Solve {l}{x/2+y/3=16}{x/3-y/4=5*(2)}

$$\left\{ \begin{array} { l } { \frac { x } { 2 } + \frac { y } { 3 } = 16 } \\ { \frac { x } { 3 } - \frac { y } { 4 } = 5 \cdot ( 2 ) } \end{array} \right.$$

Solve for x, y

$$x = \frac{528}{17} = 31\frac{1}{17} \approx 31.058823529$$

$$y = \frac{24}{17} = 1\frac{7}{17} \approx 1.411764706$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x+2y=96$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4x-3y=60\times 2$$

Multiply 60 and 2 to get 120.

$$4x-3y=120$$

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.

$$3x+2y=96,4x-3y=120$$

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

$$3x+2y=96$$

Subtract 2y from both sides of the equation.

$$3x=-2y+96$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-2y+96\right)$$

Multiply \frac{1}{3} times -2y+96.

$$x=-\frac{2}{3}y+32$$

Substitute -\frac{2y}{3}+32 for x in the other equation, 4x-3y=120.

$$4\left(-\frac{2}{3}y+32\right)-3y=120$$

Multiply 4 times -\frac{2y}{3}+32.

$$-\frac{8}{3}y+128-3y=120$$

Add -\frac{8y}{3} to -3y.

$$-\frac{17}{3}y+128=120$$

Subtract 128 from both sides of the equation.

$$-\frac{17}{3}y=-8$$

Divide both sides of the equation by -\frac{17}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{24}{17}$$

Substitute \frac{24}{17} for y in x=-\frac{2}{3}y+32. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\times \frac{24}{17}+32$$

Multiply -\frac{2}{3} times \frac{24}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{16}{17}+32$$

Add 32 to -\frac{16}{17}.

$$x=\frac{528}{17}$$

The system is now solved.

$$x=\frac{528}{17},y=\frac{24}{17}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x+2y=96$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4x-3y=60\times 2$$

Multiply 60 and 2 to get 120.

$$4x-3y=120$$

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

$$3x+2y=96,4x-3y=120$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&2\\4&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}96\\120\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\4&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&2\\4&-3\end{matrix}\right))\left(\begin{matrix}3&2\\4&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\4&-3\end{matrix}\right))\left(\begin{matrix}96\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\4&-3\end{matrix}\right))\left(\begin{matrix}96\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\4&-3\end{matrix}\right))\left(\begin{matrix}96\\120\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{3\left(-3\right)-2\times 4}&-\frac{2}{3\left(-3\right)-2\times 4}\\-\frac{4}{3\left(-3\right)-2\times 4}&\frac{3}{3\left(-3\right)-2\times 4}\end{matrix}\right)\left(\begin{matrix}96\\120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{17}&\frac{2}{17}\\\frac{4}{17}&-\frac{3}{17}\end{matrix}\right)\left(\begin{matrix}96\\120\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{17}\times 96+\frac{2}{17}\times 120\\\frac{4}{17}\times 96-\frac{3}{17}\times 120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{528}{17}\\\frac{24}{17}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{528}{17},y=\frac{24}{17}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x+2y=96$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4x-3y=60\times 2$$

Multiply 60 and 2 to get 120.

$$4x-3y=120$$

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.

$$3x+2y=96,4x-3y=120$$

To make 3x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 3.

$$4\times 3x+4\times 2y=4\times 96,3\times 4x+3\left(-3\right)y=3\times 120$$

Simplify.

$$12x+8y=384,12x-9y=360$$

Subtract 12x-9y=360 from 12x+8y=384 by subtracting like terms on each side of the equal sign.

$$12x-12x+8y+9y=384-360$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$8y+9y=384-360$$

Add 8y to 9y.

$$17y=384-360$$

Add 384 to -360.

$$17y=24$$

Divide both sides by 17.

$$y=\frac{24}{17}$$

Substitute \frac{24}{17} for y in 4x-3y=120. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x-3\times \frac{24}{17}=120$$

Multiply -3 times \frac{24}{17}.

$$4x-\frac{72}{17}=120$$

Add \frac{72}{17} to both sides of the equation.

$$4x=\frac{2112}{17}$$

Divide both sides by 4.

$$x=\frac{528}{17}$$

The system is now solved.

$$x=\frac{528}{17},y=\frac{24}{17}$$