Solve {l}{3x-5y/2+3=2x+y/5}{8-x-2y/5=x/2+frac{y}{3}}

$$\left\{ \begin{array} { l } { \frac { 3 x - 5 y } { 2 } + 3 = \frac { 2 x + y } { 5 } } \\ { 8 - \frac { x - 2 y } { 5 } = \frac { x } { 2 } + \frac { y } { 3 } } \end{array} \right.$$

Solve for x, y

$$x=12$$

$$y=6$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 2,5.

$$5\left(3x-5y\right)+30=2\left(2x+y\right)$$

Use the distributive property to multiply 5 by 3x-5y.

$$15x-25y+30=2\left(2x+y\right)$$

Use the distributive property to multiply 2 by 2x+y.

$$15x-25y+30=4x+2y$$

Subtract 4x from both sides.

$$15x-25y+30-4x=2y$$

Combine 15x and -4x to get 11x.

$$11x-25y+30=2y$$

Subtract 2y from both sides.

$$11x-25y+30-2y=0$$

Combine -25y and -2y to get -27y.

$$11x-27y+30=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$11x-27y=-30$$

Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 5,2,3.

$$240-6\left(x-2y\right)=15x+10y$$

Use the distributive property to multiply -6 by x-2y.

$$240-6x+12y=15x+10y$$

Subtract 15x from both sides.

$$240-6x+12y-15x=10y$$

Combine -6x and -15x to get -21x.

$$240-21x+12y=10y$$

Subtract 10y from both sides.

$$240-21x+12y-10y=0$$

Combine 12y and -10y to get 2y.

$$240-21x+2y=0$$

Subtract 240 from both sides. Anything subtracted from zero gives its negation.

$$-21x+2y=-240$$

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.

$$11x-27y=-30,-21x+2y=-240$$

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

$$11x-27y=-30$$

Add 27y to both sides of the equation.

$$11x=27y-30$$

Divide both sides by 11.

$$x=\frac{1}{11}\left(27y-30\right)$$

Multiply \frac{1}{11} times 27y-30.

$$x=\frac{27}{11}y-\frac{30}{11}$$

Substitute \frac{27y-30}{11} for x in the other equation, -21x+2y=-240.

$$-21\left(\frac{27}{11}y-\frac{30}{11}\right)+2y=-240$$

Multiply -21 times \frac{27y-30}{11}.

$$-\frac{567}{11}y+\frac{630}{11}+2y=-240$$

Add -\frac{567y}{11} to 2y.

$$-\frac{545}{11}y+\frac{630}{11}=-240$$

Subtract \frac{630}{11} from both sides of the equation.

$$-\frac{545}{11}y=-\frac{3270}{11}$$

Divide both sides of the equation by -\frac{545}{11}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=6$$

Substitute 6 for y in x=\frac{27}{11}y-\frac{30}{11}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{27}{11}\times 6-\frac{30}{11}$$

Multiply \frac{27}{11} times 6.

$$x=\frac{162-30}{11}$$

Add -\frac{30}{11} to \frac{162}{11} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=12$$

The system is now solved.

$$x=12,y=6$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 2,5.

$$5\left(3x-5y\right)+30=2\left(2x+y\right)$$

Use the distributive property to multiply 5 by 3x-5y.

$$15x-25y+30=2\left(2x+y\right)$$

Use the distributive property to multiply 2 by 2x+y.

$$15x-25y+30=4x+2y$$

Subtract 4x from both sides.

$$15x-25y+30-4x=2y$$

Combine 15x and -4x to get 11x.

$$11x-25y+30=2y$$

Subtract 2y from both sides.

$$11x-25y+30-2y=0$$

Combine -25y and -2y to get -27y.

$$11x-27y+30=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$11x-27y=-30$$

Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 5,2,3.

$$240-6\left(x-2y\right)=15x+10y$$

Use the distributive property to multiply -6 by x-2y.

$$240-6x+12y=15x+10y$$

Subtract 15x from both sides.

$$240-6x+12y-15x=10y$$

Combine -6x and -15x to get -21x.

$$240-21x+12y=10y$$

Subtract 10y from both sides.

