Solve {l}{15x+12y=1950}{7x+16y=1950}

$$\left\{ \begin{array} { l } { 15 x + 12 y = 1950 } \\ { 7 x + 16 y = 1950 } \end{array} \right.$$

Solve for x, y

$$x=50$$

$$y=100$$

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.

$$15x+12y=1950,7x+16y=1950$$

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

$$15x+12y=1950$$

Subtract 12y from both sides of the equation.

$$15x=-12y+1950$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(-12y+1950\right)$$

Multiply \frac{1}{15} times -12y+1950.

$$x=-\frac{4}{5}y+130$$

Substitute -\frac{4y}{5}+130 for x in the other equation, 7x+16y=1950.

$$7\left(-\frac{4}{5}y+130\right)+16y=1950$$

Multiply 7 times -\frac{4y}{5}+130.

$$-\frac{28}{5}y+910+16y=1950$$

Add -\frac{28y}{5} to 16y.

$$\frac{52}{5}y+910=1950$$

Subtract 910 from both sides of the equation.

$$\frac{52}{5}y=1040$$

Divide both sides of the equation by \frac{52}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=100$$

Substitute 100 for y in x=-\frac{4}{5}y+130. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{5}\times 100+130$$

Multiply -\frac{4}{5} times 100.

$$x=-80+130$$

Add 130 to -80.

$$x=50$$

The system is now solved.

$$x=50,y=100$$

Steps Using Matrices

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

$$15x+12y=1950,7x+16y=1950$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&12\\7&16\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1950\\1950\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&12\\7&16\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&12\\7&16\end{matrix}\right))\left(\begin{matrix}15&12\\7&16\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&12\\7&16\end{matrix}\right))\left(\begin{matrix}1950\\1950\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&12\\7&16\end{matrix}\right))\left(\begin{matrix}1950\\1950\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&12\\7&16\end{matrix}\right))\left(\begin{matrix}1950\\1950\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{16}{15\times 16-12\times 7}&-\frac{12}{15\times 16-12\times 7}\\-\frac{7}{15\times 16-12\times 7}&\frac{15}{15\times 16-12\times 7}\end{matrix}\right)\left(\begin{matrix}1950\\1950\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{39}&-\frac{1}{13}\\-\frac{7}{156}&\frac{5}{52}\end{matrix}\right)\left(\begin{matrix}1950\\1950\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{39}\times 1950-\frac{1}{13}\times 1950\\-\frac{7}{156}\times 1950+\frac{5}{52}\times 1950\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\100\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=50,y=100$$

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.

$$15x+12y=1950,7x+16y=1950$$

To make 15x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 15.

$$7\times 15x+7\times 12y=7\times 1950,15\times 7x+15\times 16y=15\times 1950$$

Simplify.

$$105x+84y=13650,105x+240y=29250$$

Subtract 105x+240y=29250 from 105x+84y=13650 by subtracting like terms on each side of the equal sign.

$$105x-105x+84y-240y=13650-29250$$

Add 105x to -105x. Terms 105x and -105x cancel out, leaving an equation with only one variable that can be solved.

$$84y-240y=13650-29250$$

Add 84y to -240y.

$$-156y=13650-29250$$

Add 13650 to -29250.

$$-156y=-15600$$

Divide both sides by -156.

$$y=100$$

Substitute 100 for y in 7x+16y=1950. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x+16\times 100=1950$$

Multiply 16 times 100.

$$7x+1600=1950$$

Subtract 1600 from both sides of the equation.

$$7x=350$$

Divide both sides by 7.

$$x=50$$

The system is now solved.

$$x=50,y=100$$