Solve {l}{1/32y=16x-150}{1/2y=2*8x}

$$\left\{ \begin{array} { l } { 1 / 32 y = 16 x - 150 } \\ { 1 / 2 y = 2 \cdot 8 x } \end{array} \right.$$

Solve for y, x

$$x=10$$

$$y=320$$

Steps Using Substitution

Consider the first equation. Subtract 16x from both sides.

$$\frac{1}{32}y-16x=-150$$

Consider the second equation. Multiply 2 and 8 to get 16.

$$\frac{1}{2}y=16x$$

Subtract 16x from both sides.

$$\frac{1}{2}y-16x=0$$

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.

$$\frac{1}{32}y-16x=-150,\frac{1}{2}y-16x=0$$

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

$$\frac{1}{32}y-16x=-150$$

Add 16x to both sides of the equation.

$$\frac{1}{32}y=16x-150$$

Multiply both sides by 32.

$$y=32\left(16x-150\right)$$

Multiply 32 times 16x-150.

$$y=512x-4800$$

Substitute 512x-4800 for y in the other equation, \frac{1}{2}y-16x=0.

$$\frac{1}{2}\left(512x-4800\right)-16x=0$$

Multiply \frac{1}{2} times 512x-4800.

$$256x-2400-16x=0$$

Add 256x to -16x.

$$240x-2400=0$$

Add 2400 to both sides of the equation.

$$240x=2400$$

Divide both sides by 240.

$$x=10$$

Substitute 10 for x in y=512x-4800. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=512\times 10-4800$$

Multiply 512 times 10.

$$y=5120-4800$$

Add -4800 to 5120.

$$y=320$$

The system is now solved.

$$y=320,x=10$$

Steps Using Matrices

Consider the first equation. Subtract 16x from both sides.

$$\frac{1}{32}y-16x=-150$$

Consider the second equation. Multiply 2 and 8 to get 16.

$$\frac{1}{2}y=16x$$

Subtract 16x from both sides.

$$\frac{1}{2}y-16x=0$$

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

$$\frac{1}{32}y-16x=-150,\frac{1}{2}y-16x=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-150\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right))\left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right))\left(\begin{matrix}-150\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right))\left(\begin{matrix}-150\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{32}&-16\\\frac{1}{2}&-16\end{matrix}\right))\left(\begin{matrix}-150\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{16}{\frac{1}{32}\left(-16\right)-\left(-16\times \frac{1}{2}\right)}&-\frac{-16}{\frac{1}{32}\left(-16\right)-\left(-16\times \frac{1}{2}\right)}\\-\frac{\frac{1}{2}}{\frac{1}{32}\left(-16\right)-\left(-16\times \frac{1}{2}\right)}&\frac{\frac{1}{32}}{\frac{1}{32}\left(-16\right)-\left(-16\times \frac{1}{2}\right)}\end{matrix}\right)\left(\begin{matrix}-150\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{32}{15}&\frac{32}{15}\\-\frac{1}{15}&\frac{1}{240}\end{matrix}\right)\left(\begin{matrix}-150\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{32}{15}\left(-150\right)\\-\frac{1}{15}\left(-150\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}320\\10\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=320,x=10$$

Steps Using Elimination

Consider the first equation. Subtract 16x from both sides.

$$\frac{1}{32}y-16x=-150$$

Consider the second equation. Multiply 2 and 8 to get 16.

$$\frac{1}{2}y=16x$$

Subtract 16x from both sides.

$$\frac{1}{2}y-16x=0$$

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.

$$\frac{1}{32}y-16x=-150,\frac{1}{2}y-16x=0$$

Subtract \frac{1}{2}y-16x=0 from \frac{1}{32}y-16x=-150 by subtracting like terms on each side of the equal sign.

$$\frac{1}{32}y-\frac{1}{2}y-16x+16x=-150$$

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

$$\frac{1}{32}y-\frac{1}{2}y=-150$$

Add \frac{y}{32} to -\frac{y}{2}.

$$-\frac{15}{32}y=-150$$

Divide both sides of the equation by -\frac{15}{32}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=320$$

Substitute 320 for y in \frac{1}{2}y-16x=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{1}{2}\times 320-16x=0$$

Multiply \frac{1}{2} times 320.

$$160-16x=0$$

Subtract 160 from both sides of the equation.

$$-16x=-160$$

Divide both sides by -16.

$$x=10$$

The system is now solved.

$$y=320,x=10$$