Solve {l}{x+y=150}{16x=43y/2}

$$\left\{ \begin{array} { l } { x + y = 150 } \\ { 16 x = \frac { 43 y } { 2 } } \end{array} \right.$$

Solve for x, y

$$x=86$$

$$y=64$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 2.

$$32x=43y$$

Subtract 43y from both sides.

$$32x-43y=0$$

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.

$$x+y=150,32x-43y=0$$

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, 32x-43y=0.

$$32\left(-y+150\right)-43y=0$$

Multiply 32 times -y+150.

$$-32y+4800-43y=0$$

Add -32y to -43y.

$$-75y+4800=0$$

Subtract 4800 from both sides of the equation.

$$-75y=-4800$$

Divide both sides by -75.

$$y=64$$

Substitute 64 for y in x=-y+150. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-64+150$$

Add 150 to -64.

$$x=86$$

The system is now solved.

$$x=86,y=64$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 2.

$$32x=43y$$

Subtract 43y from both sides.

$$32x-43y=0$$

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

$$x+y=150,32x-43y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\32&-43\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}150\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\32&-43\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\32&-43\end{matrix}\right))\left(\begin{matrix}1&1\\32&-43\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\32&-43\end{matrix}\right))\left(\begin{matrix}150\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\32&-43\end{matrix}\right))\left(\begin{matrix}150\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\32&-43\end{matrix}\right))\left(\begin{matrix}150\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{43}{-43-32}&-\frac{1}{-43-32}\\-\frac{32}{-43-32}&\frac{1}{-43-32}\end{matrix}\right)\left(\begin{matrix}150\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{43}{75}&\frac{1}{75}\\\frac{32}{75}&-\frac{1}{75}\end{matrix}\right)\left(\begin{matrix}150\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{43}{75}\times 150\\\frac{32}{75}\times 150\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}86\\64\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=86,y=64$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 2.

$$32x=43y$$

Subtract 43y from both sides.

$$32x-43y=0$$

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.

$$x+y=150,32x-43y=0$$

To make x and 32x equal, multiply all terms on each side of the first equation by 32 and all terms on each side of the second by 1.

$$32x+32y=32\times 150,32x-43y=0$$

Simplify.

$$32x+32y=4800,32x-43y=0$$

Subtract 32x-43y=0 from 32x+32y=4800 by subtracting like terms on each side of the equal sign.

$$32x-32x+32y+43y=4800$$

Add 32x to -32x. Terms 32x and -32x cancel out, leaving an equation with only one variable that can be solved.

$$32y+43y=4800$$

Add 32y to 43y.

$$75y=4800$$

Divide both sides by 75.

$$y=64$$

Substitute 64 for y in 32x-43y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$32x-43\times 64=0$$

Multiply -43 times 64.

$$32x-2752=0$$

Add 2752 to both sides of the equation.

$$32x=2752$$

Divide both sides by 32.

$$x=86$$

The system is now solved.

$$x=86,y=64$$