Solve {l}{12x+24y=60}{0.8(16x+27y)=60}

$$\left\{ \begin{array} { l } { 12 x + 24 y = 60 } \\ { 0.8 ( 16 x + 27 y ) = 60 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=1$$

Steps Using Substitution

Consider the second equation. Divide both sides by 0.8.

$$16x+27y=\frac{60}{0.8}$$

Expand \frac{60}{0.8} by multiplying both numerator and the denominator by 10.

$$16x+27y=\frac{600}{8}$$

Divide 600 by 8 to get 75.

$$16x+27y=75$$

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.

$$12x+24y=60,16x+27y=75$$

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

$$12x+24y=60$$

Subtract 24y from both sides of the equation.

$$12x=-24y+60$$

Divide both sides by 12.

$$x=\frac{1}{12}\left(-24y+60\right)$$

Multiply \frac{1}{12} times -24y+60.

$$x=-2y+5$$

Substitute -2y+5 for x in the other equation, 16x+27y=75.

$$16\left(-2y+5\right)+27y=75$$

Multiply 16 times -2y+5.

$$-32y+80+27y=75$$

Add -32y to 27y.

$$-5y+80=75$$

Subtract 80 from both sides of the equation.

$$-5y=-5$$

Divide both sides by -5.

$$y=1$$

Substitute 1 for y in x=-2y+5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2+5$$

Add 5 to -2.

$$x=3$$

The system is now solved.

$$x=3,y=1$$

Steps Using Matrices

Consider the second equation. Divide both sides by 0.8.

$$16x+27y=\frac{60}{0.8}$$

Expand \frac{60}{0.8} by multiplying both numerator and the denominator by 10.

$$16x+27y=\frac{600}{8}$$

Divide 600 by 8 to get 75.

$$16x+27y=75$$

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

$$12x+24y=60,16x+27y=75$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&24\\16&27\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}60\\75\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&24\\16&27\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&24\\16&27\end{matrix}\right))\left(\begin{matrix}12&24\\16&27\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&24\\16&27\end{matrix}\right))\left(\begin{matrix}60\\75\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&24\\16&27\end{matrix}\right))\left(\begin{matrix}60\\75\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&24\\16&27\end{matrix}\right))\left(\begin{matrix}60\\75\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{27}{12\times 27-24\times 16}&-\frac{24}{12\times 27-24\times 16}\\-\frac{16}{12\times 27-24\times 16}&\frac{12}{12\times 27-24\times 16}\end{matrix}\right)\left(\begin{matrix}60\\75\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{20}&\frac{2}{5}\\\frac{4}{15}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}60\\75\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{20}\times 60+\frac{2}{5}\times 75\\\frac{4}{15}\times 60-\frac{1}{5}\times 75\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=1$$

Steps Using Elimination

Consider the second equation. Divide both sides by 0.8.

$$16x+27y=\frac{60}{0.8}$$

Expand \frac{60}{0.8} by multiplying both numerator and the denominator by 10.

$$16x+27y=\frac{600}{8}$$

Divide 600 by 8 to get 75.

$$16x+27y=75$$

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.

$$12x+24y=60,16x+27y=75$$

To make 12x 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 12.

$$16\times 12x+16\times 24y=16\times 60,12\times 16x+12\times 27y=12\times 75$$

Simplify.

$$192x+384y=960,192x+324y=900$$

Subtract 192x+324y=900 from 192x+384y=960 by subtracting like terms on each side of the equal sign.

$$192x-192x+384y-324y=960-900$$

Add 192x to -192x. Terms 192x and -192x cancel out, leaving an equation with only one variable that can be solved.

$$384y-324y=960-900$$

Add 384y to -324y.

$$60y=960-900$$

Add 960 to -900.

$$60y=60$$

Divide both sides by 60.

$$y=1$$

Substitute 1 for y in 16x+27y=75. Because the resulting equation contains only one variable, you can solve for x directly.

$$16x+27=75$$

Subtract 27 from both sides of the equation.

$$16x=48$$

Divide both sides by 16.

$$x=3$$

The system is now solved.

$$x=3,y=1$$