Solve {l}{6y=8a-x}{2ax=6ay-2a^2}

$$\left\{ \begin{array} { l } { 6 y = 8 a - x } \\ { 2 a x = 6 a y - 2 a ^ { 2 } } \end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}\\x=2a\text{, }y=a\text{, }&\text{unconditionally}\\x=-6y\text{, }y\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$$

Steps Using Substitution

Consider the first equation. Add x to both sides.

$$6y+x=8a$$

Consider the second equation. Subtract 6ay from both sides.

$$2ax-6ay=-2a^{2}$$

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.

$$6y+x=8a,\left(-6a\right)y+2ax=-2a^{2}$$

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

$$6y+x=8a$$

Subtract x from both sides of the equation.

$$6y=-x+8a$$

Divide both sides by 6.

$$y=\frac{1}{6}\left(-x+8a\right)$$

Multiply \frac{1}{6} times -x+8a.

$$y=-\frac{1}{6}x+\frac{4a}{3}$$

Substitute -\frac{x}{6}+\frac{4a}{3} for y in the other equation, \left(-6a\right)y+2ax=-2a^{2}.

$$\left(-6a\right)\left(-\frac{1}{6}x+\frac{4a}{3}\right)+2ax=-2a^{2}$$

Multiply -6a times -\frac{x}{6}+\frac{4a}{3}.

$$ax-8a^{2}+2ax=-2a^{2}$$

Add ax to 2ax.

$$3ax-8a^{2}=-2a^{2}$$

Add 8a^{2} to both sides of the equation.

$$3ax=6a^{2}$$

Divide both sides by 3a.

$$x=2a$$

Substitute 2a for x in y=-\frac{1}{6}x+\frac{4a}{3}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{1}{6}\times 2a+\frac{4a}{3}$$

Multiply -\frac{1}{6} times 2a.

$$y=\frac{-a+4a}{3}$$

Add \frac{4a}{3} to -\frac{a}{3}.

$$y=a$$

The system is now solved.

$$y=a,x=2a$$

Steps Using Matrices

Consider the first equation. Add x to both sides.

$$6y+x=8a$$

Consider the second equation. Subtract 6ay from both sides.

$$2ax-6ay=-2a^{2}$$

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

$$6y+x=8a,\left(-6a\right)y+2ax=-2a^{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2a}{6\times 2a-\left(-6a\right)}&-\frac{1}{6\times 2a-\left(-6a\right)}\\-\frac{-6a}{6\times 2a-\left(-6a\right)}&\frac{6}{6\times 2a-\left(-6a\right)}\end{matrix}\right)\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}&-\frac{1}{18a}\\\frac{1}{3}&\frac{1}{3a}\end{matrix}\right)\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}\times 8a+\left(-\frac{1}{18a}\right)\left(-2a^{2}\right)\\\frac{1}{3}\times 8a+\frac{1}{3a}\left(-2a^{2}\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements y and x.

$$y=a,x=2a$$

Steps Using Elimination

Consider the first equation. Add x to both sides.

$$6y+x=8a$$

Consider the second equation. Subtract 6ay from both sides.

$$2ax-6ay=-2a^{2}$$

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.

$$6y+x=8a,\left(-6a\right)y+2ax=-2a^{2}$$

To make 6y and -6ay equal, multiply all terms on each side of the first equation by -6a and all terms on each side of the second by 6.

$$\left(-6a\right)\times 6y+\left(-6a\right)x=\left(-6a\right)\times 8a,6\left(-6a\right)y+6\times 2ax=6\left(-2a^{2}\right)$$

Simplify.

$$\left(-36a\right)y+\left(-6a\right)x=-48a^{2},\left(-36a\right)y+12ax=-12a^{2}$$

Subtract \left(-36a\right)y+12ax=-12a^{2} from \left(-36a\right)y+\left(-6a\right)x=-48a^{2} by subtracting like terms on each side of the equal sign.

$$\left(-36a\right)y+36ay+\left(-6a\right)x+\left(-12a\right)x=-48a^{2}+12a^{2}$$

Add -36ay to 36ay. Terms -36ay and 36ay cancel out, leaving an equation with only one variable that can be solved.

$$\left(-6a\right)x+\left(-12a\right)x=-48a^{2}+12a^{2}$$

Add -6ax to -12ax.

$$\left(-18a\right)x=-48a^{2}+12a^{2}$$

Add -48a^{2} to 12a^{2}.

$$\left(-18a\right)x=-36a^{2}$$

Divide both sides by -18a.

$$x=2a$$

Substitute 2a for x in \left(-6a\right)y+2ax=-2a^{2}. Because the resulting equation contains only one variable, you can solve for y directly.

$$\left(-6a\right)y+2a\times 2a=-2a^{2}$$

Multiply 2a times 2a.

$$\left(-6a\right)y+4a^{2}=-2a^{2}$$

Subtract 4a^{2} from both sides of the equation.

$$\left(-6a\right)y=-6a^{2}$$

Divide both sides by -6a.

$$y=a$$

The system is now solved.

