Solve {l}{2x-3y/3=3+3y-4x/4}{1/6(6y+7x)=1/5(7x+12y)-7}

$$\left. \begin{array} { l } { \frac { 2 x - 3 y } { 3 } = 3 + \frac { 3 y - 4 x } { 4 } } \\ { \frac { 1 } { 6 } ( 6 y + 7 x ) = \frac { 1 } { 5 } ( 7 x + 12 y ) - 7 } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=4$$

Steps Using Substitution

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

$$4\left(2x-3y\right)=36+3\left(3y-4x\right)$$

Use the distributive property to multiply 4 by 2x-3y.

$$8x-12y=36+3\left(3y-4x\right)$$

Use the distributive property to multiply 3 by 3y-4x.

$$8x-12y=36+9y-12x$$

Subtract 9y from both sides.

$$8x-12y-9y=36-12x$$

Combine -12y and -9y to get -21y.

$$8x-21y=36-12x$$

Add 12x to both sides.

$$8x-21y+12x=36$$

Combine 8x and 12x to get 20x.

$$20x-21y=36$$

Consider the second equation. Use the distributive property to multiply \frac{1}{6} by 6y+7x.

$$y+\frac{7}{6}x=\frac{1}{5}\left(7x+12y\right)-7$$

Use the distributive property to multiply \frac{1}{5} by 7x+12y.

$$y+\frac{7}{6}x=\frac{7}{5}x+\frac{12}{5}y-7$$

Subtract \frac{7}{5}x from both sides.

$$y+\frac{7}{6}x-\frac{7}{5}x=\frac{12}{5}y-7$$

Combine \frac{7}{6}x and -\frac{7}{5}x to get -\frac{7}{30}x.

$$y-\frac{7}{30}x=\frac{12}{5}y-7$$

Subtract \frac{12}{5}y from both sides.

$$y-\frac{7}{30}x-\frac{12}{5}y=-7$$

Combine y and -\frac{12}{5}y to get -\frac{7}{5}y.

$$-\frac{7}{5}y-\frac{7}{30}x=-7$$

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.

$$20x-21y=36,-\frac{7}{30}x-\frac{7}{5}y=-7$$

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

$$20x-21y=36$$

Add 21y to both sides of the equation.

$$20x=21y+36$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(21y+36\right)$$

Multiply \frac{1}{20} times 21y+36.

$$x=\frac{21}{20}y+\frac{9}{5}$$

Substitute \frac{21y}{20}+\frac{9}{5} for x in the other equation, -\frac{7}{30}x-\frac{7}{5}y=-7.

$$-\frac{7}{30}\left(\frac{21}{20}y+\frac{9}{5}\right)-\frac{7}{5}y=-7$$

Multiply -\frac{7}{30} times \frac{21y}{20}+\frac{9}{5}.

$$-\frac{49}{200}y-\frac{21}{50}-\frac{7}{5}y=-7$$

Add -\frac{49y}{200} to -\frac{7y}{5}.

$$-\frac{329}{200}y-\frac{21}{50}=-7$$

Add \frac{21}{50} to both sides of the equation.

$$-\frac{329}{200}y=-\frac{329}{50}$$

Divide both sides of the equation by -\frac{329}{200}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=4$$

Substitute 4 for y in x=\frac{21}{20}y+\frac{9}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{21}{20}\times 4+\frac{9}{5}$$

Multiply \frac{21}{20} times 4.

$$x=\frac{21+9}{5}$$

Add \frac{9}{5} to \frac{21}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=6$$

The system is now solved.

$$x=6,y=4$$

Steps Using Matrices

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

$$4\left(2x-3y\right)=36+3\left(3y-4x\right)$$

Use the distributive property to multiply 4 by 2x-3y.

$$8x-12y=36+3\left(3y-4x\right)$$

Use the distributive property to multiply 3 by 3y-4x.

$$8x-12y=36+9y-12x$$

Subtract 9y from both sides.

$$8x-12y-9y=36-12x$$

Combine -12y and -9y to get -21y.

$$8x-21y=36-12x$$

Add 12x to both sides.

$$8x-21y+12x=36$$

Combine 8x and 12x to get 20x.

$$20x-21y=36$$

Consider the second equation. Use the distributive property to multiply \frac{1}{6} by 6y+7x.

$$y+\frac{7}{6}x=\frac{1}{5}\left(7x+12y\right)-7$$

Use the distributive property to multiply \frac{1}{5} by 7x+12y.

$$y+\frac{7}{6}x=\frac{7}{5}x+\frac{12}{5}y-7$$

Subtract \frac{7}{5}x from both sides.

$$y+\frac{7}{6}x-\frac{7}{5}x=\frac{12}{5}y-7$$

Combine \frac{7}{6}x and -\frac{7}{5}x to get -\frac{7}{30}x.

$$y-\frac{7}{30}x=\frac{12}{5}y-7$$

Subtract \frac{12}{5}y from both sides.

$$y-\frac{7}{30}x-\frac{12}{5}y=-7$$

Combine y and -\frac{12}{5}y to get -\frac{7}{5}y.

