Solve {l}{500y+150.25x=0}{2990y+225.75x=0}

$$\left. \begin{array} { l } { 500 y + 150.25 x = 0 } \\ { 2990 y + 225.75 x = 0 } \end{array} \right.$$

Solve for y, x

$$x=0$$

$$y=0$$

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.

$$500y+150.25x=0,2990y+225.75x=0$$

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

$$500y+150.25x=0$$

Subtract \frac{601x}{4} from both sides of the equation.

$$500y=-150.25x$$

Divide both sides by 500.

$$y=\frac{1}{500}\left(-150.25\right)x$$

Multiply \frac{1}{500} times -\frac{601x}{4}.

$$y=-\frac{601}{2000}x$$

Substitute -\frac{601x}{2000} for y in the other equation, 2990y+225.75x=0.

$$2990\left(-\frac{601}{2000}\right)x+225.75x=0$$

Multiply 2990 times -\frac{601x}{2000}.

$$-\frac{179699}{200}x+225.75x=0$$

Add -\frac{179699x}{200} to \frac{903x}{4}.

$$-\frac{134549}{200}x=0$$

Divide both sides of the equation by -\frac{134549}{200}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=0$$

Substitute 0 for x in y=-\frac{601}{2000}x. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=0$$

The system is now solved.

$$y=0,x=0$$

Steps Using Matrices

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

$$500y+150.25x=0,2990y+225.75x=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right).

$$inverse(\left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right))\left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}500&150.25\\2990&225.75\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{225.75}{500\times 225.75-150.25\times 2990}&-\frac{150.25}{500\times 225.75-150.25\times 2990}\\-\frac{2990}{500\times 225.75-150.25\times 2990}&\frac{500}{500\times 225.75-150.25\times 2990}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{903}{1345490}&\frac{601}{1345490}\\\frac{1196}{134549}&-\frac{200}{134549}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=0,x=0$$

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.

$$500y+150.25x=0,2990y+225.75x=0$$

To make 500y and 2990y equal, multiply all terms on each side of the first equation by 2990 and all terms on each side of the second by 500.

$$2990\times 500y+2990\times 150.25x=0,500\times 2990y+500\times 225.75x=0$$

Simplify.

$$1495000y+449247.5x=0,1495000y+112875x=0$$

Subtract 1495000y+112875x=0 from 1495000y+449247.5x=0 by subtracting like terms on each side of the equal sign.

$$1495000y-1495000y+449247.5x-112875x=0$$

Add 1495000y to -1495000y. Terms 1495000y and -1495000y cancel out, leaving an equation with only one variable that can be solved.

$$449247.5x-112875x=0$$

Add \frac{898495x}{2} to -112875x.

$$336372.5x=0$$

Divide both sides of the equation by 336372.5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=0$$

Substitute 0 for x in 2990y+225.75x=0. Because the resulting equation contains only one variable, you can solve for y directly.

$$2990y=0$$

Divide both sides by 2990.

$$y=0$$

The system is now solved.

$$y=0,x=0$$