Solve {l}{3(x-1)/2+2(y-2)/3=13/6}{3(x+1)/2-frac{2(y+2)}{5}=frac{5}{2}}

$$\left.\begin{array} { l } { \frac { 3 ( x - 1 ) } { 2 } + \frac { 2 ( y - 2 ) } { 3 } = \frac { 13 } { 6 } } \\ { \frac { 3 ( x + 1 ) } { 2 } - \frac { 2 ( y + 2 ) } { 5 } = \frac { 5 } { 2 } } \end{array} \right\}$$

Solve for x, y

$$x=2$$

$$y=3$$

Steps Using Substitution

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

$$3\times 3\left(x-1\right)+2\times 2\left(y-2\right)=13$$

Multiply 3 and 3 to get 9.

$$9\left(x-1\right)+2\times 2\left(y-2\right)=13$$

Use the distributive property to multiply 9 by x-1.

$$9x-9+2\times 2\left(y-2\right)=13$$

Multiply 2 and 2 to get 4.

$$9x-9+4\left(y-2\right)=13$$

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

$$9x-9+4y-8=13$$

Subtract 8 from -9 to get -17.

$$9x-17+4y=13$$

Add 17 to both sides.

$$9x+4y=13+17$$

Add 13 and 17 to get 30.

$$9x+4y=30$$

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

$$5\times 3\left(x+1\right)-2\times 2\left(y+2\right)=25$$

Multiply 5 and 3 to get 15.

$$15\left(x+1\right)-2\times 2\left(y+2\right)=25$$

Use the distributive property to multiply 15 by x+1.

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

Multiply -2 and 2 to get -4.

$$15x+15-4\left(y+2\right)=25$$

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

$$15x+15-4y-8=25$$

Subtract 8 from 15 to get 7.

$$15x+7-4y=25$$

Subtract 7 from both sides.

$$15x-4y=25-7$$

Subtract 7 from 25 to get 18.

$$15x-4y=18$$

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.

$$9x+4y=30,15x-4y=18$$

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

$$9x+4y=30$$

Subtract 4y from both sides of the equation.

$$9x=-4y+30$$

Divide both sides by 9.

$$x=\frac{1}{9}\left(-4y+30\right)$$

Multiply \frac{1}{9} times -4y+30.

$$x=-\frac{4}{9}y+\frac{10}{3}$$

Substitute -\frac{4y}{9}+\frac{10}{3} for x in the other equation, 15x-4y=18.

$$15\left(-\frac{4}{9}y+\frac{10}{3}\right)-4y=18$$

Multiply 15 times -\frac{4y}{9}+\frac{10}{3}.

$$-\frac{20}{3}y+50-4y=18$$

Add -\frac{20y}{3} to -4y.

$$-\frac{32}{3}y+50=18$$

Subtract 50 from both sides of the equation.

$$-\frac{32}{3}y=-32$$

Divide both sides of the equation by -\frac{32}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=3$$

Substitute 3 for y in x=-\frac{4}{9}y+\frac{10}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{9}\times 3+\frac{10}{3}$$

Multiply -\frac{4}{9} times 3.

$$x=\frac{-4+10}{3}$$

Add \frac{10}{3} to -\frac{4}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=2$$

The system is now solved.

$$x=2,y=3$$

Steps Using Matrices

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

$$3\times 3\left(x-1\right)+2\times 2\left(y-2\right)=13$$

Multiply 3 and 3 to get 9.

$$9\left(x-1\right)+2\times 2\left(y-2\right)=13$$

Use the distributive property to multiply 9 by x-1.

$$9x-9+2\times 2\left(y-2\right)=13$$

Multiply 2 and 2 to get 4.

$$9x-9+4\left(y-2\right)=13$$

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

$$9x-9+4y-8=13$$

Subtract 8 from -9 to get -17.

$$9x-17+4y=13$$

Add 17 to both sides.

$$9x+4y=13+17$$

Add 13 and 17 to get 30.

$$9x+4y=30$$

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

$$5\times 3\left(x+1\right)-2\times 2\left(y+2\right)=25$$

Multiply 5 and 3 to get 15.

$$15\left(x+1\right)-2\times 2\left(y+2\right)=25$$

Use the distributive property to multiply 15 by x+1.

