Solve {l}{x-2/3-y-1/2=19/6}{2x+2/2+frac{5y-2}{3}=0}

$$\left\{ \begin{array} { l } { \frac { x - 2 } { 3 } - \frac { y - 1 } { 2 } = \frac { 19 } { 6 } } \\ { \frac { 2 x + 2 } { 2 } + \frac { 5 y - 2 } { 3 } = 0 } \end{array} \right.$$

Solve for x, y

$$x = \frac{97}{19} = 5\frac{2}{19} \approx 5.105263158$$

$$y = -\frac{62}{19} = -3\frac{5}{19} \approx -3.263157895$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2,6.

$$2\left(x-2\right)-3\left(y-1\right)=19$$

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

$$2x-4-3\left(y-1\right)=19$$

Use the distributive property to multiply -3 by y-1.

$$2x-4-3y+3=19$$

Add -4 and 3 to get -1.

$$2x-1-3y=19$$

Add 1 to both sides.

$$2x-3y=19+1$$

Add 19 and 1 to get 20.

$$2x-3y=20$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply 3 by 2x+2.

$$6x+6+2\left(5y-2\right)=0$$

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

$$6x+6+10y-4=0$$

Subtract 4 from 6 to get 2.

$$6x+2+10y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$6x+10y=-2$$

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.

$$2x-3y=20,6x+10y=-2$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$2x-3y=20$$

Add 3y to both sides of the equation.

$$2x=3y+20$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(3y+20\right)$$

Multiply \frac{1}{2} times 3y+20.

$$x=\frac{3}{2}y+10$$

Substitute \frac{3y}{2}+10 for x in the other equation, 6x+10y=-2.

$$6\left(\frac{3}{2}y+10\right)+10y=-2$$

Multiply 6 times \frac{3y}{2}+10.

$$9y+60+10y=-2$$

Add 9y to 10y.

$$19y+60=-2$$

Subtract 60 from both sides of the equation.

$$19y=-62$$

Divide both sides by 19.

$$y=-\frac{62}{19}$$

Substitute -\frac{62}{19} for y in x=\frac{3}{2}y+10. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{2}\left(-\frac{62}{19}\right)+10$$

Multiply \frac{3}{2} times -\frac{62}{19} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{93}{19}+10$$

Add 10 to -\frac{93}{19}.

$$x=\frac{97}{19}$$

The system is now solved.

$$x=\frac{97}{19},y=-\frac{62}{19}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2,6.

$$2\left(x-2\right)-3\left(y-1\right)=19$$

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

$$2x-4-3\left(y-1\right)=19$$

Use the distributive property to multiply -3 by y-1.

$$2x-4-3y+3=19$$

Add -4 and 3 to get -1.

$$2x-1-3y=19$$

Add 1 to both sides.

$$2x-3y=19+1$$

Add 19 and 1 to get 20.

$$2x-3y=20$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply 3 by 2x+2.

$$6x+6+2\left(5y-2\right)=0$$

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

$$6x+6+10y-4=0$$

Subtract 4 from 6 to get 2.

$$6x+2+10y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$6x+10y=-2$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$2x-3y=20,6x+10y=-2$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\6&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\-2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\6&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\6&10\end{matrix}\right))\left(\begin{matrix}2&-3\\6&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\6&10\end{matrix}\right))\left(\begin{matrix}20\\-2\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\6&10\end{matrix}\right))\left(\begin{matrix}20\\-2\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\6&10\end{matrix}\right))\left(\begin{matrix}20\\-2\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{2\times 10-\left(-3\times 6\right)}&-\frac{-3}{2\times 10-\left(-3\times 6\right)}\\-\frac{6}{2\times 10-\left(-3\times 6\right)}&\frac{2}{2\times 10-\left(-3\times 6\right)}\end{matrix}\right)\left(\begin{matrix}20\\-2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}&\frac{3}{38}\\-\frac{3}{19}&\frac{1}{19}\end{matrix}\right)\left(\begin{matrix}20\\-2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{19}\times 20+\frac{3}{38}\left(-2\right)\\-\frac{3}{19}\times 20+\frac{1}{19}\left(-2\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{97}{19}\\-\frac{62}{19}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{97}{19},y=-\frac{62}{19}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2,6.

$$2\left(x-2\right)-3\left(y-1\right)=19$$

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

$$2x-4-3\left(y-1\right)=19$$

Use the distributive property to multiply -3 by y-1.

$$2x-4-3y+3=19$$

Add -4 and 3 to get -1.

$$2x-1-3y=19$$

Add 1 to both sides.

$$2x-3y=19+1$$

Add 19 and 1 to get 20.

$$2x-3y=20$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply 3 by 2x+2.

$$6x+6+2\left(5y-2\right)=0$$

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

$$6x+6+10y-4=0$$

Subtract 4 from 6 to get 2.

$$6x+2+10y=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$6x+10y=-2$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$2x-3y=20,6x+10y=-2$$

To make 2x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 2.

$$6\times 2x+6\left(-3\right)y=6\times 20,2\times 6x+2\times 10y=2\left(-2\right)$$

Simplify.

$$12x-18y=120,12x+20y=-4$$

Subtract 12x+20y=-4 from 12x-18y=120 by subtracting like terms on each side of the equal sign.

$$12x-12x-18y-20y=120+4$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$-18y-20y=120+4$$

Add -18y to -20y.

$$-38y=120+4$$

Add 120 to 4.

$$-38y=124$$

Divide both sides by -38.

$$y=-\frac{62}{19}$$

Substitute -\frac{62}{19} for y in 6x+10y=-2. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x+10\left(-\frac{62}{19}\right)=-2$$

Multiply 10 times -\frac{62}{19}.

$$6x-\frac{620}{19}=-2$$

Add \frac{620}{19} to both sides of the equation.

$$6x=\frac{582}{19}$$

Divide both sides by 6.

$$x=\frac{97}{19}$$

The system is now solved.

$$x=\frac{97}{19},y=-\frac{62}{19}$$