Solve {l}{5x/2+5y/3=55/3}{x/4+frac{y}{3}=frac{19}{6}}

$$\left. \begin{array} { l } { \frac { 5 x } { 2 } + \frac { 5 y } { 3 } = \frac { 55 } { 3 } } \\ { \frac { x } { 4 } + \frac { y } { 3 } = \frac { 19 } { 6 } } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=8$$

Steps Using Substitution

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

$$3\times 5x+2\times 5y=110$$

Multiply 3 and 5 to get 15.

$$15x+2\times 5y=110$$

Multiply 2 and 5 to get 10.

$$15x+10y=110$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3,6.

$$3x+4y=38$$

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.

$$15x+10y=110,3x+4y=38$$

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

$$15x+10y=110$$

Subtract 10y from both sides of the equation.

$$15x=-10y+110$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(-10y+110\right)$$

Multiply \frac{1}{15} times -10y+110.

$$x=-\frac{2}{3}y+\frac{22}{3}$$

Substitute \frac{-2y+22}{3} for x in the other equation, 3x+4y=38.

$$3\left(-\frac{2}{3}y+\frac{22}{3}\right)+4y=38$$

Multiply 3 times \frac{-2y+22}{3}.

$$-2y+22+4y=38$$

Add -2y to 4y.

$$2y+22=38$$

Subtract 22 from both sides of the equation.

$$2y=16$$

Divide both sides by 2.

$$y=8$$

Substitute 8 for y in x=-\frac{2}{3}y+\frac{22}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\times 8+\frac{22}{3}$$

Multiply -\frac{2}{3} times 8.

$$x=\frac{-16+22}{3}$$

Add \frac{22}{3} to -\frac{16}{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.

$$x=2,y=8$$

Steps Using Matrices

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

$$3\times 5x+2\times 5y=110$$

Multiply 3 and 5 to get 15.

$$15x+2\times 5y=110$$

Multiply 2 and 5 to get 10.

$$15x+10y=110$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3,6.

$$3x+4y=38$$

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

$$15x+10y=110,3x+4y=38$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&10\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}110\\38\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&10\\3&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&10\\3&4\end{matrix}\right))\left(\begin{matrix}15&10\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&10\\3&4\end{matrix}\right))\left(\begin{matrix}110\\38\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&10\\3&4\end{matrix}\right))\left(\begin{matrix}110\\38\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&10\\3&4\end{matrix}\right))\left(\begin{matrix}110\\38\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{15\times 4-10\times 3}&-\frac{10}{15\times 4-10\times 3}\\-\frac{3}{15\times 4-10\times 3}&\frac{15}{15\times 4-10\times 3}\end{matrix}\right)\left(\begin{matrix}110\\38\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{15}&-\frac{1}{3}\\-\frac{1}{10}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}110\\38\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{15}\times 110-\frac{1}{3}\times 38\\-\frac{1}{10}\times 110+\frac{1}{2}\times 38\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\8\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2,y=8$$

Steps Using Elimination

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

$$3\times 5x+2\times 5y=110$$

Multiply 3 and 5 to get 15.

$$15x+2\times 5y=110$$

Multiply 2 and 5 to get 10.

$$15x+10y=110$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3,6.

$$3x+4y=38$$

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.

$$15x+10y=110,3x+4y=38$$

To make 15x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 15.

$$3\times 15x+3\times 10y=3\times 110,15\times 3x+15\times 4y=15\times 38$$

Simplify.

$$45x+30y=330,45x+60y=570$$

Subtract 45x+60y=570 from 45x+30y=330 by subtracting like terms on each side of the equal sign.

$$45x-45x+30y-60y=330-570$$

Add 45x to -45x. Terms 45x and -45x cancel out, leaving an equation with only one variable that can be solved.

$$30y-60y=330-570$$

Add 30y to -60y.

$$-30y=330-570$$

Add 330 to -570.

$$-30y=-240$$

Divide both sides by -30.

$$y=8$$

Substitute 8 for y in 3x+4y=38. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x+4\times 8=38$$

Multiply 4 times 8.

$$3x+32=38$$

Subtract 32 from both sides of the equation.

$$3x=6$$

Divide both sides by 3.

$$x=2$$

The system is now solved.

$$x=2,y=8$$