Solve {l}{15y=y-x+4}{15y=2x-8}

$$\left. \begin{array} { l } { 15 y = y - x + 4 } \\ { 15 y = 2 x - 8 } \end{array} \right.$$

Solve for y, x

$$x=4$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Subtract y from both sides.

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

Combine 15y and -y to get 14y.

$$14y=-x+4$$

Divide both sides by 14.

$$y=\frac{1}{14}\left(-x+4\right)$$

Multiply \frac{1}{14} times -x+4.

$$y=-\frac{1}{14}x+\frac{2}{7}$$

Substitute -\frac{x}{14}+\frac{2}{7} for y in the other equation, 15y-2x=-8.

$$15\left(-\frac{1}{14}x+\frac{2}{7}\right)-2x=-8$$

Multiply 15 times -\frac{x}{14}+\frac{2}{7}.

$$-\frac{15}{14}x+\frac{30}{7}-2x=-8$$

Add -\frac{15x}{14} to -2x.

$$-\frac{43}{14}x+\frac{30}{7}=-8$$

Subtract \frac{30}{7} from both sides of the equation.

$$-\frac{43}{14}x=-\frac{86}{7}$$

Divide both sides of the equation by -\frac{43}{14}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=4$$

Substitute 4 for x in y=-\frac{1}{14}x+\frac{2}{7}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{1}{14}\times 4+\frac{2}{7}$$

Multiply -\frac{1}{14} times 4.

$$y=\frac{-2+2}{7}$$

Add \frac{2}{7} to -\frac{2}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=0$$

The system is now solved.

$$y=0,x=4$$

Steps Using Matrices

Consider the first equation. Subtract y from both sides.

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

Combine 15y and -y to get 14y.

$$14y=-x+4$$

Add x to both sides.

$$14y+x=4$$

Consider the second equation. Subtract 2x from both sides.

$$15y-2x=-8$$

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

$$14y+x=4,15y-2x=-8$$

Write the equations in matrix form.

$$\left(\begin{matrix}14&1\\15&-2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}4\\-8\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}14&1\\15&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}14&1\\15&-2\end{matrix}\right))\left(\begin{matrix}14&1\\15&-2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}14&1\\15&-2\end{matrix}\right))\left(\begin{matrix}4\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}14&1\\15&-2\end{matrix}\right))\left(\begin{matrix}4\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}14&1\\15&-2\end{matrix}\right))\left(\begin{matrix}4\\-8\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{14\left(-2\right)-15}&-\frac{1}{14\left(-2\right)-15}\\-\frac{15}{14\left(-2\right)-15}&\frac{14}{14\left(-2\right)-15}\end{matrix}\right)\left(\begin{matrix}4\\-8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{43}&\frac{1}{43}\\\frac{15}{43}&-\frac{14}{43}\end{matrix}\right)\left(\begin{matrix}4\\-8\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{43}\times 4+\frac{1}{43}\left(-8\right)\\\frac{15}{43}\times 4-\frac{14}{43}\left(-8\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\4\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=0,x=4$$

Steps Using Elimination

Consider the first equation. Subtract y from both sides.

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

Combine 15y and -y to get 14y.

$$14y=-x+4$$

Add x to both sides.

$$14y+x=4$$

Consider the second equation. Subtract 2x from both sides.

$$15y-2x=-8$$

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.

$$14y+x=4,15y-2x=-8$$

To make 14y and 15y equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 14.

$$15\times 14y+15x=15\times 4,14\times 15y+14\left(-2\right)x=14\left(-8\right)$$

Simplify.

$$210y+15x=60,210y-28x=-112$$

Subtract 210y-28x=-112 from 210y+15x=60 by subtracting like terms on each side of the equal sign.

$$210y-210y+15x+28x=60+112$$

Add 210y to -210y. Terms 210y and -210y cancel out, leaving an equation with only one variable that can be solved.

$$15x+28x=60+112$$

Add 15x to 28x.

$$43x=60+112$$

Add 60 to 112.

$$43x=172$$

Divide both sides by 43.

$$x=4$$

Substitute 4 for x in 15y-2x=-8. Because the resulting equation contains only one variable, you can solve for y directly.

$$15y-2\times 4=-8$$

Multiply -2 times 4.

$$15y-8=-8$$

Add 8 to both sides of the equation.

$$15y=0$$

Divide both sides by 15.

$$y=0$$

The system is now solved.

$$y=0,x=4$$