Solve {l}{2x-1/5+3y2/4=2}{3x+1/5-3y+2/4=0}

$$\left\{ \begin{array} { l } { \frac { 2 x - 1 } { 5 } + \frac { 3 y 2 } { 4 } = 2 } \\ { \frac { 3 x + 1 } { 5 } - \frac { 3 y + 2 } { 4 } = 0 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{15y_{2}}{8}+\frac{11}{2}$$

$$y=-\frac{3y_{2}}{2}+4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4\left(2x-1\right)+5\times 3y_{2}=40$$

Use the distributive property to multiply 4 by 2x-1.

$$8x-4+5\times 3y_{2}=40$$

Multiply 5 and 3 to get 15.

$$8x-4+15y_{2}=40$$

Add 4 to both sides.

$$8x+15y_{2}=40+4$$

Add 40 and 4 to get 44.

$$8x+15y_{2}=44$$

Subtract 15y_{2} from both sides.

$$8x=44-15y_{2}$$

Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4\left(3x+1\right)-5\left(3y+2\right)=0$$

Use the distributive property to multiply 4 by 3x+1.

$$12x+4-5\left(3y+2\right)=0$$

Use the distributive property to multiply -5 by 3y+2.

$$12x+4-15y-10=0$$

Subtract 10 from 4 to get -6.

$$12x-6-15y=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$12x-15y=6$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$8x=44-15y_{2},12x-15y=6$$

Pick one of the two equations which is more simple to solve for x by isolating x on the left hand side of the equal sign.

$$8x=44-15y_{2}$$

Divide both sides by 8.

$$x=-\frac{15y_{2}}{8}+\frac{11}{2}$$

Substitute \frac{11}{2}-\frac{15y_{2}}{8} for x in the other equation, 12x-15y=6.

$$12\left(-\frac{15y_{2}}{8}+\frac{11}{2}\right)-15y=6$$

Multiply 12 times \frac{11}{2}-\frac{15y_{2}}{8}.

$$-\frac{45y_{2}}{2}+66-15y=6$$

Subtract 66-\frac{45y_{2}}{2} from both sides of the equation.

$$-15y=\frac{45y_{2}}{2}-60$$

Divide both sides by -15.

$$y=-\frac{3y_{2}}{2}+4$$

The system is now solved.

$$x=-\frac{15y_{2}}{8}+\frac{11}{2},y=-\frac{3y_{2}}{2}+4$$