3}{2}\right)-y\right)=186$$

Multiply -1 and -\frac{3}{2} to get \frac{3}{2}.

$$20y-18+15\left(11,5+\frac{3}{2}-y\right)=186$$

Add 11,5 and \frac{3}{2} to get 13.

$$20y-18+15\left(13-y\right)=186$$

Use the distributive property to multiply 15 by 13-y.

$$20y-18+195-15y=186$$

Add -18 and 195 to get 177.

$$20y+177-15y=186$$

Combine 20y and -15y to get 5y.

$$5y+177=186$$

Subtract 177 from both sides.

$$5y=186-177$$

Subtract 177 from 186 to get 9.

$$5y=9$$

Divide both sides by 5.

$$y=\frac{9}{5}$$

The system is now solved.

$$x=-\frac{3}{2} y=\frac{9}{5}$$