Solve {l}{2x+20y=0}{14x+91y=98}

$$\left. \begin{array} { l } { 2 x + 20 y = 0 } \\ { 14 x + 91 y = 98 } \end{array} \right.$$

Solve for x, y

$$x=20$$

$$y=-2$$

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.

$$2x+20y=0,14x+91y=98$$

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

$$2x+20y=0$$

Subtract 20y from both sides of the equation.

$$2x=-20y$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-20\right)y$$

Multiply \frac{1}{2} times -20y.

$$x=-10y$$

Substitute -10y for x in the other equation, 14x+91y=98.

$$14\left(-10\right)y+91y=98$$

Multiply 14 times -10y.

$$-140y+91y=98$$

Add -140y to 91y.

$$-49y=98$$

Divide both sides by -49.

$$y=-2$$

Substitute -2 for y in x=-10y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-10\left(-2\right)$$

Multiply -10 times -2.

$$x=20$$

The system is now solved.

$$x=20,y=-2$$

Steps Using Matrices

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

$$2x+20y=0,14x+91y=98$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&20\\14&91\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\98\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&20\\14&91\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&20\\14&91\end{matrix}\right))\left(\begin{matrix}2&20\\14&91\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&20\\14&91\end{matrix}\right))\left(\begin{matrix}0\\98\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&20\\14&91\end{matrix}\right))\left(\begin{matrix}0\\98\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&20\\14&91\end{matrix}\right))\left(\begin{matrix}0\\98\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{91}{2\times 91-20\times 14}&-\frac{20}{2\times 91-20\times 14}\\-\frac{14}{2\times 91-20\times 14}&\frac{2}{2\times 91-20\times 14}\end{matrix}\right)\left(\begin{matrix}0\\98\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{14}&\frac{10}{49}\\\frac{1}{7}&-\frac{1}{49}\end{matrix}\right)\left(\begin{matrix}0\\98\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{49}\times 98\\-\frac{1}{49}\times 98\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\-2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=20,y=-2$$

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.

$$2x+20y=0,14x+91y=98$$

To make 2x and 14x equal, multiply all terms on each side of the first equation by 14 and all terms on each side of the second by 2.

$$14\times 2x+14\times 20y=0,2\times 14x+2\times 91y=2\times 98$$

Simplify.

$$28x+280y=0,28x+182y=196$$

Subtract 28x+182y=196 from 28x+280y=0 by subtracting like terms on each side of the equal sign.

$$28x-28x+280y-182y=-196$$

Add 28x to -28x. Terms 28x and -28x cancel out, leaving an equation with only one variable that can be solved.

$$280y-182y=-196$$

Add 280y to -182y.

$$98y=-196$$

Divide both sides by 98.

$$y=-2$$

Substitute -2 for y in 14x+91y=98. Because the resulting equation contains only one variable, you can solve for x directly.

$$14x+91\left(-2\right)=98$$

Multiply 91 times -2.

$$14x-182=98$$

Add 182 to both sides of the equation.

$$14x=280$$

Divide both sides by 14.

$$x=20$$

The system is now solved.

$$x=20,y=-2$$