Solve {l}{2x+24y=2110}{12x+6y=30}

$$\left\{ \begin{array}{l}{ 2 x + 24 y = 2110 }\\{ 12 x + 6 y = 30 }\end{array} \right.$$

Solve for x, y

$$x = -\frac{995}{23} = -43\frac{6}{23} \approx -43.260869565$$

$$y = \frac{2105}{23} = 91\frac{12}{23} \approx 91.52173913$$

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.

$$2x+24y=2110,12x+6y=30$$

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

$$2x+24y=2110$$

Subtract 24y from both sides of the equation.

$$2x=-24y+2110$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-24y+2110\right)$$

Multiply \frac{1}{2} times -24y+2110.

$$x=-12y+1055$$

Substitute -12y+1055 for x in the other equation, 12x+6y=30.

$$12\left(-12y+1055\right)+6y=30$$

Multiply 12 times -12y+1055.

$$-144y+12660+6y=30$$

Add -144y to 6y.

$$-138y+12660=30$$

Subtract 12660 from both sides of the equation.

$$-138y=-12630$$

Divide both sides by -138.

$$y=\frac{2105}{23}$$

Substitute \frac{2105}{23} for y in x=-12y+1055. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-12\times \frac{2105}{23}+1055$$

Multiply -12 times \frac{2105}{23}.

$$x=-\frac{25260}{23}+1055$$

Add 1055 to -\frac{25260}{23}.

$$x=-\frac{995}{23}$$

The system is now solved.

$$x=-\frac{995}{23},y=\frac{2105}{23}$$

Steps Using Matrices

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

$$2x+24y=2110,12x+6y=30$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&24\\12&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2110\\30\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&24\\12&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&24\\12&6\end{matrix}\right))\left(\begin{matrix}2&24\\12&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&24\\12&6\end{matrix}\right))\left(\begin{matrix}2110\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&24\\12&6\end{matrix}\right))\left(\begin{matrix}2110\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&24\\12&6\end{matrix}\right))\left(\begin{matrix}2110\\30\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{2\times 6-24\times 12}&-\frac{24}{2\times 6-24\times 12}\\-\frac{12}{2\times 6-24\times 12}&\frac{2}{2\times 6-24\times 12}\end{matrix}\right)\left(\begin{matrix}2110\\30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{46}&\frac{2}{23}\\\frac{1}{23}&-\frac{1}{138}\end{matrix}\right)\left(\begin{matrix}2110\\30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{46}\times 2110+\frac{2}{23}\times 30\\\frac{1}{23}\times 2110-\frac{1}{138}\times 30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{995}{23}\\\frac{2105}{23}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{995}{23},y=\frac{2105}{23}$$

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.

$$2x+24y=2110,12x+6y=30$$

To make 2x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 2.

$$12\times 2x+12\times 24y=12\times 2110,2\times 12x+2\times 6y=2\times 30$$

Simplify.

$$24x+288y=25320,24x+12y=60$$

Subtract 24x+12y=60 from 24x+288y=25320 by subtracting like terms on each side of the equal sign.

$$24x-24x+288y-12y=25320-60$$

Add 24x to -24x. Terms 24x and -24x cancel out, leaving an equation with only one variable that can be solved.

$$288y-12y=25320-60$$

Add 288y to -12y.

$$276y=25320-60$$

Add 25320 to -60.

$$276y=25260$$

Divide both sides by 276.

$$y=\frac{2105}{23}$$

Substitute \frac{2105}{23} for y in 12x+6y=30. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x+6\times \frac{2105}{23}=30$$

Multiply 6 times \frac{2105}{23}.

$$12x+\frac{12630}{23}=30$$

Subtract \frac{12630}{23} from both sides of the equation.

$$12x=-\frac{11940}{23}$$

Divide both sides by 12.

$$x=-\frac{995}{23}$$

The system is now solved.

$$x=-\frac{995}{23},y=\frac{2105}{23}$$