Solve {l}{6x+5y=300}{15x+80y=4800}

$$\left. \begin{array} { l } { 6 x + 5 y = 300 } \\ { 15 x + 80 y = 4800 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=60$$

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.

$$6x+5y=300,15x+80y=4800$$

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

$$6x+5y=300$$

Subtract 5y from both sides of the equation.

$$6x=-5y+300$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(-5y+300\right)$$

Multiply \frac{1}{6} times -5y+300.

$$x=-\frac{5}{6}y+50$$

Substitute -\frac{5y}{6}+50 for x in the other equation, 15x+80y=4800.

$$15\left(-\frac{5}{6}y+50\right)+80y=4800$$

Multiply 15 times -\frac{5y}{6}+50.

$$-\frac{25}{2}y+750+80y=4800$$

Add -\frac{25y}{2} to 80y.

$$\frac{135}{2}y+750=4800$$

Subtract 750 from both sides of the equation.

$$\frac{135}{2}y=4050$$

Divide both sides of the equation by \frac{135}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=60$$

Substitute 60 for y in x=-\frac{5}{6}y+50. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{6}\times 60+50$$

Multiply -\frac{5}{6} times 60.

$$x=-50+50$$

Add 50 to -50.

$$x=0$$

The system is now solved.

$$x=0,y=60$$

Steps Using Matrices

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

$$6x+5y=300,15x+80y=4800$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&5\\15&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\4800\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&5\\15&80\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&5\\15&80\end{matrix}\right))\left(\begin{matrix}6&5\\15&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\15&80\end{matrix}\right))\left(\begin{matrix}300\\4800\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\15&80\end{matrix}\right))\left(\begin{matrix}300\\4800\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\15&80\end{matrix}\right))\left(\begin{matrix}300\\4800\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{6\times 80-5\times 15}&-\frac{5}{6\times 80-5\times 15}\\-\frac{15}{6\times 80-5\times 15}&\frac{6}{6\times 80-5\times 15}\end{matrix}\right)\left(\begin{matrix}300\\4800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{16}{81}&-\frac{1}{81}\\-\frac{1}{27}&\frac{2}{135}\end{matrix}\right)\left(\begin{matrix}300\\4800\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{16}{81}\times 300-\frac{1}{81}\times 4800\\-\frac{1}{27}\times 300+\frac{2}{135}\times 4800\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\60\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=60$$

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.

$$6x+5y=300,15x+80y=4800$$

To make 6x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 6.

$$15\times 6x+15\times 5y=15\times 300,6\times 15x+6\times 80y=6\times 4800$$

Simplify.

$$90x+75y=4500,90x+480y=28800$$

Subtract 90x+480y=28800 from 90x+75y=4500 by subtracting like terms on each side of the equal sign.

$$90x-90x+75y-480y=4500-28800$$

Add 90x to -90x. Terms 90x and -90x cancel out, leaving an equation with only one variable that can be solved.

$$75y-480y=4500-28800$$

Add 75y to -480y.

$$-405y=4500-28800$$

Add 4500 to -28800.

$$-405y=-24300$$

Divide both sides by -405.

$$y=60$$

Substitute 60 for y in 15x+80y=4800. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x+80\times 60=4800$$

Multiply 80 times 60.

$$15x+4800=4800$$

Subtract 4800 from both sides of the equation.

$$15x=0$$

Divide both sides by 15.

$$x=0$$

The system is now solved.

$$x=0,y=60$$