Solve {l}{10x+10y=1646}{30x+12y=3894}

$$\left\{ \begin{array} { l } { 10 x + 10 y = 1646 } \\ { 30 x + 12 y = 3894 } \end{array} \right.$$

Solve for x, y

$$x = \frac{533}{5} = 106\frac{3}{5} = 106.6$$

$$y=58$$

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.

$$10x+10y=1646,30x+12y=3894$$

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

$$10x+10y=1646$$

Subtract 10y from both sides of the equation.

$$10x=-10y+1646$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(-10y+1646\right)$$

Multiply \frac{1}{10} times -10y+1646.

$$x=-y+\frac{823}{5}$$

Substitute -y+\frac{823}{5} for x in the other equation, 30x+12y=3894.

$$30\left(-y+\frac{823}{5}\right)+12y=3894$$

Multiply 30 times -y+\frac{823}{5}.

$$-30y+4938+12y=3894$$

Add -30y to 12y.

$$-18y+4938=3894$$

Subtract 4938 from both sides of the equation.

$$-18y=-1044$$

Divide both sides by -18.

$$y=58$$

Substitute 58 for y in x=-y+\frac{823}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-58+\frac{823}{5}$$

Add \frac{823}{5} to -58.

$$x=\frac{533}{5}$$

The system is now solved.

$$x=\frac{533}{5},y=58$$

Steps Using Matrices

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

$$10x+10y=1646,30x+12y=3894$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&10\\30&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1646\\3894\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&10\\30&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&10\\30&12\end{matrix}\right))\left(\begin{matrix}10&10\\30&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&10\\30&12\end{matrix}\right))\left(\begin{matrix}1646\\3894\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&10\\30&12\end{matrix}\right))\left(\begin{matrix}1646\\3894\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&10\\30&12\end{matrix}\right))\left(\begin{matrix}1646\\3894\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{12}{10\times 12-10\times 30}&-\frac{10}{10\times 12-10\times 30}\\-\frac{30}{10\times 12-10\times 30}&\frac{10}{10\times 12-10\times 30}\end{matrix}\right)\left(\begin{matrix}1646\\3894\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{15}&\frac{1}{18}\\\frac{1}{6}&-\frac{1}{18}\end{matrix}\right)\left(\begin{matrix}1646\\3894\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{15}\times 1646+\frac{1}{18}\times 3894\\\frac{1}{6}\times 1646-\frac{1}{18}\times 3894\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{533}{5}\\58\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{533}{5},y=58$$

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.

$$10x+10y=1646,30x+12y=3894$$

To make 10x and 30x equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 10.

$$30\times 10x+30\times 10y=30\times 1646,10\times 30x+10\times 12y=10\times 3894$$

Simplify.

$$300x+300y=49380,300x+120y=38940$$

Subtract 300x+120y=38940 from 300x+300y=49380 by subtracting like terms on each side of the equal sign.

$$300x-300x+300y-120y=49380-38940$$

Add 300x to -300x. Terms 300x and -300x cancel out, leaving an equation with only one variable that can be solved.

$$300y-120y=49380-38940$$

Add 300y to -120y.

$$180y=49380-38940$$

Add 49380 to -38940.

$$180y=10440$$

Divide both sides by 180.

$$y=58$$

Substitute 58 for y in 30x+12y=3894. Because the resulting equation contains only one variable, you can solve for x directly.

$$30x+12\times 58=3894$$

Multiply 12 times 58.

$$30x+696=3894$$

Subtract 696 from both sides of the equation.

$$30x=3198$$

Divide both sides by 30.

$$x=\frac{533}{5}$$

The system is now solved.

$$x=\frac{533}{5},y=58$$