$$-\frac{7}{5}y-\frac{7}{30}x=-7$$

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

$$20x-21y=36,-\frac{7}{30}x-\frac{7}{5}y=-7$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36\\-7\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right))\left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right))\left(\begin{matrix}36\\-7\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right))\left(\begin{matrix}36\\-7\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-21\\-\frac{7}{30}&-\frac{7}{5}\end{matrix}\right))\left(\begin{matrix}36\\-7\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{7}{5}}{20\left(-\frac{7}{5}\right)-\left(-21\left(-\frac{7}{30}\right)\right)}&-\frac{-21}{20\left(-\frac{7}{5}\right)-\left(-21\left(-\frac{7}{30}\right)\right)}\\-\frac{-\frac{7}{30}}{20\left(-\frac{7}{5}\right)-\left(-21\left(-\frac{7}{30}\right)\right)}&\frac{20}{20\left(-\frac{7}{5}\right)-\left(-21\left(-\frac{7}{30}\right)\right)}\end{matrix}\right)\left(\begin{matrix}36\\-7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{47}&-\frac{30}{47}\\-\frac{1}{141}&-\frac{200}{329}\end{matrix}\right)\left(\begin{matrix}36\\-7\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{47}\times 36-\frac{30}{47}\left(-7\right)\\-\frac{1}{141}\times 36-\frac{200}{329}\left(-7\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=4$$

Steps Using Elimination

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

$$4\left(2x-3y\right)=36+3\left(3y-4x\right)$$

Use the distributive property to multiply 4 by 2x-3y.

$$8x-12y=36+3\left(3y-4x\right)$$

Use the distributive property to multiply 3 by 3y-4x.

$$8x-12y=36+9y-12x$$

Subtract 9y from both sides.

$$8x-12y-9y=36-12x$$

Combine -12y and -9y to get -21y.

$$8x-21y=36-12x$$

Add 12x to both sides.

$$8x-21y+12x=36$$

Combine 8x and 12x to get 20x.

$$20x-21y=36$$

Consider the second equation. Use the distributive property to multiply \frac{1}{6} by 6y+7x.

$$y+\frac{7}{6}x=\frac{1}{5}\left(7x+12y\right)-7$$

Use the distributive property to multiply \frac{1}{5} by 7x+12y.

$$y+\frac{7}{6}x=\frac{7}{5}x+\frac{12}{5}y-7$$

Subtract \frac{7}{5}x from both sides.

$$y+\frac{7}{6}x-\frac{7}{5}x=\frac{12}{5}y-7$$

Combine \frac{7}{6}x and -\frac{7}{5}x to get -\frac{7}{30}x.

$$y-\frac{7}{30}x=\frac{12}{5}y-7$$

Subtract \frac{12}{5}y from both sides.

$$y-\frac{7}{30}x-\frac{12}{5}y=-7$$

Combine y and -\frac{12}{5}y to get -\frac{7}{5}y.

$$-\frac{7}{5}y-\frac{7}{30}x=-7$$

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.

$$20x-21y=36,-\frac{7}{30}x-\frac{7}{5}y=-7$$

To make 20x and -\frac{7x}{30} equal, multiply all terms on each side of the first equation by -\frac{7}{30} and all terms on each side of the second by 20.

$$-\frac{7}{30}\times 20x-\frac{7}{30}\left(-21\right)y=-\frac{7}{30}\times 36,20\left(-\frac{7}{30}\right)x+20\left(-\frac{7}{5}\right)y=20\left(-7\right)$$

Simplify.

$$-\frac{14}{3}x+\frac{49}{10}y=-\frac{42}{5},-\frac{14}{3}x-28y=-140$$

Subtract -\frac{14}{3}x-28y=-140 from -\frac{14}{3}x+\frac{49}{10}y=-\frac{42}{5} by subtracting like terms on each side of the equal sign.

$$-\frac{14}{3}x+\frac{14}{3}x+\frac{49}{10}y+28y=-\frac{42}{5}+140$$

Add -\frac{14x}{3} to \frac{14x}{3}. Terms -\frac{14x}{3} and \frac{14x}{3} cancel out, leaving an equation with only one variable that can be solved.

$$\frac{49}{10}y+28y=-\frac{42}{5}+140$$

Add \frac{49y}{10} to 28y.

$$\frac{329}{10}y=-\frac{42}{5}+140$$

Add -\frac{42}{5} to 140.

$$\frac{329}{10}y=\frac{658}{5}$$

Divide both sides of the equation by \frac{329}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=4$$

Substitute 4 for y in -\frac{7}{30}x-\frac{7}{5}y=-7. Because the resulting equation contains only one variable, you can solve for x directly.

$$-\frac{7}{30}x-\frac{7}{5}\times 4=-7$$

Multiply -\frac{7}{5} times 4.

$$-\frac{7}{30}x-\frac{28}{5}=-7$$

Add \frac{28}{5} to both sides of the equation.

$$-\frac{7}{30}x=-\frac{7}{5}$$

Divide both sides of the equation by -\frac{7}{30}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=6$$

The system is now solved.

$$x=6,y=4$$