Solve {l}{100x=120y}{121x=55y}

$$\left. \begin{array} { l } { 100 x = 120 y } \\ { 121 x = 55 y } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Subtract 120y from both sides.

$$100x-120y=0$$

Consider the second equation. Subtract 55y from both sides.

$$121x-55y=0$$

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.

$$100x-120y=0,121x-55y=0$$

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

$$100x-120y=0$$

Add 120y to both sides of the equation.

$$100x=120y$$

Divide both sides by 100.

$$x=\frac{1}{100}\times 120y$$

Multiply \frac{1}{100} times 120y.

$$x=\frac{6}{5}y$$

Substitute \frac{6y}{5} for x in the other equation, 121x-55y=0.

$$121\times \frac{6}{5}y-55y=0$$

Multiply 121 times \frac{6y}{5}.

$$\frac{726}{5}y-55y=0$$

Add \frac{726y}{5} to -55y.

$$\frac{451}{5}y=0$$

Divide both sides of the equation by \frac{451}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=0$$

Substitute 0 for y in x=\frac{6}{5}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=0$$

The system is now solved.

$$x=0,y=0$$

Steps Using Matrices

Consider the first equation. Subtract 120y from both sides.

$$100x-120y=0$$

Consider the second equation. Subtract 55y from both sides.

$$121x-55y=0$$

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

$$100x-120y=0,121x-55y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}100&-120\\121&-55\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}100&-120\\121&-55\end{matrix}\right).

$$inverse(\left(\begin{matrix}100&-120\\121&-55\end{matrix}\right))\left(\begin{matrix}100&-120\\121&-55\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}100&-120\\121&-55\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}100&-120\\121&-55\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}100&-120\\121&-55\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{55}{100\left(-55\right)-\left(-120\times 121\right)}&-\frac{-120}{100\left(-55\right)-\left(-120\times 121\right)}\\-\frac{121}{100\left(-55\right)-\left(-120\times 121\right)}&\frac{100}{100\left(-55\right)-\left(-120\times 121\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{164}&\frac{6}{451}\\-\frac{11}{820}&\frac{5}{451}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

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

Extract the matrix elements x and y.

$$x=0,y=0$$

Steps Using Elimination

Consider the first equation. Subtract 120y from both sides.

$$100x-120y=0$$

Consider the second equation. Subtract 55y from both sides.

$$121x-55y=0$$

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.

$$100x-120y=0,121x-55y=0$$

To make 100x and 121x equal, multiply all terms on each side of the first equation by 121 and all terms on each side of the second by 100.

$$121\times 100x+121\left(-120\right)y=0,100\times 121x+100\left(-55\right)y=0$$

Simplify.

$$12100x-14520y=0,12100x-5500y=0$$

Subtract 12100x-5500y=0 from 12100x-14520y=0 by subtracting like terms on each side of the equal sign.

$$12100x-12100x-14520y+5500y=0$$

Add 12100x to -12100x. Terms 12100x and -12100x cancel out, leaving an equation with only one variable that can be solved.

$$-14520y+5500y=0$$

Add -14520y to 5500y.

$$-9020y=0$$

Divide both sides by -9020.

$$y=0$$

Substitute 0 for y in 121x-55y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$121x=0$$

Divide both sides by 121.

$$x=0$$

The system is now solved.

$$x=0,y=0$$