$$240-21x+12y-10y=0$$

Combine 12y and -10y to get 2y.

$$240-21x+2y=0$$

Subtract 240 from both sides. Anything subtracted from zero gives its negation.

$$-21x+2y=-240$$

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

$$11x-27y=-30,-21x+2y=-240$$

Write the equations in matrix form.

$$\left(\begin{matrix}11&-27\\-21&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-30\\-240\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}11&-27\\-21&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}11&-27\\-21&2\end{matrix}\right))\left(\begin{matrix}11&-27\\-21&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}11&-27\\-21&2\end{matrix}\right))\left(\begin{matrix}-30\\-240\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}11&-27\\-21&2\end{matrix}\right))\left(\begin{matrix}-30\\-240\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}11&-27\\-21&2\end{matrix}\right))\left(\begin{matrix}-30\\-240\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{11\times 2-\left(-27\left(-21\right)\right)}&-\frac{-27}{11\times 2-\left(-27\left(-21\right)\right)}\\-\frac{-21}{11\times 2-\left(-27\left(-21\right)\right)}&\frac{11}{11\times 2-\left(-27\left(-21\right)\right)}\end{matrix}\right)\left(\begin{matrix}-30\\-240\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{545}&-\frac{27}{545}\\-\frac{21}{545}&-\frac{11}{545}\end{matrix}\right)\left(\begin{matrix}-30\\-240\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{545}\left(-30\right)-\frac{27}{545}\left(-240\right)\\-\frac{21}{545}\left(-30\right)-\frac{11}{545}\left(-240\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\6\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=12,y=6$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 2,5.

$$5\left(3x-5y\right)+30=2\left(2x+y\right)$$

Use the distributive property to multiply 5 by 3x-5y.

$$15x-25y+30=2\left(2x+y\right)$$

Use the distributive property to multiply 2 by 2x+y.

$$15x-25y+30=4x+2y$$

Subtract 4x from both sides.

$$15x-25y+30-4x=2y$$

Combine 15x and -4x to get 11x.

$$11x-25y+30=2y$$

Subtract 2y from both sides.

$$11x-25y+30-2y=0$$

Combine -25y and -2y to get -27y.

$$11x-27y+30=0$$

Subtract 30 from both sides. Anything subtracted from zero gives its negation.

$$11x-27y=-30$$

Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 5,2,3.

$$240-6\left(x-2y\right)=15x+10y$$

Use the distributive property to multiply -6 by x-2y.

$$240-6x+12y=15x+10y$$

Subtract 15x from both sides.

$$240-6x+12y-15x=10y$$

Combine -6x and -15x to get -21x.

$$240-21x+12y=10y$$

Subtract 10y from both sides.

$$240-21x+12y-10y=0$$

Combine 12y and -10y to get 2y.

$$240-21x+2y=0$$

Subtract 240 from both sides. Anything subtracted from zero gives its negation.

$$-21x+2y=-240$$

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.

$$11x-27y=-30,-21x+2y=-240$$

To make 11x and -21x equal, multiply all terms on each side of the first equation by -21 and all terms on each side of the second by 11.

$$-21\times 11x-21\left(-27\right)y=-21\left(-30\right),11\left(-21\right)x+11\times 2y=11\left(-240\right)$$

Simplify.

$$-231x+567y=630,-231x+22y=-2640$$

Subtract -231x+22y=-2640 from -231x+567y=630 by subtracting like terms on each side of the equal sign.

$$-231x+231x+567y-22y=630+2640$$

Add -231x to 231x. Terms -231x and 231x cancel out, leaving an equation with only one variable that can be solved.

$$567y-22y=630+2640$$

Add 567y to -22y.

$$545y=630+2640$$

Add 630 to 2640.

$$545y=3270$$

Divide both sides by 545.

$$y=6$$

Substitute 6 for y in -21x+2y=-240. Because the resulting equation contains only one variable, you can solve for x directly.

$$-21x+2\times 6=-240$$

Multiply 2 times 6.

$$-21x+12=-240$$

Subtract 12 from both sides of the equation.

$$-21x=-252$$

Divide both sides by -21.

$$x=12$$

The system is now solved.

$$x=12,y=6$$