Solve {l}{2bx+ay=2ab}{bx-ay=4ab}

$$\left. \begin{array} { l } { 2 b x + a y = 2 a b } \\ { b x - a y = 4 a b } \end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}\\x=2a\text{, }y=-2b\text{, }&\text{unconditionally}\\x=0\text{, }y\in \mathrm{C}\text{, }&a=0\\x\in \mathrm{C}\text{, }y=0\text{, }&b=0\\x\in \mathrm{C}\text{, }y\in \mathrm{C}\text{, }&b=0\text{ and }a=0\end{matrix}\right.$$

Steps Using Substitution

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.

$$2bx+ay=2ab,bx+\left(-a\right)y=4ab$$

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

$$2bx+ay=2ab$$

Subtract ay from both sides of the equation.

$$2bx=\left(-a\right)y+2ab$$

Divide both sides by 2b.

$$x=\frac{1}{2b}\left(\left(-a\right)y+2ab\right)$$

Multiply \frac{1}{2b} times a\left(-y+2b\right).

$$x=\left(-\frac{a}{2b}\right)y+a$$

Substitute a-\frac{ay}{2b} for x in the other equation, bx+\left(-a\right)y=4ab.

$$b\left(\left(-\frac{a}{2b}\right)y+a\right)+\left(-a\right)y=4ab$$

Multiply b times a-\frac{ay}{2b}.

$$\left(-\frac{a}{2}\right)y+ab+\left(-a\right)y=4ab$$

Add -\frac{ay}{2} to -ay.

$$\left(-\frac{3a}{2}\right)y+ab=4ab$$

Subtract ba from both sides of the equation.

$$\left(-\frac{3a}{2}\right)y=3ab$$

Divide both sides by -\frac{3a}{2}.

$$y=-2b$$

Substitute -2b for y in x=\left(-\frac{a}{2b}\right)y+a. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\left(-\frac{a}{2b}\right)\left(-2b\right)+a$$

Multiply -\frac{a}{2b} times -2b.

$$x=a+a$$

Add a to a.

$$x=2a$$

The system is now solved.

$$x=2a,y=-2b$$

Steps Using Matrices

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

$$2bx+ay=2ab,bx+\left(-a\right)y=4ab$$

Write the equations in matrix form.

$$\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right))\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right))\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

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

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

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right))\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{a}{2b\left(-a\right)-ab}&-\frac{a}{2b\left(-a\right)-ab}\\-\frac{b}{2b\left(-a\right)-ab}&\frac{2b}{2b\left(-a\right)-ab}\end{matrix}\right)\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3b}&\frac{1}{3b}\\\frac{1}{3a}&-\frac{2}{3a}\end{matrix}\right)\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3b}\times 2ab+\frac{1}{3b}\times 4ab\\\frac{1}{3a}\times 2ab+\left(-\frac{2}{3a}\right)\times 4ab\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=2a,y=-2b$$

Steps Using Elimination

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.

$$2bx+ay=2ab,bx+\left(-a\right)y=4ab$$

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

$$b\times 2bx+bay=b\times 2ab,2bbx+2b\left(-a\right)y=2b\times 4ab$$

Simplify.

$$2b^{2}x+aby=2ab^{2},2b^{2}x+\left(-2ab\right)y=8ab^{2}$$

Subtract 2b^{2}x+\left(-2ab\right)y=8ab^{2} from 2b^{2}x+aby=2ab^{2} by subtracting like terms on each side of the equal sign.

$$2b^{2}x+\left(-2b^{2}\right)x+aby+2aby=2ab^{2}-8ab^{2}$$

Add 2b^{2}x to -2b^{2}x. Terms 2b^{2}x and -2b^{2}x cancel out, leaving an equation with only one variable that can be solved.

$$aby+2aby=2ab^{2}-8ab^{2}$$

Add bay to 2bay.

$$3aby=2ab^{2}-8ab^{2}$$

Add 2ab^{2} to -8ab^{2}.

$$3aby=-6ab^{2}$$

Divide both sides by 3ba.

$$y=-2b$$

Substitute -2b for y in bx+\left(-a\right)y=4ab. Because the resulting equation contains only one variable, you can solve for x directly.

$$bx+\left(-a\right)\left(-2b\right)=4ab$$

Multiply -a times -2b.

$$bx+2ab=4ab$$

Subtract 2ba from both sides of the equation.

$$bx=2ab$$

Divide both sides by b.

$$x=2a$$

The system is now solved.