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

Multiply -2 and 2 to get -4.

$$15x+15-4\left(y+2\right)=25$$

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

$$15x+15-4y-8=25$$

Subtract 8 from 15 to get 7.

$$15x+7-4y=25$$

Subtract 7 from both sides.

$$15x-4y=25-7$$

Subtract 7 from 25 to get 18.

$$15x-4y=18$$

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

$$9x+4y=30,15x-4y=18$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&4\\15&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}30\\18\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&4\\15&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&4\\15&-4\end{matrix}\right))\left(\begin{matrix}9&4\\15&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&4\\15&-4\end{matrix}\right))\left(\begin{matrix}30\\18\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&4\\15&-4\end{matrix}\right))\left(\begin{matrix}30\\18\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&4\\15&-4\end{matrix}\right))\left(\begin{matrix}30\\18\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{9\left(-4\right)-4\times 15}&-\frac{4}{9\left(-4\right)-4\times 15}\\-\frac{15}{9\left(-4\right)-4\times 15}&\frac{9}{9\left(-4\right)-4\times 15}\end{matrix}\right)\left(\begin{matrix}30\\18\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{24}&\frac{1}{24}\\\frac{5}{32}&-\frac{3}{32}\end{matrix}\right)\left(\begin{matrix}30\\18\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{24}\times 30+\frac{1}{24}\times 18\\\frac{5}{32}\times 30-\frac{3}{32}\times 18\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2,y=3$$

Steps Using Elimination

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

$$3\times 3\left(x-1\right)+2\times 2\left(y-2\right)=13$$

Multiply 3 and 3 to get 9.

$$9\left(x-1\right)+2\times 2\left(y-2\right)=13$$

Use the distributive property to multiply 9 by x-1.

$$9x-9+2\times 2\left(y-2\right)=13$$

Multiply 2 and 2 to get 4.

$$9x-9+4\left(y-2\right)=13$$

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

$$9x-9+4y-8=13$$

Subtract 8 from -9 to get -17.

$$9x-17+4y=13$$

Add 17 to both sides.

$$9x+4y=13+17$$

Add 13 and 17 to get 30.

$$9x+4y=30$$

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

$$5\times 3\left(x+1\right)-2\times 2\left(y+2\right)=25$$

Multiply 5 and 3 to get 15.

$$15\left(x+1\right)-2\times 2\left(y+2\right)=25$$

Use the distributive property to multiply 15 by x+1.

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

Multiply -2 and 2 to get -4.

$$15x+15-4\left(y+2\right)=25$$

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

$$15x+15-4y-8=25$$

Subtract 8 from 15 to get 7.

$$15x+7-4y=25$$

Subtract 7 from both sides.

$$15x-4y=25-7$$

Subtract 7 from 25 to get 18.

$$15x-4y=18$$

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.

$$9x+4y=30,15x-4y=18$$

To make 9x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 9.

$$15\times 9x+15\times 4y=15\times 30,9\times 15x+9\left(-4\right)y=9\times 18$$

Simplify.

$$135x+60y=450,135x-36y=162$$

Subtract 135x-36y=162 from 135x+60y=450 by subtracting like terms on each side of the equal sign.

$$135x-135x+60y+36y=450-162$$

Add 135x to -135x. Terms 135x and -135x cancel out, leaving an equation with only one variable that can be solved.

$$60y+36y=450-162$$

Add 60y to 36y.

$$96y=450-162$$

Add 450 to -162.

$$96y=288$$

Divide both sides by 96.

$$y=3$$

Substitute 3 for y in 15x-4y=18. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x-4\times 3=18$$

Multiply -4 times 3.

$$15x-12=18$$

Add 12 to both sides of the equation.

$$15x=30$$

Divide both sides by 15.

$$x=2$$

The system is now solved.

$$x=2,y=3$$