Solve {l}{y_{1}+6y_{3}=2}{y_{1}+2y_{3}=1}

$$\left. \begin{array} { l } { y _ { 1 } + 6 y _ { 3 } = 2 } \\ { y _ { 1 } + 2 y _ { 3 } = 1 } \end{array} \right.$$

Solve for y_1, y_3

$$y_{1}=\frac{1}{2}=0.5$$

$$y_{3}=\frac{1}{4}=0.25$$

Steps Using Substitution

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.

$$y_{1}+6y_{3}=2,y_{1}+2y_{3}=1$$

Choose one of the equations and solve it for y_{1} by isolating y_{1} on the left hand side of the equal sign.

$$y_{1}+6y_{3}=2$$

Subtract 6y_{3} from both sides of the equation.

$$y_{1}=-6y_{3}+2$$

Substitute -6y_{3}+2 for y_{1} in the other equation, y_{1}+2y_{3}=1.

$$-6y_{3}+2+2y_{3}=1$$

Add -6y_{3} to 2y_{3}.

$$-4y_{3}+2=1$$

Subtract 2 from both sides of the equation.

$$-4y_{3}=-1$$

Divide both sides by -4.

$$y_{3}=\frac{1}{4}$$

Substitute \frac{1}{4} for y_{3} in y_{1}=-6y_{3}+2. Because the resulting equation contains only one variable, you can solve for y_{1} directly.

$$y_{1}=-6\times \frac{1}{4}+2$$

Multiply -6 times \frac{1}{4}.

$$y_{1}=-\frac{3}{2}+2$$

Add 2 to -\frac{3}{2}.

$$y_{1}=\frac{1}{2}$$

The system is now solved.

$$y_{1}=\frac{1}{2},y_{3}=\frac{1}{4}$$

Steps Using Matrices

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

$$y_{1}+6y_{3}=2,y_{1}+2y_{3}=1$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&6\\1&2\end{matrix}\right)\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=\left(\begin{matrix}2\\1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&6\\1&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&6\\1&2\end{matrix}\right))\left(\begin{matrix}1&6\\1&2\end{matrix}\right)\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=inverse(\left(\begin{matrix}1&6\\1&2\end{matrix}\right))\left(\begin{matrix}2\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=inverse(\left(\begin{matrix}1&6\\1&2\end{matrix}\right))\left(\begin{matrix}2\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=inverse(\left(\begin{matrix}1&6\\1&2\end{matrix}\right))\left(\begin{matrix}2\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2-6}&-\frac{6}{2-6}\\-\frac{1}{2-6}&\frac{1}{2-6}\end{matrix}\right)\left(\begin{matrix}2\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&\frac{3}{2}\\\frac{1}{4}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}2\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 2+\frac{3}{2}\\\frac{1}{4}\times 2-\frac{1}{4}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y_{1}\\y_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\\\frac{1}{4}\end{matrix}\right)$$

Extract the matrix elements y_{1} and y_{3}.

$$y_{1}=\frac{1}{2},y_{3}=\frac{1}{4}$$

Steps Using Elimination

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.

$$y_{1}+6y_{3}=2,y_{1}+2y_{3}=1$$

Subtract y_{1}+2y_{3}=1 from y_{1}+6y_{3}=2 by subtracting like terms on each side of the equal sign.

$$y_{1}-y_{1}+6y_{3}-2y_{3}=2-1$$

Add y_{1} to -y_{1}. Terms y_{1} and -y_{1} cancel out, leaving an equation with only one variable that can be solved.

$$6y_{3}-2y_{3}=2-1$$

Add 6y_{3} to -2y_{3}.

$$4y_{3}=2-1$$

Add 2 to -1.

$$4y_{3}=1$$

Divide both sides by 4.

$$y_{3}=\frac{1}{4}$$

Substitute \frac{1}{4} for y_{3} in y_{1}+2y_{3}=1. Because the resulting equation contains only one variable, you can solve for y_{1} directly.

$$y_{1}+2\times \frac{1}{4}=1$$

Multiply 2 times \frac{1}{4}.

$$y_{1}+\frac{1}{2}=1$$

Subtract \frac{1}{2} from both sides of the equation.

$$y_{1}=\frac{1}{2}$$

The system is now solved.

$$y_{1}=\frac{1}{2},y_{3}=\frac{1}{4}$$