Solve {l}{4x+3y=2x+21y}{x+y+3x=4y+2x+7y-2x}

$$\left. \begin{array} { l } { 4 x + 3 y = 2 x + 21 y } \\ { x + y + 3 x = 4 y + 2 x + 7 y - 2 x } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Subtract 2x from both sides.

$$4x+3y-2x=21y$$

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

$$2x+3y=21y$$

Subtract 21y from both sides.

$$2x+3y-21y=0$$

Combine 3y and -21y to get -18y.

$$2x-18y=0$$

Consider the second equation. Combine x and 3x to get 4x.

$$4x+y=4y+2x+7y-2x$$

Combine 4y and 7y to get 11y.

$$4x+y=11y+2x-2x$$

Combine 2x and -2x to get 0.

$$4x+y=11y$$

Subtract 11y from both sides.

$$4x+y-11y=0$$

Combine y and -11y to get -10y.

$$4x-10y=0$$

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.

$$2x-18y=0,4x-10y=0$$

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

$$2x-18y=0$$

Add 18y to both sides of the equation.

$$2x=18y$$

Divide both sides by 2.

$$x=\frac{1}{2}\times 18y$$

Multiply \frac{1}{2} times 18y.

$$x=9y$$

Substitute 9y for x in the other equation, 4x-10y=0.

$$4\times 9y-10y=0$$

Multiply 4 times 9y.

$$36y-10y=0$$

Add 36y to -10y.

$$26y=0$$

Divide both sides by 26.

$$y=0$$

Substitute 0 for y in x=9y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=0$$

The system is now solved.

$$x=0,y=0$$

Steps Using Matrices

Consider the first equation. Subtract 2x from both sides.

$$4x+3y-2x=21y$$

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

$$2x+3y=21y$$

Subtract 21y from both sides.

$$2x+3y-21y=0$$

Combine 3y and -21y to get -18y.

$$2x-18y=0$$

Consider the second equation. Combine x and 3x to get 4x.

$$4x+y=4y+2x+7y-2x$$

Combine 4y and 7y to get 11y.

$$4x+y=11y+2x-2x$$

Combine 2x and -2x to get 0.

$$4x+y=11y$$

Subtract 11y from both sides.

$$4x+y-11y=0$$

Combine y and -11y to get -10y.

$$4x-10y=0$$

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

$$2x-18y=0,4x-10y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-18\\4&-10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-18\\4&-10\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-18\\4&-10\end{matrix}\right))\left(\begin{matrix}2&-18\\4&-10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-18\\4&-10\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-18\\4&-10\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-18\\4&-10\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{10}{2\left(-10\right)-\left(-18\times 4\right)}&-\frac{-18}{2\left(-10\right)-\left(-18\times 4\right)}\\-\frac{4}{2\left(-10\right)-\left(-18\times 4\right)}&\frac{2}{2\left(-10\right)-\left(-18\times 4\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{26}&\frac{9}{26}\\-\frac{1}{13}&\frac{1}{26}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

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

Extract the matrix elements x and y.

$$x=0,y=0$$

Steps Using Elimination

Consider the first equation. Subtract 2x from both sides.

$$4x+3y-2x=21y$$

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

$$2x+3y=21y$$

Subtract 21y from both sides.

$$2x+3y-21y=0$$

Combine 3y and -21y to get -18y.

$$2x-18y=0$$

Consider the second equation. Combine x and 3x to get 4x.

$$4x+y=4y+2x+7y-2x$$

Combine 4y and 7y to get 11y.

$$4x+y=11y+2x-2x$$

Combine 2x and -2x to get 0.

$$4x+y=11y$$

Subtract 11y from both sides.

$$4x+y-11y=0$$

Combine y and -11y to get -10y.

$$4x-10y=0$$

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.

$$2x-18y=0,4x-10y=0$$

To make 2x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 2.

$$4\times 2x+4\left(-18\right)y=0,2\times 4x+2\left(-10\right)y=0$$

Simplify.

$$8x-72y=0,8x-20y=0$$

Subtract 8x-20y=0 from 8x-72y=0 by subtracting like terms on each side of the equal sign.

$$8x-8x-72y+20y=0$$

Add 8x to -8x. Terms 8x and -8x cancel out, leaving an equation with only one variable that can be solved.

$$-72y+20y=0$$

Add -72y to 20y.

$$-52y=0$$

Divide both sides by -52.

$$y=0$$

Substitute 0 for y in 4x-10y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x=0$$

Divide both sides by 4.

$$x=0$$

The system is now solved.

$$x=0,y=0$$