Solve {l}{x/5=y/2}{3x-2y=22}

$$\left\{ \begin{array} { l } { \frac { x } { 5 } = \frac { y } { 2 } } \\ { 3 x - 2 y = 22 } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=4$$

Steps Using Substitution

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

$$2x=5y$$

Divide both sides by 2.

$$x=\frac{1}{2}\times 5y$$

Multiply \frac{1}{2} times 5y.

$$x=\frac{5}{2}y$$

Substitute \frac{5y}{2} for x in the other equation, 3x-2y=22.

$$3\times \frac{5}{2}y-2y=22$$

Multiply 3 times \frac{5y}{2}.

$$\frac{15}{2}y-2y=22$$

Add \frac{15y}{2} to -2y.

$$\frac{11}{2}y=22$$

Divide both sides of the equation by \frac{11}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=4$$

Substitute 4 for y in x=\frac{5}{2}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{2}\times 4$$

Multiply \frac{5}{2} times 4.

$$x=10$$

The system is now solved.

$$x=10,y=4$$

Steps Using Matrices

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

$$2x=5y$$

Subtract 5y from both sides.

$$2x-5y=0$$

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

$$2x-5y=0,3x-2y=22$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-5\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\22\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-5\\3&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-5\\3&-2\end{matrix}\right))\left(\begin{matrix}2&-5\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\3&-2\end{matrix}\right))\left(\begin{matrix}0\\22\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\3&-2\end{matrix}\right))\left(\begin{matrix}0\\22\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\3&-2\end{matrix}\right))\left(\begin{matrix}0\\22\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{2\left(-2\right)-\left(-5\times 3\right)}&-\frac{-5}{2\left(-2\right)-\left(-5\times 3\right)}\\-\frac{3}{2\left(-2\right)-\left(-5\times 3\right)}&\frac{2}{2\left(-2\right)-\left(-5\times 3\right)}\end{matrix}\right)\left(\begin{matrix}0\\22\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{11}&\frac{5}{11}\\-\frac{3}{11}&\frac{2}{11}\end{matrix}\right)\left(\begin{matrix}0\\22\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{11}\times 22\\\frac{2}{11}\times 22\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=4$$

Steps Using Elimination

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

$$2x=5y$$

Subtract 5y from both sides.

$$2x-5y=0$$

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.

$$2x-5y=0,3x-2y=22$$

To make 2x 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 2.

$$3\times 2x+3\left(-5\right)y=0,2\times 3x+2\left(-2\right)y=2\times 22$$

Simplify.

$$6x-15y=0,6x-4y=44$$

Subtract 6x-4y=44 from 6x-15y=0 by subtracting like terms on each side of the equal sign.

$$6x-6x-15y+4y=-44$$

Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.

$$-15y+4y=-44$$

Add -15y to 4y.

$$-11y=-44$$

Divide both sides by -11.

$$y=4$$

Substitute 4 for y in 3x-2y=22. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x-2\times 4=22$$

Multiply -2 times 4.

$$3x-8=22$$

Add 8 to both sides of the equation.

$$3x=30$$

Divide both sides by 3.

$$x=10$$

The system is now solved.

$$x=10,y=4$$