Solve {l}{5x-14=3y}{7(3x-2y)=35}

$$\left. \begin{array} { l } { 5 x - 14 = 3 y } \\ { 7 ( 3 x - 2 y ) = 35 } \end{array} \right.$$

Solve for x, y

$$x=13$$

$$y=17$$

Steps Using Substitution

Consider the first equation. Subtract 3y from both sides.

$$5x-14-3y=0$$

Add 14 to both sides. Anything plus zero gives itself.

$$5x-3y=14$$

Consider the second equation. Divide both sides by 7.

$$3x-2y=\frac{35}{7}$$

Divide 35 by 7 to get 5.

$$3x-2y=5$$

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.

$$5x-3y=14,3x-2y=5$$

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

$$5x-3y=14$$

Add 3y to both sides of the equation.

$$5x=3y+14$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(3y+14\right)$$

Multiply \frac{1}{5} times 3y+14.

$$x=\frac{3}{5}y+\frac{14}{5}$$

Substitute \frac{3y+14}{5} for x in the other equation, 3x-2y=5.

$$3\left(\frac{3}{5}y+\frac{14}{5}\right)-2y=5$$

Multiply 3 times \frac{3y+14}{5}.

$$\frac{9}{5}y+\frac{42}{5}-2y=5$$

Add \frac{9y}{5} to -2y.

$$-\frac{1}{5}y+\frac{42}{5}=5$$

Subtract \frac{42}{5} from both sides of the equation.

$$-\frac{1}{5}y=-\frac{17}{5}$$

Multiply both sides by -5.

$$y=17$$

Substitute 17 for y in x=\frac{3}{5}y+\frac{14}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{5}\times 17+\frac{14}{5}$$

Multiply \frac{3}{5} times 17.

$$x=\frac{51+14}{5}$$

Add \frac{14}{5} to \frac{51}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=13$$

The system is now solved.

$$x=13,y=17$$

Steps Using Matrices

Consider the first equation. Subtract 3y from both sides.

$$5x-14-3y=0$$

Add 14 to both sides. Anything plus zero gives itself.

$$5x-3y=14$$

Consider the second equation. Divide both sides by 7.

$$3x-2y=\frac{35}{7}$$

Divide 35 by 7 to get 5.

$$3x-2y=5$$

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

$$5x-3y=14,3x-2y=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-3\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\5\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}5&-3\\3&-2\end{matrix}\right))\left(\begin{matrix}5&-3\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\3&-2\end{matrix}\right))\left(\begin{matrix}14\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\3&-2\end{matrix}\right))\left(\begin{matrix}14\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\3&-2\end{matrix}\right))\left(\begin{matrix}14\\5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{5\left(-2\right)-\left(-3\times 3\right)}&-\frac{-3}{5\left(-2\right)-\left(-3\times 3\right)}\\-\frac{3}{5\left(-2\right)-\left(-3\times 3\right)}&\frac{5}{5\left(-2\right)-\left(-3\times 3\right)}\end{matrix}\right)\left(\begin{matrix}14\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2&-3\\3&-5\end{matrix}\right)\left(\begin{matrix}14\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\times 14-3\times 5\\3\times 14-5\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}13\\17\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=13,y=17$$

Steps Using Elimination

Consider the first equation. Subtract 3y from both sides.

$$5x-14-3y=0$$

Add 14 to both sides. Anything plus zero gives itself.

$$5x-3y=14$$

Consider the second equation. Divide both sides by 7.

$$3x-2y=\frac{35}{7}$$

Divide 35 by 7 to get 5.

$$3x-2y=5$$

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.

$$5x-3y=14,3x-2y=5$$

To make 5x 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 5.

$$3\times 5x+3\left(-3\right)y=3\times 14,5\times 3x+5\left(-2\right)y=5\times 5$$

Simplify.

$$15x-9y=42,15x-10y=25$$

Subtract 15x-10y=25 from 15x-9y=42 by subtracting like terms on each side of the equal sign.

$$15x-15x-9y+10y=42-25$$

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

$$-9y+10y=42-25$$

Add -9y to 10y.

$$y=42-25$$

Add 42 to -25.

$$y=17$$

Substitute 17 for y in 3x-2y=5. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x-2\times 17=5$$

Multiply -2 times 17.

$$3x-34=5$$

Add 34 to both sides of the equation.

$$3x=39$$

Divide both sides by 3.

$$x=13$$

The system is now solved.

$$x=13,y=17$$