Solve {l}{11/4x+3/5=-67/3y}{12/5y-7=(frac{1}{2}-frac{3}{4}x)}

$$\left\{ \begin{array} { l } { 1 \frac { 1 } { 4 } x + \frac { 3 } { 5 } = - \frac { 67 } { 3 } y } \\ { \frac { 12 } { 5 } y - 7 = ( \frac { 1 } { 2 } - \frac { 3 } { 4 } x ) } \end{array} \right.$$

Solve for x, y

$$x = \frac{16894}{1375} = 12\frac{394}{1375} \approx 12.286545455$$

$$y=-\frac{393}{550}\approx -0.714545455$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 60, the least common multiple of 4,5,3.

$$15\left(1\times 4+1\right)x+36=-1340y$$

Multiply 1 and 4 to get 4.

$$15\left(4+1\right)x+36=-1340y$$

Add 4 and 1 to get 5.

$$15\times 5x+36=-1340y$$

Multiply 15 and 5 to get 75.

$$75x+36=-1340y$$

Add 1340y to both sides.

$$75x+36+1340y=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$75x+1340y=-36$$

Consider the second equation. Add \frac{3}{4}x to both sides.

$$\frac{12}{5}y-7+\frac{3}{4}x=\frac{1}{2}$$

Add 7 to both sides.

$$\frac{12}{5}y+\frac{3}{4}x=\frac{1}{2}+7$$

Add \frac{1}{2} and 7 to get \frac{15}{2}.

$$\frac{12}{5}y+\frac{3}{4}x=\frac{15}{2}$$

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.

$$75x+1340y=-36,\frac{3}{4}x+\frac{12}{5}y=\frac{15}{2}$$

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

$$75x+1340y=-36$$

Subtract 1340y from both sides of the equation.

$$75x=-1340y-36$$

Divide both sides by 75.

$$x=\frac{1}{75}\left(-1340y-36\right)$$

Multiply \frac{1}{75} times -1340y-36.

$$x=-\frac{268}{15}y-\frac{12}{25}$$

Substitute -\frac{268y}{15}-\frac{12}{25} for x in the other equation, \frac{3}{4}x+\frac{12}{5}y=\frac{15}{2}.

$$\frac{3}{4}\left(-\frac{268}{15}y-\frac{12}{25}\right)+\frac{12}{5}y=\frac{15}{2}$$

Multiply \frac{3}{4} times -\frac{268y}{15}-\frac{12}{25}.

$$-\frac{67}{5}y-\frac{9}{25}+\frac{12}{5}y=\frac{15}{2}$$

Add -\frac{67y}{5} to \frac{12y}{5}.

$$-11y-\frac{9}{25}=\frac{15}{2}$$

Add \frac{9}{25} to both sides of the equation.

$$-11y=\frac{393}{50}$$

Divide both sides by -11.

$$y=-\frac{393}{550}$$

Substitute -\frac{393}{550} for y in x=-\frac{268}{15}y-\frac{12}{25}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{268}{15}\left(-\frac{393}{550}\right)-\frac{12}{25}$$

Multiply -\frac{268}{15} times -\frac{393}{550} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{17554}{1375}-\frac{12}{25}$$

Add -\frac{12}{25} to \frac{17554}{1375} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{16894}{1375}$$

The system is now solved.

$$x=\frac{16894}{1375},y=-\frac{393}{550}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 60, the least common multiple of 4,5,3.

$$15\left(1\times 4+1\right)x+36=-1340y$$

Multiply 1 and 4 to get 4.

$$15\left(4+1\right)x+36=-1340y$$

Add 4 and 1 to get 5.

$$15\times 5x+36=-1340y$$

Multiply 15 and 5 to get 75.

$$75x+36=-1340y$$

Add 1340y to both sides.

$$75x+36+1340y=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$75x+1340y=-36$$

Consider the second equation. Add \frac{3}{4}x to both sides.

$$\frac{12}{5}y-7+\frac{3}{4}x=\frac{1}{2}$$

Add 7 to both sides.

$$\frac{12}{5}y+\frac{3}{4}x=\frac{1}{2}+7$$

Add \frac{1}{2} and 7 to get \frac{15}{2}.

$$\frac{12}{5}y+\frac{3}{4}x=\frac{15}{2}$$

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

$$75x+1340y=-36,\frac{3}{4}x+\frac{12}{5}y=\frac{15}{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-36\\\frac{15}{2}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right).

