Solve {l}{2x/5+3y/4=1/5}{5x/6+frac{5y}{2}=2}

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

Solve for x, y

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$y = \frac{76}{45} = 1\frac{31}{45} \approx 1.688888889$$

Steps Using Substitution

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

$$4\times 2x+5\times 3y=4$$

Multiply 4 and 2 to get 8.

$$8x+5\times 3y=4$$

Multiply 5 and 3 to get 15.

$$8x+15y=4$$

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

$$5x+3\times 5y=12$$

Multiply 3 and 5 to get 15.

$$5x+15y=12$$

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.

$$8x+15y=4,5x+15y=12$$

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

$$8x+15y=4$$

Subtract 15y from both sides of the equation.

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

Divide both sides by 8.

$$x=\frac{1}{8}\left(-15y+4\right)$$

Multiply \frac{1}{8} times -15y+4.

$$x=-\frac{15}{8}y+\frac{1}{2}$$

Substitute -\frac{15y}{8}+\frac{1}{2} for x in the other equation, 5x+15y=12.

$$5\left(-\frac{15}{8}y+\frac{1}{2}\right)+15y=12$$

Multiply 5 times -\frac{15y}{8}+\frac{1}{2}.

$$-\frac{75}{8}y+\frac{5}{2}+15y=12$$

Add -\frac{75y}{8} to 15y.

$$\frac{45}{8}y+\frac{5}{2}=12$$

Subtract \frac{5}{2} from both sides of the equation.

$$\frac{45}{8}y=\frac{19}{2}$$

Divide both sides of the equation by \frac{45}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{76}{45}$$

Substitute \frac{76}{45} for y in x=-\frac{15}{8}y+\frac{1}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{15}{8}\times \frac{76}{45}+\frac{1}{2}$$

Multiply -\frac{15}{8} times \frac{76}{45} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{19}{6}+\frac{1}{2}$$

Add \frac{1}{2} to -\frac{19}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{8}{3}$$

The system is now solved.

$$x=-\frac{8}{3},y=\frac{76}{45}$$

Steps Using Matrices

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

$$4\times 2x+5\times 3y=4$$

Multiply 4 and 2 to get 8.

$$8x+5\times 3y=4$$

Multiply 5 and 3 to get 15.

$$8x+15y=4$$

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

$$5x+3\times 5y=12$$

Multiply 3 and 5 to get 15.

$$5x+15y=12$$

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

$$8x+15y=4,5x+15y=12$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&15\\5&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\12\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&15\\5&15\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&15\\5&15\end{matrix}\right))\left(\begin{matrix}8&15\\5&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&15\\5&15\end{matrix}\right))\left(\begin{matrix}4\\12\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&15\\5&15\end{matrix}\right))\left(\begin{matrix}4\\12\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&15\\5&15\end{matrix}\right))\left(\begin{matrix}4\\12\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{8\times 15-15\times 5}&-\frac{15}{8\times 15-15\times 5}\\-\frac{5}{8\times 15-15\times 5}&\frac{8}{8\times 15-15\times 5}\end{matrix}\right)\left(\begin{matrix}4\\12\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}&-\frac{1}{3}\\-\frac{1}{9}&\frac{8}{45}\end{matrix}\right)\left(\begin{matrix}4\\12\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 4-\frac{1}{3}\times 12\\-\frac{1}{9}\times 4+\frac{8}{45}\times 12\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{3}\\\frac{76}{45}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{8}{3},y=\frac{76}{45}$$

Steps Using Elimination

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

$$4\times 2x+5\times 3y=4$$

Multiply 4 and 2 to get 8.

$$8x+5\times 3y=4$$

Multiply 5 and 3 to get 15.

$$8x+15y=4$$

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

$$5x+3\times 5y=12$$

Multiply 3 and 5 to get 15.

$$5x+15y=12$$

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.

$$8x+15y=4,5x+15y=12$$

Subtract 5x+15y=12 from 8x+15y=4 by subtracting like terms on each side of the equal sign.

$$8x-5x+15y-15y=4-12$$

Add 15y to -15y. Terms 15y and -15y cancel out, leaving an equation with only one variable that can be solved.

$$8x-5x=4-12$$

Add 8x to -5x.

$$3x=4-12$$

Add 4 to -12.

$$3x=-8$$

Divide both sides by 3.

$$x=-\frac{8}{3}$$

Substitute -\frac{8}{3} for x in 5x+15y=12. Because the resulting equation contains only one variable, you can solve for y directly.

$$5\left(-\frac{8}{3}\right)+15y=12$$

Multiply 5 times -\frac{8}{3}.

$$-\frac{40}{3}+15y=12$$

Add \frac{40}{3} to both sides of the equation.

$$15y=\frac{76}{3}$$

Divide both sides by 15.

$$y=\frac{76}{45}$$

The system is now solved.

$$x=-\frac{8}{3},y=\frac{76}{45}$$