Solve {l}{2x-9y=1}{px+90y=-10}

$$\left\{ \begin{array} { l } { 2 x - 9 y = 1 } \\ { p x + 90 y = - 10 } \end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}\\x=0\text{, }y=-\frac{1}{9}\approx -0.111111111\text{, }&\text{unconditionally}\\x=\frac{9y+1}{2}\text{, }y\in \mathrm{C}\text{, }&p=-20\end{matrix}\right.$$

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.

$$2x-9y=1,px+90y=-10$$

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

$$2x-9y=1$$

Add 9y to both sides of the equation.

$$2x=9y+1$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(9y+1\right)$$

Multiply \frac{1}{2} times 9y+1.

$$x=\frac{9}{2}y+\frac{1}{2}$$

Substitute \frac{9y+1}{2} for x in the other equation, px+90y=-10.

$$p\left(\frac{9}{2}y+\frac{1}{2}\right)+90y=-10$$

Multiply p times \frac{9y+1}{2}.

$$\frac{9p}{2}y+\frac{p}{2}+90y=-10$$

Add \frac{9py}{2} to 90y.

$$\left(\frac{9p}{2}+90\right)y+\frac{p}{2}=-10$$

Subtract \frac{p}{2} from both sides of the equation.

$$\left(\frac{9p}{2}+90\right)y=-\frac{p}{2}-10$$

Divide both sides by 90+\frac{9p}{2}.

$$y=-\frac{1}{9}$$

Substitute -\frac{1}{9} for y in x=\frac{9}{2}y+\frac{1}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{9}{2}\left(-\frac{1}{9}\right)+\frac{1}{2}$$

Multiply \frac{9}{2} times -\frac{1}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-1+1}{2}$$

Add \frac{1}{2} to -\frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=0$$

The system is now solved.

$$x=0,y=-\frac{1}{9}$$

Steps Using Matrices

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

$$2x-9y=1,px+90y=-10$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-9\\p&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\-10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-9\\p&90\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}2&-9\\p&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}1\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}1\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}1\\-10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{90}{2\times 90-\left(-9p\right)}&-\frac{-9}{2\times 90-\left(-9p\right)}\\-\frac{p}{2\times 90-\left(-9p\right)}&\frac{2}{2\times 90-\left(-9p\right)}\end{matrix}\right)\left(\begin{matrix}1\\-10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{p+20}&\frac{1}{p+20}\\-\frac{p}{9\left(p+20\right)}&\frac{2}{9\left(p+20\right)}\end{matrix}\right)\left(\begin{matrix}1\\-10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{p+20}+\frac{1}{p+20}\left(-10\right)\\-\frac{p}{9\left(p+20\right)}+\frac{2}{9\left(p+20\right)}\left(-10\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-\frac{1}{9}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=-\frac{1}{9}$$

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.

$$2x-9y=1,px+90y=-10$$

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

$$p\times 2x+p\left(-9\right)y=p,2px+2\times 90y=2\left(-10\right)$$

Simplify.

$$2px+\left(-9p\right)y=p,2px+180y=-20$$

Subtract 2px+180y=-20 from 2px+\left(-9p\right)y=p by subtracting like terms on each side of the equal sign.

$$2px+\left(-2p\right)x+\left(-9p\right)y-180y=p+20$$

Add 2px to -2px. Terms 2px and -2px cancel out, leaving an equation with only one variable that can be solved.

$$\left(-9p\right)y-180y=p+20$$

Add -9py to -180y.

$$\left(-9p-180\right)y=p+20$$

Divide both sides by -9p-180.

$$y=-\frac{1}{9}$$

Substitute -\frac{1}{9} for y in px+90y=-10. Because the resulting equation contains only one variable, you can solve for x directly.

$$px+90\left(-\frac{1}{9}\right)=-10$$

Multiply 90 times -\frac{1}{9}.

$$px-10=-10$$

Add 10 to both sides of the equation.

$$px=0$$

Divide both sides by p.

$$x=0$$

The system is now solved.