$$inverse(\left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right))\left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right))\left(\begin{matrix}-36\\\frac{15}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right))\left(\begin{matrix}-36\\\frac{15}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}75&1340\\\frac{3}{4}&\frac{12}{5}\end{matrix}\right))\left(\begin{matrix}-36\\\frac{15}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{12}{5}}{75\times \frac{12}{5}-1340\times \frac{3}{4}}&-\frac{1340}{75\times \frac{12}{5}-1340\times \frac{3}{4}}\\-\frac{\frac{3}{4}}{75\times \frac{12}{5}-1340\times \frac{3}{4}}&\frac{75}{75\times \frac{12}{5}-1340\times \frac{3}{4}}\end{matrix}\right)\left(\begin{matrix}-36\\\frac{15}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{1375}&\frac{268}{165}\\\frac{1}{1100}&-\frac{1}{11}\end{matrix}\right)\left(\begin{matrix}-36\\\frac{15}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{1375}\left(-36\right)+\frac{268}{165}\times \frac{15}{2}\\\frac{1}{1100}\left(-36\right)-\frac{1}{11}\times \frac{15}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{16894}{1375}\\-\frac{393}{550}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{16894}{1375},y=-\frac{393}{550}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 60, the least common multiple of 4,5,3.

$$15\left(1\times 4+1\right)x+36=-1340y$$

Multiply 1 and 4 to get 4.

$$15\left(4+1\right)x+36=-1340y$$

Add 4 and 1 to get 5.

$$15\times 5x+36=-1340y$$

Multiply 15 and 5 to get 75.

$$75x+36=-1340y$$

Add 1340y to both sides.

$$75x+36+1340y=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$75x+1340y=-36$$

Consider the second equation. Add \frac{3}{4}x to both sides.

$$\frac{12}{5}y-7+\frac{3}{4}x=\frac{1}{2}$$

Add 7 to both sides.

$$\frac{12}{5}y+\frac{3}{4}x=\frac{1}{2}+7$$

Add \frac{1}{2} and 7 to get \frac{15}{2}.

$$\frac{12}{5}y+\frac{3}{4}x=\frac{15}{2}$$

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.

$$75x+1340y=-36,\frac{3}{4}x+\frac{12}{5}y=\frac{15}{2}$$

To make 75x and \frac{3x}{4} equal, multiply all terms on each side of the first equation by \frac{3}{4} and all terms on each side of the second by 75.

$$\frac{3}{4}\times 75x+\frac{3}{4}\times 1340y=\frac{3}{4}\left(-36\right),75\times \frac{3}{4}x+75\times \frac{12}{5}y=75\times \frac{15}{2}$$

Simplify.

$$\frac{225}{4}x+1005y=-27,\frac{225}{4}x+180y=\frac{1125}{2}$$

Subtract \frac{225}{4}x+180y=\frac{1125}{2} from \frac{225}{4}x+1005y=-27 by subtracting like terms on each side of the equal sign.

$$\frac{225}{4}x-\frac{225}{4}x+1005y-180y=-27-\frac{1125}{2}$$

Add \frac{225x}{4} to -\frac{225x}{4}. Terms \frac{225x}{4} and -\frac{225x}{4} cancel out, leaving an equation with only one variable that can be solved.

$$1005y-180y=-27-\frac{1125}{2}$$

Add 1005y to -180y.

$$825y=-27-\frac{1125}{2}$$

Add -27 to -\frac{1125}{2}.

$$825y=-\frac{1179}{2}$$

Divide both sides by 825.

$$y=-\frac{393}{550}$$

Substitute -\frac{393}{550} for y in \frac{3}{4}x+\frac{12}{5}y=\frac{15}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$\frac{3}{4}x+\frac{12}{5}\left(-\frac{393}{550}\right)=\frac{15}{2}$$

Multiply \frac{12}{5} times -\frac{393}{550} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$\frac{3}{4}x-\frac{2358}{1375}=\frac{15}{2}$$

Add \frac{2358}{1375} to both sides of the equation.

$$\frac{3}{4}x=\frac{25341}{2750}$$

Divide both sides of the equation by \frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{16894}{1375}$$

The system is now solved.

$$x=\frac{16894}{1375},y=-\frac{393}{550}$$