Solve {l}{a+x=2400}{15a+10x=16002}

$$\left. \begin{array} { l } { a + x = 2400 } \\ { 15 a + 10 x = 16002 } \end{array} \right.$$

Solve for a, x

$$x = \frac{19998}{5} = 3999\frac{3}{5} = 3999.6$$

$$a = -\frac{7998}{5} = -1599\frac{3}{5} = -1599.6$$

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.

$$a+x=2400,15a+10x=16002$$

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

$$a+x=2400$$

Subtract x from both sides of the equation.

$$a=-x+2400$$

Substitute -x+2400 for a in the other equation, 15a+10x=16002.

$$15\left(-x+2400\right)+10x=16002$$

Multiply 15 times -x+2400.

$$-15x+36000+10x=16002$$

Add -15x to 10x.

$$-5x+36000=16002$$

Subtract 36000 from both sides of the equation.

$$-5x=-19998$$

Divide both sides by -5.

$$x=\frac{19998}{5}$$

Substitute \frac{19998}{5} for x in a=-x+2400. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{19998}{5}+2400$$

Add 2400 to -\frac{19998}{5}.

$$a=-\frac{7998}{5}$$

The system is now solved.

$$a=-\frac{7998}{5},x=\frac{19998}{5}$$

Steps Using Matrices

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

$$a+x=2400,15a+10x=16002$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\15&10\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}2400\\16002\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\15&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\15&10\end{matrix}\right))\left(\begin{matrix}1&1\\15&10\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&10\end{matrix}\right))\left(\begin{matrix}2400\\16002\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&10\end{matrix}\right))\left(\begin{matrix}2400\\16002\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\15&10\end{matrix}\right))\left(\begin{matrix}2400\\16002\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}\frac{10}{10-15}&-\frac{1}{10-15}\\-\frac{15}{10-15}&\frac{1}{10-15}\end{matrix}\right)\left(\begin{matrix}2400\\16002\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}-2&\frac{1}{5}\\3&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}2400\\16002\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}-2\times 2400+\frac{1}{5}\times 16002\\3\times 2400-\frac{1}{5}\times 16002\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{7998}{5}\\\frac{19998}{5}\end{matrix}\right)$$

Extract the matrix elements a and x.

$$a=-\frac{7998}{5},x=\frac{19998}{5}$$

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.

$$a+x=2400,15a+10x=16002$$

To make a and 15a equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 1.

$$15a+15x=15\times 2400,15a+10x=16002$$

Simplify.

$$15a+15x=36000,15a+10x=16002$$

Subtract 15a+10x=16002 from 15a+15x=36000 by subtracting like terms on each side of the equal sign.

$$15a-15a+15x-10x=36000-16002$$

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

$$15x-10x=36000-16002$$

Add 15x to -10x.

$$5x=36000-16002$$

Add 36000 to -16002.

$$5x=19998$$

Divide both sides by 5.

$$x=\frac{19998}{5}$$

Substitute \frac{19998}{5} for x in 15a+10x=16002. Because the resulting equation contains only one variable, you can solve for a directly.

$$15a+10\times \frac{19998}{5}=16002$$

Multiply 10 times \frac{19998}{5}.

$$15a+39996=16002$$

Subtract 39996 from both sides of the equation.

$$15a=-23994$$

Divide both sides by 15.

$$a=-\frac{7998}{5}$$

The system is now solved.

$$a=-\frac{7998}{5},x=\frac{19998}{5}$$