Solve {l}{30x+435y=300}{40x+780y=400}

$$\left\{ \begin{array} { l } { 30 x + 435 y = 300 } \\ { 40 x + 780 y = 400 } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=0$$

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.

$$30x+435y=300,40x+780y=400$$

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

$$30x+435y=300$$

Subtract 435y from both sides of the equation.

$$30x=-435y+300$$

Divide both sides by 30.

$$x=\frac{1}{30}\left(-435y+300\right)$$

Multiply \frac{1}{30} times -435y+300.

$$x=-\frac{29}{2}y+10$$

Substitute -\frac{29y}{2}+10 for x in the other equation, 40x+780y=400.

$$40\left(-\frac{29}{2}y+10\right)+780y=400$$

Multiply 40 times -\frac{29y}{2}+10.

$$-580y+400+780y=400$$

Add -580y to 780y.

$$200y+400=400$$

Subtract 400 from both sides of the equation.

$$200y=0$$

Divide both sides by 200.

$$y=0$$

Substitute 0 for y in x=-\frac{29}{2}y+10. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=10$$

The system is now solved.

$$x=10,y=0$$

Steps Using Matrices

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

$$30x+435y=300,40x+780y=400$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&435\\40&780\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\400\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&435\\40&780\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&435\\40&780\end{matrix}\right))\left(\begin{matrix}30&435\\40&780\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&435\\40&780\end{matrix}\right))\left(\begin{matrix}300\\400\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&435\\40&780\end{matrix}\right))\left(\begin{matrix}300\\400\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&435\\40&780\end{matrix}\right))\left(\begin{matrix}300\\400\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{780}{30\times 780-435\times 40}&-\frac{435}{30\times 780-435\times 40}\\-\frac{40}{30\times 780-435\times 40}&\frac{30}{30\times 780-435\times 40}\end{matrix}\right)\left(\begin{matrix}300\\400\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{13}{100}&-\frac{29}{400}\\-\frac{1}{150}&\frac{1}{200}\end{matrix}\right)\left(\begin{matrix}300\\400\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{13}{100}\times 300-\frac{29}{400}\times 400\\-\frac{1}{150}\times 300+\frac{1}{200}\times 400\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=0$$

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.

$$30x+435y=300,40x+780y=400$$

To make 30x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 30.

$$40\times 30x+40\times 435y=40\times 300,30\times 40x+30\times 780y=30\times 400$$

Simplify.

$$1200x+17400y=12000,1200x+23400y=12000$$

Subtract 1200x+23400y=12000 from 1200x+17400y=12000 by subtracting like terms on each side of the equal sign.

$$1200x-1200x+17400y-23400y=12000-12000$$

Add 1200x to -1200x. Terms 1200x and -1200x cancel out, leaving an equation with only one variable that can be solved.

$$17400y-23400y=12000-12000$$

Add 17400y to -23400y.

$$-6000y=12000-12000$$

Add 12000 to -12000.

$$-6000y=0$$

Divide both sides by -6000.

$$y=0$$

Substitute 0 for y in 40x+780y=400. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x=400$$

Divide both sides by 40.

$$x=10$$

The system is now solved.

$$x=10,y=0$$