Solve {l}{y=14x+160}{y=26x+16.TI}

$$\left. \begin{array} { l } { y = 14 x + 160 } \\ { y = 26 x + 16 . TI } \end{array} \right.$$

Solve for y, x

$$x=12$$

$$y=328$$

Steps Using Substitution

Consider the first equation. Subtract 14x from both sides.

$$y-14x=160$$

Consider the second equation. Subtract 26x from both sides.

$$y-26x=16$$

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.

$$y-14x=160,y-26x=16$$

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

$$y-14x=160$$

Add 14x to both sides of the equation.

$$y=14x+160$$

Substitute 14x+160 for y in the other equation, y-26x=16.

$$14x+160-26x=16$$

Add 14x to -26x.

$$-12x+160=16$$

Subtract 160 from both sides of the equation.

$$-12x=-144$$

Divide both sides by -12.

$$x=12$$

Substitute 12 for x in y=14x+160. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=14\times 12+160$$

Multiply 14 times 12.

$$y=168+160$$

Add 160 to 168.

$$y=328$$

The system is now solved.

$$y=328,x=12$$

Steps Using Matrices

Consider the first equation. Subtract 14x from both sides.

$$y-14x=160$$

Consider the second equation. Subtract 26x from both sides.

$$y-26x=16$$

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

$$y-14x=160,y-26x=16$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-14\\1&-26\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}160\\16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-14\\1&-26\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-14\\1&-26\end{matrix}\right))\left(\begin{matrix}1&-14\\1&-26\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-14\\1&-26\end{matrix}\right))\left(\begin{matrix}160\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-14\\1&-26\end{matrix}\right))\left(\begin{matrix}160\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-14\\1&-26\end{matrix}\right))\left(\begin{matrix}160\\16\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{26}{-26-\left(-14\right)}&-\frac{-14}{-26-\left(-14\right)}\\-\frac{1}{-26-\left(-14\right)}&\frac{1}{-26-\left(-14\right)}\end{matrix}\right)\left(\begin{matrix}160\\16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{13}{6}&-\frac{7}{6}\\\frac{1}{12}&-\frac{1}{12}\end{matrix}\right)\left(\begin{matrix}160\\16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{13}{6}\times 160-\frac{7}{6}\times 16\\\frac{1}{12}\times 160-\frac{1}{12}\times 16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}328\\12\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=328,x=12$$

Steps Using Elimination

Consider the first equation. Subtract 14x from both sides.

$$y-14x=160$$

Consider the second equation. Subtract 26x from both sides.

$$y-26x=16$$

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.

$$y-14x=160,y-26x=16$$

Subtract y-26x=16 from y-14x=160 by subtracting like terms on each side of the equal sign.

$$y-y-14x+26x=160-16$$

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

$$-14x+26x=160-16$$

Add -14x to 26x.

$$12x=160-16$$

Add 160 to -16.

$$12x=144$$

Divide both sides by 12.

$$x=12$$

Substitute 12 for x in y-26x=16. Because the resulting equation contains only one variable, you can solve for y directly.

$$y-26\times 12=16$$

Multiply -26 times 12.

$$y-312=16$$

Add 312 to both sides of the equation.

$$y=328$$

The system is now solved.

$$y=328,x=12$$