$$y=a,x=2a$$

Solve for x, y

$$\left\{\begin{matrix}\\x=2a\text{, }y=a\text{, }&\text{unconditionally}\\x=-6y\text{, }y\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps Using Substitution

Consider the first equation. Add x to both sides.

$$6y+x=8a$$

Consider the second equation. Subtract 6ay from both sides.

$$2ax-6ay=-2a^{2}$$

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.

$$6y+x=8a,\left(-6a\right)y+2ax=-2a^{2}$$

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

$$6y+x=8a$$

Subtract x from both sides of the equation.

$$6y=-x+8a$$

Divide both sides by 6.

$$y=\frac{1}{6}\left(-x+8a\right)$$

Multiply \frac{1}{6} times -x+8a.

$$y=-\frac{1}{6}x+\frac{4a}{3}$$

Substitute -\frac{x}{6}+\frac{4a}{3} for y in the other equation, \left(-6a\right)y+2ax=-2a^{2}.

$$\left(-6a\right)\left(-\frac{1}{6}x+\frac{4a}{3}\right)+2ax=-2a^{2}$$

Multiply -6a times -\frac{x}{6}+\frac{4a}{3}.

$$ax-8a^{2}+2ax=-2a^{2}$$

Add ax to 2ax.

$$3ax-8a^{2}=-2a^{2}$$

Add 8a^{2} to both sides of the equation.

$$3ax=6a^{2}$$

Divide both sides by 3a.

$$x=2a$$

Substitute 2a for x in y=-\frac{1}{6}x+\frac{4a}{3}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{1}{6}\times 2a+\frac{4a}{3}$$

Multiply -\frac{1}{6} times 2a.

$$y=\frac{-a+4a}{3}$$

Add \frac{4a}{3} to -\frac{a}{3}.

$$y=a$$

The system is now solved.

$$y=a,x=2a$$

Steps Using Matrices

Consider the first equation. Add x to both sides.

$$6y+x=8a$$

Consider the second equation. Subtract 6ay from both sides.

$$2ax-6ay=-2a^{2}$$

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

$$6y+x=8a,\left(-6a\right)y+2ax=-2a^{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&1\\-6a&2a\end{matrix}\right))\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2a}{6\times 2a-\left(-6a\right)}&-\frac{1}{6\times 2a-\left(-6a\right)}\\-\frac{-6a}{6\times 2a-\left(-6a\right)}&\frac{6}{6\times 2a-\left(-6a\right)}\end{matrix}\right)\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}&-\frac{1}{18a}\\\frac{1}{3}&\frac{1}{3a}\end{matrix}\right)\left(\begin{matrix}8a\\-2a^{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}\times 8a+\left(-\frac{1}{18a}\right)\left(-2a^{2}\right)\\\frac{1}{3}\times 8a+\frac{1}{3a}\left(-2a^{2}\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements y and x.

$$y=a,x=2a$$

Steps Using Elimination

Consider the first equation. Add x to both sides.

$$6y+x=8a$$

Consider the second equation. Subtract 6ay from both sides.

$$2ax-6ay=-2a^{2}$$

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.

$$6y+x=8a,\left(-6a\right)y+2ax=-2a^{2}$$

To make 6y and -6ay equal, multiply all terms on each side of the first equation by -6a and all terms on each side of the second by 6.

$$\left(-6a\right)\times 6y+\left(-6a\right)x=\left(-6a\right)\times 8a,6\left(-6a\right)y+6\times 2ax=6\left(-2a^{2}\right)$$

Simplify.

$$\left(-36a\right)y+\left(-6a\right)x=-48a^{2},\left(-36a\right)y+12ax=-12a^{2}$$

Subtract \left(-36a\right)y+12ax=-12a^{2} from \left(-36a\right)y+\left(-6a\right)x=-48a^{2} by subtracting like terms on each side of the equal sign.

$$\left(-36a\right)y+36ay+\left(-6a\right)x+\left(-12a\right)x=-48a^{2}+12a^{2}$$

Add -36ay to 36ay. Terms -36ay and 36ay cancel out, leaving an equation with only one variable that can be solved.

$$\left(-6a\right)x+\left(-12a\right)x=-48a^{2}+12a^{2}$$

Add -6ax to -12ax.

$$\left(-18a\right)x=-48a^{2}+12a^{2}$$

Add -48a^{2} to 12a^{2}.

$$\left(-18a\right)x=-36a^{2}$$

Divide both sides by -18a.

$$x=2a$$

Substitute 2a for x in \left(-6a\right)y+2ax=-2a^{2}. Because the resulting equation contains only one variable, you can solve for y directly.

$$\left(-6a\right)y+2a\times 2a=-2a^{2}$$

Multiply 2a times 2a.

$$\left(-6a\right)y+4a^{2}=-2a^{2}$$

Subtract 4a^{2} from both sides of the equation.

$$\left(-6a\right)y=-6a^{2}$$

Divide both sides by -6a.

$$y=a$$

The system is now solved.

$$y=a,x=2a$$