Solve {l}{29x+92y=1924}{29x+30y=1924}

$$\left. \begin{array} { l } { 29 x + 92 y = 1924 } \\ { 29 x + 30 y = 1924 } \end{array} \right.$$

Solve for x, y

$$x = \frac{1924}{29} = 66\frac{10}{29} \approx 66.344827586$$

$$y=0$$

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.

$$29x+92y=1924,29x+30y=1924$$

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

$$29x+92y=1924$$

Subtract 92y from both sides of the equation.

$$29x=-92y+1924$$

Divide both sides by 29.

$$x=\frac{1}{29}\left(-92y+1924\right)$$

Multiply \frac{1}{29} times -92y+1924.

$$x=-\frac{92}{29}y+\frac{1924}{29}$$

Substitute \frac{-92y+1924}{29} for x in the other equation, 29x+30y=1924.

$$29\left(-\frac{92}{29}y+\frac{1924}{29}\right)+30y=1924$$

Multiply 29 times \frac{-92y+1924}{29}.

$$-92y+1924+30y=1924$$

Add -92y to 30y.

$$-62y+1924=1924$$

Subtract 1924 from both sides of the equation.

$$-62y=0$$

Divide both sides by -62.

$$y=0$$

Substitute 0 for y in x=-\frac{92}{29}y+\frac{1924}{29}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1924}{29}$$

The system is now solved.

$$x=\frac{1924}{29},y=0$$

Steps Using Matrices

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

$$29x+92y=1924,29x+30y=1924$$

Write the equations in matrix form.

$$\left(\begin{matrix}29&92\\29&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1924\\1924\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}29&92\\29&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}29&92\\29&30\end{matrix}\right))\left(\begin{matrix}29&92\\29&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}29&92\\29&30\end{matrix}\right))\left(\begin{matrix}1924\\1924\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}29&92\\29&30\end{matrix}\right))\left(\begin{matrix}1924\\1924\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}29&92\\29&30\end{matrix}\right))\left(\begin{matrix}1924\\1924\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{29\times 30-92\times 29}&-\frac{92}{29\times 30-92\times 29}\\-\frac{29}{29\times 30-92\times 29}&\frac{29}{29\times 30-92\times 29}\end{matrix}\right)\left(\begin{matrix}1924\\1924\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{899}&\frac{46}{899}\\\frac{1}{62}&-\frac{1}{62}\end{matrix}\right)\left(\begin{matrix}1924\\1924\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{899}\times 1924+\frac{46}{899}\times 1924\\\frac{1}{62}\times 1924-\frac{1}{62}\times 1924\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1924}{29}\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1924}{29},y=0$$

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.

$$29x+92y=1924,29x+30y=1924$$

Subtract 29x+30y=1924 from 29x+92y=1924 by subtracting like terms on each side of the equal sign.

$$29x-29x+92y-30y=1924-1924$$

Add 29x to -29x. Terms 29x and -29x cancel out, leaving an equation with only one variable that can be solved.

$$92y-30y=1924-1924$$

Add 92y to -30y.

$$62y=1924-1924$$

Add 1924 to -1924.

$$62y=0$$

Divide both sides by 62.

$$y=0$$

Substitute 0 for y in 29x+30y=1924. Because the resulting equation contains only one variable, you can solve for x directly.

$$29x=1924$$

Divide both sides by 29.

$$x=\frac{1924}{29}$$

The system is now solved.

$$x=\frac{1924}{29},y=0$$