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

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

Solve for x, y

$$x=-2$$

$$y=40$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 10.

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

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

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

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

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

To find the opposite of y-10, find the opposite of each term.

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

Add 10 and 10 to get 20.

$$-10x+20-y=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-10x-y=-20$$

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.

$$-10x-y=-20,40x+3y=40$$

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

$$-10x-y=-20$$

Add y to both sides of the equation.

$$-10x=y-20$$

Divide both sides by -10.

$$x=-\frac{1}{10}\left(y-20\right)$$

Multiply -\frac{1}{10} times y-20.

$$x=-\frac{1}{10}y+2$$

Substitute -\frac{y}{10}+2 for x in the other equation, 40x+3y=40.

$$40\left(-\frac{1}{10}y+2\right)+3y=40$$

Multiply 40 times -\frac{y}{10}+2.

$$-4y+80+3y=40$$

Add -4y to 3y.

$$-y+80=40$$

Subtract 80 from both sides of the equation.

$$-y=-40$$

Divide both sides by -1.

$$y=40$$

Substitute 40 for y in x=-\frac{1}{10}y+2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{10}\times 40+2$$

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

$$x=-4+2$$

Add 2 to -4.

$$x=-2$$

The system is now solved.

$$x=-2,y=40$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 10.

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

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

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

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

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

To find the opposite of y-10, find the opposite of each term.

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

Add 10 and 10 to get 20.

$$-10x+20-y=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-10x-y=-20$$

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.

$$-10x-y=-20,40x+3y=40$$

Write the equations in matrix form.

$$\left(\begin{matrix}-10&-1\\40&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-20\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-10&-1\\40&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}-10&-1\\40&3\end{matrix}\right))\left(\begin{matrix}-10&-1\\40&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-10&-1\\40&3\end{matrix}\right))\left(\begin{matrix}-20\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-10&-1\\40&3\end{matrix}\right))\left(\begin{matrix}-20\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-10&-1\\40&3\end{matrix}\right))\left(\begin{matrix}-20\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{-10\times 3-\left(-40\right)}&-\frac{-1}{-10\times 3-\left(-40\right)}\\-\frac{40}{-10\times 3-\left(-40\right)}&-\frac{10}{-10\times 3-\left(-40\right)}\end{matrix}\right)\left(\begin{matrix}-20\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}&\frac{1}{10}\\-4&-1\end{matrix}\right)\left(\begin{matrix}-20\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\left(-20\right)+\frac{1}{10}\times 40\\-4\left(-20\right)-40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\\40\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-2,y=40$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 10.

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

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

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

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

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

To find the opposite of y-10, find the opposite of each term.

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

Add 10 and 10 to get 20.

$$-10x+20-y=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-10x-y=-20$$

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.

$$-10x-y=-20,40x+3y=40$$

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.

$$40\left(-10\right)x+40\left(-1\right)y=40\left(-20\right),-10\times 40x-10\times 3y=-10\times 40$$

Simplify.

$$-400x-40y=-800,-400x-30y=-400$$

Subtract -400x-30y=-400 from -400x-40y=-800 by subtracting like terms on each side of the equal sign.

$$-400x+400x-40y+30y=-800+400$$

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

$$-40y+30y=-800+400$$

Add -40y to 30y.

$$-10y=-800+400$$

Add -800 to 400.

$$-10y=-400$$

Divide both sides by -10.

$$y=40$$

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

$$40x+3\times 40=40$$

Multiply 3 times 40.

$$40x+120=40$$

Subtract 120 from both sides of the equation.

$$40x=-80$$

Divide both sides by 40.

$$x=-2$$

The system is now solved.

$$x=-2,y=40$$