$$x=0,y=-\frac{1}{9}$$

Solve for x, y

$$\left\{\begin{matrix}\\x=0\text{, }y=-\frac{1}{9}\approx -0.111111111\text{, }&\text{unconditionally}\\x=\frac{9y+1}{2}\text{, }y\in \mathrm{R}\text{, }&p=-20\end{matrix}\right.$$

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.

$$2x-9y=1,px+90y=-10$$

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

$$2x-9y=1$$

Add 9y to both sides of the equation.

$$2x=9y+1$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(9y+1\right)$$

Multiply \frac{1}{2} times 9y+1.

$$x=\frac{9}{2}y+\frac{1}{2}$$

Substitute \frac{9y+1}{2} for x in the other equation, px+90y=-10.

$$p\left(\frac{9}{2}y+\frac{1}{2}\right)+90y=-10$$

Multiply p times \frac{9y+1}{2}.

$$\frac{9p}{2}y+\frac{p}{2}+90y=-10$$

Add \frac{9py}{2} to 90y.

$$\left(\frac{9p}{2}+90\right)y+\frac{p}{2}=-10$$

Subtract \frac{p}{2} from both sides of the equation.

$$\left(\frac{9p}{2}+90\right)y=-\frac{p}{2}-10$$

Divide both sides by 90+\frac{9p}{2}.

$$y=-\frac{1}{9}$$

Substitute -\frac{1}{9} for y in x=\frac{9}{2}y+\frac{1}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{9}{2}\left(-\frac{1}{9}\right)+\frac{1}{2}$$

Multiply \frac{9}{2} times -\frac{1}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-1+1}{2}$$

Add \frac{1}{2} to -\frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=0$$

The system is now solved.

$$x=0,y=-\frac{1}{9}$$

Steps Using Matrices

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

$$2x-9y=1,px+90y=-10$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-9\\p&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\-10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-9\\p&90\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}2&-9\\p&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}1\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}1\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-9\\p&90\end{matrix}\right))\left(\begin{matrix}1\\-10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{90}{2\times 90-\left(-9p\right)}&-\frac{-9}{2\times 90-\left(-9p\right)}\\-\frac{p}{2\times 90-\left(-9p\right)}&\frac{2}{2\times 90-\left(-9p\right)}\end{matrix}\right)\left(\begin{matrix}1\\-10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{p+20}&\frac{1}{p+20}\\-\frac{p}{9\left(p+20\right)}&\frac{2}{9\left(p+20\right)}\end{matrix}\right)\left(\begin{matrix}1\\-10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{p+20}+\frac{1}{p+20}\left(-10\right)\\-\frac{p}{9\left(p+20\right)}+\frac{2}{9\left(p+20\right)}\left(-10\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-\frac{1}{9}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=-\frac{1}{9}$$

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.

$$2x-9y=1,px+90y=-10$$

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

$$p\times 2x+p\left(-9\right)y=p,2px+2\times 90y=2\left(-10\right)$$

Simplify.

$$2px+\left(-9p\right)y=p,2px+180y=-20$$

Subtract 2px+180y=-20 from 2px+\left(-9p\right)y=p by subtracting like terms on each side of the equal sign.

$$2px+\left(-2p\right)x+\left(-9p\right)y-180y=p+20$$

Add 2px to -2px. Terms 2px and -2px cancel out, leaving an equation with only one variable that can be solved.

$$\left(-9p\right)y-180y=p+20$$

Add -9py to -180y.

$$\left(-9p-180\right)y=p+20$$

Divide both sides by -9p-180.

$$y=-\frac{1}{9}$$

Substitute -\frac{1}{9} for y in px+90y=-10. Because the resulting equation contains only one variable, you can solve for x directly.

$$px+90\left(-\frac{1}{9}\right)=-10$$

Multiply 90 times -\frac{1}{9}.

$$px-10=-10$$

Add 10 to both sides of the equation.

$$px=0$$

Divide both sides by p.

$$x=0$$

The system is now solved.

$$x=0,y=-\frac{1}{9}$$