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

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

Solve for x, y

$$x=2$$

$$y=4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 4,5.

$$5\times 3\left(x-2\right)+4\times 2\left(y-3\right)=8$$

Multiply 5 and 3 to get 15.

$$15\left(x-2\right)+4\times 2\left(y-3\right)=8$$

Use the distributive property to multiply 15 by x-2.

$$15x-30+4\times 2\left(y-3\right)=8$$

Multiply 4 and 2 to get 8.

$$15x-30+8\left(y-3\right)=8$$

Use the distributive property to multiply 8 by y-3.

$$15x-30+8y-24=8$$

Subtract 24 from -30 to get -54.

$$15x-54+8y=8$$

Add 54 to both sides.

$$15x+8y=8+54$$

Add 8 and 54 to get 62.

$$15x+8y=62$$

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

$$2\times 2\left(y-4\right)+3\times 3\left(x-1\right)=9$$

Multiply 2 and 2 to get 4.

$$4\left(y-4\right)+3\times 3\left(x-1\right)=9$$

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

$$4y-16+3\times 3\left(x-1\right)=9$$

Multiply 3 and 3 to get 9.

$$4y-16+9\left(x-1\right)=9$$

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

$$4y-16+9x-9=9$$

Subtract 9 from -16 to get -25.

$$4y-25+9x=9$$

Add 25 to both sides.

$$4y+9x=9+25$$

Add 9 and 25 to get 34.

$$4y+9x=34$$

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.

$$15x+8y=62,9x+4y=34$$

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

$$15x+8y=62$$

Subtract 8y from both sides of the equation.

$$15x=-8y+62$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(-8y+62\right)$$

Multiply \frac{1}{15} times -8y+62.

$$x=-\frac{8}{15}y+\frac{62}{15}$$

Substitute \frac{-8y+62}{15} for x in the other equation, 9x+4y=34.

$$9\left(-\frac{8}{15}y+\frac{62}{15}\right)+4y=34$$

Multiply 9 times \frac{-8y+62}{15}.

$$-\frac{24}{5}y+\frac{186}{5}+4y=34$$

Add -\frac{24y}{5} to 4y.

$$-\frac{4}{5}y+\frac{186}{5}=34$$

Subtract \frac{186}{5} from both sides of the equation.

$$-\frac{4}{5}y=-\frac{16}{5}$$

Divide both sides of the equation by -\frac{4}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=4$$

Substitute 4 for y in x=-\frac{8}{15}y+\frac{62}{15}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{8}{15}\times 4+\frac{62}{15}$$

Multiply -\frac{8}{15} times 4.

$$x=\frac{-32+62}{15}$$

Add \frac{62}{15} to -\frac{32}{15} 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.

$$x=2,y=4$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 4,5.

$$5\times 3\left(x-2\right)+4\times 2\left(y-3\right)=8$$

Multiply 5 and 3 to get 15.

$$15\left(x-2\right)+4\times 2\left(y-3\right)=8$$

Use the distributive property to multiply 15 by x-2.

$$15x-30+4\times 2\left(y-3\right)=8$$

Multiply 4 and 2 to get 8.

$$15x-30+8\left(y-3\right)=8$$

Use the distributive property to multiply 8 by y-3.

$$15x-30+8y-24=8$$

Subtract 24 from -30 to get -54.

$$15x-54+8y=8$$

Add 54 to both sides.

$$15x+8y=8+54$$

Add 8 and 54 to get 62.

$$15x+8y=62$$

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

$$2\times 2\left(y-4\right)+3\times 3\left(x-1\right)=9$$

Multiply 2 and 2 to get 4.

$$4\left(y-4\right)+3\times 3\left(x-1\right)=9$$

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

$$4y-16+3\times 3\left(x-1\right)=9$$

Multiply 3 and 3 to get 9.

$$4y-16+9\left(x-1\right)=9$$

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

$$4y-16+9x-9=9$$

Subtract 9 from -16 to get -25.

$$4y-25+9x=9$$

Add 25 to both sides.

$$4y+9x=9+25$$

Add 9 and 25 to get 34.

$$4y+9x=34$$

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

$$15x+8y=62,9x+4y=34$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&8\\9&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}62\\34\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&8\\9&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&8\\9&4\end{matrix}\right))\left(\begin{matrix}15&8\\9&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&8\\9&4\end{matrix}\right))\left(\begin{matrix}62\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&8\\9&4\end{matrix}\right))\left(\begin{matrix}62\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&8\\9&4\end{matrix}\right))\left(\begin{matrix}62\\34\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{15\times 4-8\times 9}&-\frac{8}{15\times 4-8\times 9}\\-\frac{9}{15\times 4-8\times 9}&\frac{15}{15\times 4-8\times 9}\end{matrix}\right)\left(\begin{matrix}62\\34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}&\frac{2}{3}\\\frac{3}{4}&-\frac{5}{4}\end{matrix}\right)\left(\begin{matrix}62\\34\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 62+\frac{2}{3}\times 34\\\frac{3}{4}\times 62-\frac{5}{4}\times 34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2,y=4$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 4,5.

$$5\times 3\left(x-2\right)+4\times 2\left(y-3\right)=8$$

Multiply 5 and 3 to get 15.

$$15\left(x-2\right)+4\times 2\left(y-3\right)=8$$

Use the distributive property to multiply 15 by x-2.

$$15x-30+4\times 2\left(y-3\right)=8$$

Multiply 4 and 2 to get 8.

$$15x-30+8\left(y-3\right)=8$$

Use the distributive property to multiply 8 by y-3.

$$15x-30+8y-24=8$$

Subtract 24 from -30 to get -54.

$$15x-54+8y=8$$

Add 54 to both sides.

$$15x+8y=8+54$$

Add 8 and 54 to get 62.

$$15x+8y=62$$

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

$$2\times 2\left(y-4\right)+3\times 3\left(x-1\right)=9$$

Multiply 2 and 2 to get 4.

$$4\left(y-4\right)+3\times 3\left(x-1\right)=9$$

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

$$4y-16+3\times 3\left(x-1\right)=9$$

Multiply 3 and 3 to get 9.

$$4y-16+9\left(x-1\right)=9$$

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

$$4y-16+9x-9=9$$

Subtract 9 from -16 to get -25.

$$4y-25+9x=9$$

Add 25 to both sides.

$$4y+9x=9+25$$

Add 9 and 25 to get 34.

$$4y+9x=34$$

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.

$$15x+8y=62,9x+4y=34$$

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

$$9\times 15x+9\times 8y=9\times 62,15\times 9x+15\times 4y=15\times 34$$

Simplify.

$$135x+72y=558,135x+60y=510$$

Subtract 135x+60y=510 from 135x+72y=558 by subtracting like terms on each side of the equal sign.

$$135x-135x+72y-60y=558-510$$

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

$$72y-60y=558-510$$

Add 72y to -60y.

$$12y=558-510$$

Add 558 to -510.

$$12y=48$$

Divide both sides by 12.

$$y=4$$

Substitute 4 for y in 9x+4y=34. Because the resulting equation contains only one variable, you can solve for x directly.

$$9x+4\times 4=34$$

Multiply 4 times 4.

$$9x+16=34$$

Subtract 16 from both sides of the equation.

$$9x=18$$

Divide both sides by 9.

$$x=2$$

The system is now solved.

$$x=2,y=4$$