Solve {l}{5x+2y=120}{16x-4y=228}

$$\left. \begin{array} { l } { 5 x + 2 y = 120 } \\ { 16 x - 4 y = 228 } \end{array} \right.$$

Solve for x, y

$$x=18$$

$$y=15$$

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.

$$5x+2y=120,16x-4y=228$$

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

$$5x+2y=120$$

Subtract 2y from both sides of the equation.

$$5x=-2y+120$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-2y+120\right)$$

Multiply \frac{1}{5} times -2y+120.

$$x=-\frac{2}{5}y+24$$

Substitute -\frac{2y}{5}+24 for x in the other equation, 16x-4y=228.

$$16\left(-\frac{2}{5}y+24\right)-4y=228$$

Multiply 16 times -\frac{2y}{5}+24.

$$-\frac{32}{5}y+384-4y=228$$

Add -\frac{32y}{5} to -4y.

$$-\frac{52}{5}y+384=228$$

Subtract 384 from both sides of the equation.

$$-\frac{52}{5}y=-156$$

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

$$y=15$$

Substitute 15 for y in x=-\frac{2}{5}y+24. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{5}\times 15+24$$

Multiply -\frac{2}{5} times 15.

$$x=-6+24$$

Add 24 to -6.

$$x=18$$

The system is now solved.

$$x=18,y=15$$

Steps Using Matrices

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

$$5x+2y=120,16x-4y=228$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&2\\16&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\228\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&2\\16&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&2\\16&-4\end{matrix}\right))\left(\begin{matrix}5&2\\16&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&2\\16&-4\end{matrix}\right))\left(\begin{matrix}120\\228\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&2\\16&-4\end{matrix}\right))\left(\begin{matrix}120\\228\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&2\\16&-4\end{matrix}\right))\left(\begin{matrix}120\\228\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5\left(-4\right)-2\times 16}&-\frac{2}{5\left(-4\right)-2\times 16}\\-\frac{16}{5\left(-4\right)-2\times 16}&\frac{5}{5\left(-4\right)-2\times 16}\end{matrix}\right)\left(\begin{matrix}120\\228\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{13}&\frac{1}{26}\\\frac{4}{13}&-\frac{5}{52}\end{matrix}\right)\left(\begin{matrix}120\\228\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{13}\times 120+\frac{1}{26}\times 228\\\frac{4}{13}\times 120-\frac{5}{52}\times 228\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=18,y=15$$

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.

$$5x+2y=120,16x-4y=228$$

To make 5x and 16x equal, multiply all terms on each side of the first equation by 16 and all terms on each side of the second by 5.

$$16\times 5x+16\times 2y=16\times 120,5\times 16x+5\left(-4\right)y=5\times 228$$

Simplify.

$$80x+32y=1920,80x-20y=1140$$

Subtract 80x-20y=1140 from 80x+32y=1920 by subtracting like terms on each side of the equal sign.

$$80x-80x+32y+20y=1920-1140$$

Add 80x to -80x. Terms 80x and -80x cancel out, leaving an equation with only one variable that can be solved.

$$32y+20y=1920-1140$$

Add 32y to 20y.

$$52y=1920-1140$$

Add 1920 to -1140.

$$52y=780$$

Divide both sides by 52.

$$y=15$$

Substitute 15 for y in 16x-4y=228. Because the resulting equation contains only one variable, you can solve for x directly.

$$16x-4\times 15=228$$

Multiply -4 times 15.

$$16x-60=228$$

Add 60 to both sides of the equation.

$$16x=288$$

Divide both sides by 16.

$$x=18$$

The system is now solved.

$$x=18,y=15$$