Solve {l}{3a_{1}+9x=84}{a_{1}+8x=73}

$$\left\{ \begin{array} { l } { 3 a _ { 1 } + 9 x = 84 } \\ { a _ { 1 } + 8 x = 73 } \end{array} \right.$$

Solve for a_1, x

$$x=9$$

$$a_{1}=1$$

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.

$$3a_{1}+9x=84,a_{1}+8x=73$$

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

$$3a_{1}+9x=84$$

Subtract 9x from both sides of the equation.

$$3a_{1}=-9x+84$$

Divide both sides by 3.

$$a_{1}=\frac{1}{3}\left(-9x+84\right)$$

Multiply \frac{1}{3} times -9x+84.

$$a_{1}=-3x+28$$

Substitute -3x+28 for a_{1} in the other equation, a_{1}+8x=73.

$$-3x+28+8x=73$$

Add -3x to 8x.

$$5x+28=73$$

Subtract 28 from both sides of the equation.

$$5x=45$$

Divide both sides by 5.

$$x=9$$

Substitute 9 for x in a_{1}=-3x+28. Because the resulting equation contains only one variable, you can solve for a_{1} directly.

$$a_{1}=-3\times 9+28$$

Multiply -3 times 9.

$$a_{1}=-27+28$$

Add 28 to -27.

$$a_{1}=1$$

The system is now solved.

$$a_{1}=1,x=9$$

Steps Using Matrices

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

$$3a_{1}+9x=84,a_{1}+8x=73$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&9\\1&8\end{matrix}\right)\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=\left(\begin{matrix}84\\73\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&9\\1&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&9\\1&8\end{matrix}\right))\left(\begin{matrix}3&9\\1&8\end{matrix}\right)\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&9\\1&8\end{matrix}\right))\left(\begin{matrix}84\\73\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&9\\1&8\end{matrix}\right))\left(\begin{matrix}84\\73\end{matrix}\right)$$

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

$$\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&9\\1&8\end{matrix}\right))\left(\begin{matrix}84\\73\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=\left(\begin{matrix}\frac{8}{3\times 8-9}&-\frac{9}{3\times 8-9}\\-\frac{1}{3\times 8-9}&\frac{3}{3\times 8-9}\end{matrix}\right)\left(\begin{matrix}84\\73\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=\left(\begin{matrix}\frac{8}{15}&-\frac{3}{5}\\-\frac{1}{15}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}84\\73\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=\left(\begin{matrix}\frac{8}{15}\times 84-\frac{3}{5}\times 73\\-\frac{1}{15}\times 84+\frac{1}{5}\times 73\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a_{1}\\x\end{matrix}\right)=\left(\begin{matrix}1\\9\end{matrix}\right)$$

Extract the matrix elements a_{1} and x.

$$a_{1}=1,x=9$$

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.

$$3a_{1}+9x=84,a_{1}+8x=73$$

To make 3a_{1} and a_{1} equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 3.

$$3a_{1}+9x=84,3a_{1}+3\times 8x=3\times 73$$

Simplify.

$$3a_{1}+9x=84,3a_{1}+24x=219$$

Subtract 3a_{1}+24x=219 from 3a_{1}+9x=84 by subtracting like terms on each side of the equal sign.

$$3a_{1}-3a_{1}+9x-24x=84-219$$

Add 3a_{1} to -3a_{1}. Terms 3a_{1} and -3a_{1} cancel out, leaving an equation with only one variable that can be solved.

$$9x-24x=84-219$$

Add 9x to -24x.

$$-15x=84-219$$

Add 84 to -219.

$$-15x=-135$$

Divide both sides by -15.

$$x=9$$

Substitute 9 for x in a_{1}+8x=73. Because the resulting equation contains only one variable, you can solve for a_{1} directly.

$$a_{1}+8\times 9=73$$

Multiply 8 times 9.

$$a_{1}+72=73$$

Subtract 72 from both sides of the equation.

$$a_{1}=1$$

The system is now solved.

$$a_{1}=1,x=9$$