Solve {l}{a=200x+400}{a=250x-750}

$$\left. \begin{array} { l } { a = 200 x + 400 } \\ { a = 250 x - 750 } \end{array} \right.$$

Solve for a, x

$$x=23$$

$$a=5000$$

Steps Using Substitution

Consider the first equation. Subtract 200x from both sides.

$$a-200x=400$$

Consider the second equation. Subtract 250x from both sides.

$$a-250x=-750$$

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.

$$a-200x=400,a-250x=-750$$

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

$$a-200x=400$$

Add 200x to both sides of the equation.

$$a=200x+400$$

Substitute 400+200x for a in the other equation, a-250x=-750.

$$200x+400-250x=-750$$

Add 200x to -250x.

$$-50x+400=-750$$

Subtract 400 from both sides of the equation.

$$-50x=-1150$$

Divide both sides by -50.

$$x=23$$

Substitute 23 for x in a=200x+400. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=200\times 23+400$$

Multiply 200 times 23.

$$a=4600+400$$

Add 400 to 4600.

$$a=5000$$

The system is now solved.

$$a=5000,x=23$$

Steps Using Matrices

Consider the first equation. Subtract 200x from both sides.

$$a-200x=400$$

Consider the second equation. Subtract 250x from both sides.

$$a-250x=-750$$

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

$$a-200x=400,a-250x=-750$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-200\\1&-250\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}400\\-750\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-200\\1&-250\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-200\\1&-250\end{matrix}\right))\left(\begin{matrix}1&-200\\1&-250\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-200\\1&-250\end{matrix}\right))\left(\begin{matrix}400\\-750\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-200\\1&-250\end{matrix}\right))\left(\begin{matrix}400\\-750\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&-200\\1&-250\end{matrix}\right))\left(\begin{matrix}400\\-750\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{250}{-250-\left(-200\right)}&-\frac{-200}{-250-\left(-200\right)}\\-\frac{1}{-250-\left(-200\right)}&\frac{1}{-250-\left(-200\right)}\end{matrix}\right)\left(\begin{matrix}400\\-750\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}5&-4\\\frac{1}{50}&-\frac{1}{50}\end{matrix}\right)\left(\begin{matrix}400\\-750\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}5\times 400-4\left(-750\right)\\\frac{1}{50}\times 400-\frac{1}{50}\left(-750\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\x\end{matrix}\right)=\left(\begin{matrix}5000\\23\end{matrix}\right)$$

Extract the matrix elements a and x.

$$a=5000,x=23$$

Steps Using Elimination

Consider the first equation. Subtract 200x from both sides.

$$a-200x=400$$

Consider the second equation. Subtract 250x from both sides.

$$a-250x=-750$$

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.

$$a-200x=400,a-250x=-750$$

Subtract a-250x=-750 from a-200x=400 by subtracting like terms on each side of the equal sign.

$$a-a-200x+250x=400+750$$

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

$$-200x+250x=400+750$$

Add -200x to 250x.

$$50x=400+750$$

Add 400 to 750.

$$50x=1150$$

Divide both sides by 50.

$$x=23$$

Substitute 23 for x in a-250x=-750. Because the resulting equation contains only one variable, you can solve for a directly.

$$a-250\times 23=-750$$

Multiply -250 times 23.

$$a-5750=-750$$

Add 5750 to both sides of the equation.

$$a=5000$$

The system is now solved.

$$a=5000,x=23$$