Solve {l}{x-20x-10/30-y-10/10=0}{20x-y/20+y-10/10+frac{y}{40}=0}

$$\left. \begin{array} { l } { x - \frac { 20 x - 10 } { 30 } - \frac { y - 10 } { 10 } = 0 } \\ { \frac { 20 x - y } { 20 } + \frac { y - 10 } { 10 } + \frac { y } { 40 } = 0 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y = \frac{40}{3} = 13\frac{1}{3} \approx 13.333333333$$

Steps Using Substitution

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

$$30x-\left(20x-10\right)-3\left(y-10\right)=0$$

To find the opposite of 20x-10, find the opposite of each term.

$$30x-20x+10-3\left(y-10\right)=0$$

Combine 30x and -20x to get 10x.

$$10x+10-3\left(y-10\right)=0$$

Use the distributive property to multiply -3 by y-10.

$$10x+10-3y+30=0$$

Add 10 and 30 to get 40.

$$10x+40-3y=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$10x-3y=-40$$

Consider the second equation. Multiply both sides of the equation by 40, the least common multiple of 20,10,40.

$$2\left(20x-y\right)+4\left(y-10\right)+y=0$$

Use the distributive property to multiply 2 by 20x-y.

$$40x-2y+4\left(y-10\right)+y=0$$

Use the distributive property to multiply 4 by y-10.

$$40x-2y+4y-40+y=0$$

Combine -2y and 4y to get 2y.

$$40x+2y-40+y=0$$

Combine 2y and y to get 3y.

$$40x+3y-40=0$$

Add 40 to both sides. Anything plus zero gives itself.

$$40x+3y=40$$

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.

$$10x-3y=-40,40x+3y=40$$

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

$$10x-3y=-40$$

Add 3y to both sides of the equation.

$$10x=3y-40$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(3y-40\right)$$

Multiply \frac{1}{10} times 3y-40.

$$x=\frac{3}{10}y-4$$

Substitute \frac{3y}{10}-4 for x in the other equation, 40x+3y=40.

$$40\left(\frac{3}{10}y-4\right)+3y=40$$

Multiply 40 times \frac{3y}{10}-4.

$$12y-160+3y=40$$

Add 12y to 3y.

$$15y-160=40$$

Add 160 to both sides of the equation.

$$15y=200$$

Divide both sides by 15.

$$y=\frac{40}{3}$$

Substitute \frac{40}{3} for y in x=\frac{3}{10}y-4. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{10}\times \frac{40}{3}-4$$

Multiply \frac{3}{10} times \frac{40}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=4-4$$

Add -4 to 4.

$$x=0$$

The system is now solved.

$$x=0,y=\frac{40}{3}$$

Steps Using Matrices

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

$$30x-\left(20x-10\right)-3\left(y-10\right)=0$$

To find the opposite of 20x-10, find the opposite of each term.

$$30x-20x+10-3\left(y-10\right)=0$$

Combine 30x and -20x to get 10x.

$$10x+10-3\left(y-10\right)=0$$

Use the distributive property to multiply -3 by y-10.

$$10x+10-3y+30=0$$

Add 10 and 30 to get 40.

$$10x+40-3y=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$10x-3y=-40$$

Consider the second equation. Multiply both sides of the equation by 40, the least common multiple of 20,10,40.

$$2\left(20x-y\right)+4\left(y-10\right)+y=0$$

Use the distributive property to multiply 2 by 20x-y.

$$40x-2y+4\left(y-10\right)+y=0$$

Use the distributive property to multiply 4 by y-10.

$$40x-2y+4y-40+y=0$$

Combine -2y and 4y to get 2y.

$$40x+2y-40+y=0$$

Combine 2y and y to get 3y.

$$40x+3y-40=0$$

Add 40 to both sides. Anything plus zero gives itself.

$$40x+3y=40$$

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

$$10x-3y=-40,40x+3y=40$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&-3\\40&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-40\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&-3\\40&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&-3\\40&3\end{matrix}\right))\left(\begin{matrix}10&-3\\40&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&-3\\40&3\end{matrix}\right))\left(\begin{matrix}-40\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&-3\\40&3\end{matrix}\right))\left(\begin{matrix}-40\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&-3\\40&3\end{matrix}\right))\left(\begin{matrix}-40\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10\times 3-\left(-3\times 40\right)}&-\frac{-3}{10\times 3-\left(-3\times 40\right)}\\-\frac{40}{10\times 3-\left(-3\times 40\right)}&\frac{10}{10\times 3-\left(-3\times 40\right)}\end{matrix}\right)\left(\begin{matrix}-40\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{50}&\frac{1}{50}\\-\frac{4}{15}&\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}-40\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{50}\left(-40\right)+\frac{1}{50}\times 40\\-\frac{4}{15}\left(-40\right)+\frac{1}{15}\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\\frac{40}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=\frac{40}{3}$$

Steps Using Elimination

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

$$30x-\left(20x-10\right)-3\left(y-10\right)=0$$

To find the opposite of 20x-10, find the opposite of each term.

$$30x-20x+10-3\left(y-10\right)=0$$

Combine 30x and -20x to get 10x.

$$10x+10-3\left(y-10\right)=0$$

Use the distributive property to multiply -3 by y-10.

$$10x+10-3y+30=0$$

Add 10 and 30 to get 40.

$$10x+40-3y=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$10x-3y=-40$$

Consider the second equation. Multiply both sides of the equation by 40, the least common multiple of 20,10,40.

$$2\left(20x-y\right)+4\left(y-10\right)+y=0$$

Use the distributive property to multiply 2 by 20x-y.

$$40x-2y+4\left(y-10\right)+y=0$$

Use the distributive property to multiply 4 by y-10.

$$40x-2y+4y-40+y=0$$

Combine -2y and 4y to get 2y.

$$40x+2y-40+y=0$$

Combine 2y and y to get 3y.

$$40x+3y-40=0$$

Add 40 to both sides. Anything plus zero gives itself.

$$40x+3y=40$$

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.

$$10x-3y=-40,40x+3y=40$$

To make 10x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 10.

$$40\times 10x+40\left(-3\right)y=40\left(-40\right),10\times 40x+10\times 3y=10\times 40$$

Simplify.

$$400x-120y=-1600,400x+30y=400$$

Subtract 400x+30y=400 from 400x-120y=-1600 by subtracting like terms on each side of the equal sign.

$$400x-400x-120y-30y=-1600-400$$

Add 400x to -400x. Terms 400x and -400x cancel out, leaving an equation with only one variable that can be solved.

$$-120y-30y=-1600-400$$

Add -120y to -30y.

$$-150y=-1600-400$$

Add -1600 to -400.

$$-150y=-2000$$

Divide both sides by -150.

$$y=\frac{40}{3}$$

Substitute \frac{40}{3} for y in 40x+3y=40. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x+3\times \frac{40}{3}=40$$

Multiply 3 times \frac{40}{3}.

$$40x+40=40$$

Subtract 40 from both sides of the equation.

$$40x=0$$

Divide both sides by 40.

$$x=0$$

The system is now solved.

$$x=0,y=\frac{40}{3}$$