Solve {l}{10x+6y=40}{20x+3y=20}

$$\left. \begin{array} { l } { 10 x + 6 y = 40 } \\ { 20 x + 3 y = 20 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y = \frac{20}{3} = 6\frac{2}{3} \approx 6.666666667$$

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.

$$10x+6y=40,20x+3y=20$$

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

$$10x+6y=40$$

Subtract 6y from both sides of the equation.

$$10x=-6y+40$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(-6y+40\right)$$

Multiply \frac{1}{10} times -6y+40.

$$x=-\frac{3}{5}y+4$$

Substitute -\frac{3y}{5}+4 for x in the other equation, 20x+3y=20.

$$20\left(-\frac{3}{5}y+4\right)+3y=20$$

Multiply 20 times -\frac{3y}{5}+4.

$$-12y+80+3y=20$$

Add -12y to 3y.

$$-9y+80=20$$

Subtract 80 from both sides of the equation.

$$-9y=-60$$

Divide both sides by -9.

$$y=\frac{20}{3}$$

Substitute \frac{20}{3} for y in x=-\frac{3}{5}y+4. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{5}\times \frac{20}{3}+4$$

Multiply -\frac{3}{5} times \frac{20}{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.

$$x=0,y=\frac{20}{3}$$

Steps Using Matrices

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

$$10x+6y=40,20x+3y=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&6\\20&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}40\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&6\\20&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&6\\20&3\end{matrix}\right))\left(\begin{matrix}10&6\\20&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&6\\20&3\end{matrix}\right))\left(\begin{matrix}40\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&6\\20&3\end{matrix}\right))\left(\begin{matrix}40\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&6\\20&3\end{matrix}\right))\left(\begin{matrix}40\\20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10\times 3-6\times 20}&-\frac{6}{10\times 3-6\times 20}\\-\frac{20}{10\times 3-6\times 20}&\frac{10}{10\times 3-6\times 20}\end{matrix}\right)\left(\begin{matrix}40\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30}&\frac{1}{15}\\\frac{2}{9}&-\frac{1}{9}\end{matrix}\right)\left(\begin{matrix}40\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30}\times 40+\frac{1}{15}\times 20\\\frac{2}{9}\times 40-\frac{1}{9}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\\frac{20}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=\frac{20}{3}$$

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.

$$10x+6y=40,20x+3y=20$$

To make 10x and 20x equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 10.

$$20\times 10x+20\times 6y=20\times 40,10\times 20x+10\times 3y=10\times 20$$

Simplify.

$$200x+120y=800,200x+30y=200$$

Subtract 200x+30y=200 from 200x+120y=800 by subtracting like terms on each side of the equal sign.

$$200x-200x+120y-30y=800-200$$

Add 200x to -200x. Terms 200x and -200x cancel out, leaving an equation with only one variable that can be solved.

$$120y-30y=800-200$$

Add 120y to -30y.

$$90y=800-200$$

Add 800 to -200.

$$90y=600$$

Divide both sides by 90.

$$y=\frac{20}{3}$$

Substitute \frac{20}{3} for y in 20x+3y=20. Because the resulting equation contains only one variable, you can solve for x directly.

$$20x+3\times \frac{20}{3}=20$$

Multiply 3 times \frac{20}{3}.

$$20x+20=20$$

Subtract 20 from both sides of the equation.

$$20x=0$$

Divide both sides by 20.

$$x=0$$

The system is now solved.

$$x=0,y=\frac{20}{3}$$