$$x=2a,y=-2b$$

Solve for x, y

$$\left\{\begin{matrix}\\x=2a\text{, }y=-2b\text{, }&\text{unconditionally}\\x=0\text{, }y\in \mathrm{R}\text{, }&a=0\\x\in \mathrm{R}\text{, }y=0\text{, }&b=0\\x\in \mathrm{R}\text{, }y\in \mathrm{R}\text{, }&b=0\text{ and }a=0\end{matrix}\right.$$

Steps Using Substitution

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.

$$2bx+ay=2ab,bx+\left(-a\right)y=4ab$$

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

$$2bx+ay=2ab$$

Subtract ay from both sides of the equation.

$$2bx=\left(-a\right)y+2ab$$

Divide both sides by 2b.

$$x=\frac{1}{2b}\left(\left(-a\right)y+2ab\right)$$

Multiply \frac{1}{2b} times a\left(-y+2b\right).

$$x=\left(-\frac{a}{2b}\right)y+a$$

Substitute a-\frac{ay}{2b} for x in the other equation, bx+\left(-a\right)y=4ab.

$$b\left(\left(-\frac{a}{2b}\right)y+a\right)+\left(-a\right)y=4ab$$

Multiply b times a-\frac{ay}{2b}.

$$\left(-\frac{a}{2}\right)y+ab+\left(-a\right)y=4ab$$

Add -\frac{ay}{2} to -ay.

$$\left(-\frac{3a}{2}\right)y+ab=4ab$$

Subtract ba from both sides of the equation.

$$\left(-\frac{3a}{2}\right)y=3ab$$

Divide both sides by -\frac{3a}{2}.

$$y=-2b$$

Substitute -2b for y in x=\left(-\frac{a}{2b}\right)y+a. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\left(-\frac{a}{2b}\right)\left(-2b\right)+a$$

Multiply -\frac{a}{2b} times -2b.

$$x=a+a$$

Add a to a.

$$x=2a$$

The system is now solved.

$$x=2a,y=-2b$$

Steps Using Matrices

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

$$2bx+ay=2ab,bx+\left(-a\right)y=4ab$$

Write the equations in matrix form.

$$\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right))\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right))\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

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

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

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2b&a\\b&-a\end{matrix}\right))\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{a}{2b\left(-a\right)-ab}&-\frac{a}{2b\left(-a\right)-ab}\\-\frac{b}{2b\left(-a\right)-ab}&\frac{2b}{2b\left(-a\right)-ab}\end{matrix}\right)\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3b}&\frac{1}{3b}\\\frac{1}{3a}&-\frac{2}{3a}\end{matrix}\right)\left(\begin{matrix}2ab\\4ab\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3b}\times 2ab+\frac{1}{3b}\times 4ab\\\frac{1}{3a}\times 2ab+\left(-\frac{2}{3a}\right)\times 4ab\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=2a,y=-2b$$

Steps Using Elimination

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.

$$2bx+ay=2ab,bx+\left(-a\right)y=4ab$$

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

$$b\times 2bx+bay=b\times 2ab,2bbx+2b\left(-a\right)y=2b\times 4ab$$

Simplify.

$$2b^{2}x+aby=2ab^{2},2b^{2}x+\left(-2ab\right)y=8ab^{2}$$

Subtract 2b^{2}x+\left(-2ab\right)y=8ab^{2} from 2b^{2}x+aby=2ab^{2} by subtracting like terms on each side of the equal sign.

$$2b^{2}x+\left(-2b^{2}\right)x+aby+2aby=2ab^{2}-8ab^{2}$$

Add 2b^{2}x to -2b^{2}x. Terms 2b^{2}x and -2b^{2}x cancel out, leaving an equation with only one variable that can be solved.

$$aby+2aby=2ab^{2}-8ab^{2}$$

Add bay to 2bay.

$$3aby=2ab^{2}-8ab^{2}$$

Add 2ab^{2} to -8ab^{2}.

$$3aby=-6ab^{2}$$

Divide both sides by 3ba.

$$y=-2b$$

Substitute -2b for y in bx+\left(-a\right)y=4ab. Because the resulting equation contains only one variable, you can solve for x directly.

$$bx+\left(-a\right)\left(-2b\right)=4ab$$

Multiply -a times -2b.

$$bx+2ab=4ab$$

Subtract 2ba from both sides of the equation.

$$bx=2ab$$

Divide both sides by b.

$$x=2a$$

The system is now solved.

$$x=2a,y=-2b$$