Solve {l}{x+y=150}{16x=22y}

$$\left. \begin{array} { l } { x + y = 150 } \\ { 16 x = 22 y } \end{array} \right.$$

Solve for x, y

$$x = \frac{1650}{19} = 86\frac{16}{19} \approx 86.842105263$$

$$y = \frac{1200}{19} = 63\frac{3}{19} \approx 63.157894737$$

Steps Using Substitution

Consider the second equation. Subtract 22y from both sides.

$$16x-22y=0$$

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.

$$x+y=150,16x-22y=0$$

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

$$x+y=150$$

Subtract y from both sides of the equation.

$$x=-y+150$$

Substitute -y+150 for x in the other equation, 16x-22y=0.

$$16\left(-y+150\right)-22y=0$$

Multiply 16 times -y+150.

$$-16y+2400-22y=0$$

Add -16y to -22y.

$$-38y+2400=0$$

Subtract 2400 from both sides of the equation.

$$-38y=-2400$$

Divide both sides by -38.

$$y=\frac{1200}{19}$$

Substitute \frac{1200}{19} for y in x=-y+150. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1200}{19}+150$$

Add 150 to -\frac{1200}{19}.

$$x=\frac{1650}{19}$$

The system is now solved.

$$x=\frac{1650}{19},y=\frac{1200}{19}$$

Steps Using Matrices

Consider the second equation. Subtract 22y from both sides.

$$16x-22y=0$$

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

$$x+y=150,16x-22y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\16&-22\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}150\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\16&-22\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\16&-22\end{matrix}\right))\left(\begin{matrix}1&1\\16&-22\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\16&-22\end{matrix}\right))\left(\begin{matrix}150\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\16&-22\end{matrix}\right))\left(\begin{matrix}150\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\16&-22\end{matrix}\right))\left(\begin{matrix}150\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{22}{-22-16}&-\frac{1}{-22-16}\\-\frac{16}{-22-16}&\frac{1}{-22-16}\end{matrix}\right)\left(\begin{matrix}150\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{19}&\frac{1}{38}\\\frac{8}{19}&-\frac{1}{38}\end{matrix}\right)\left(\begin{matrix}150\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{19}\times 150\\\frac{8}{19}\times 150\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1650}{19}\\\frac{1200}{19}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1650}{19},y=\frac{1200}{19}$$

Steps Using Elimination

Consider the second equation. Subtract 22y from both sides.

$$16x-22y=0$$

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.

$$x+y=150,16x-22y=0$$

To make x 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 1.

$$16x+16y=16\times 150,16x-22y=0$$

Simplify.

$$16x+16y=2400,16x-22y=0$$

Subtract 16x-22y=0 from 16x+16y=2400 by subtracting like terms on each side of the equal sign.

$$16x-16x+16y+22y=2400$$

Add 16x to -16x. Terms 16x and -16x cancel out, leaving an equation with only one variable that can be solved.

$$16y+22y=2400$$

Add 16y to 22y.

$$38y=2400$$

Divide both sides by 38.

$$y=\frac{1200}{19}$$

Substitute \frac{1200}{19} for y in 16x-22y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$16x-22\times \frac{1200}{19}=0$$

Multiply -22 times \frac{1200}{19}.

$$16x-\frac{26400}{19}=0$$

Add \frac{26400}{19} to both sides of the equation.

$$16x=\frac{26400}{19}$$

Divide both sides by 16.

$$x=\frac{1650}{19}$$

The system is now solved.

$$x=\frac{1650}{19},y=\frac{1200}{19}$$