Solve {l}{9x+10y=1810}{11x+8y=1790}

$$\left\{ \begin{array} { l } { 9 x + 10 y = 1810 } \\ { 11 x + 8 y = 1790 } \end{array} \right.$$

Solve for x, y

$$x=90$$

$$y=100$$

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.

$$9x+10y=1810,11x+8y=1790$$

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

$$9x+10y=1810$$

Subtract 10y from both sides of the equation.

$$9x=-10y+1810$$

Divide both sides by 9.

$$x=\frac{1}{9}\left(-10y+1810\right)$$

Multiply \frac{1}{9} times -10y+1810.

$$x=-\frac{10}{9}y+\frac{1810}{9}$$

Substitute \frac{-10y+1810}{9} for x in the other equation, 11x+8y=1790.

$$11\left(-\frac{10}{9}y+\frac{1810}{9}\right)+8y=1790$$

Multiply 11 times \frac{-10y+1810}{9}.

$$-\frac{110}{9}y+\frac{19910}{9}+8y=1790$$

Add -\frac{110y}{9} to 8y.

$$-\frac{38}{9}y+\frac{19910}{9}=1790$$

Subtract \frac{19910}{9} from both sides of the equation.

$$-\frac{38}{9}y=-\frac{3800}{9}$$

Divide both sides of the equation by -\frac{38}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=100$$

Substitute 100 for y in x=-\frac{10}{9}y+\frac{1810}{9}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{10}{9}\times 100+\frac{1810}{9}$$

Multiply -\frac{10}{9} times 100.

$$x=\frac{-1000+1810}{9}$$

Add \frac{1810}{9} to -\frac{1000}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=90$$

The system is now solved.

$$x=90,y=100$$

Steps Using Matrices

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

$$9x+10y=1810,11x+8y=1790$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&10\\11&8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1810\\1790\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&10\\11&8\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&10\\11&8\end{matrix}\right))\left(\begin{matrix}9&10\\11&8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&10\\11&8\end{matrix}\right))\left(\begin{matrix}1810\\1790\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&10\\11&8\end{matrix}\right))\left(\begin{matrix}1810\\1790\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&10\\11&8\end{matrix}\right))\left(\begin{matrix}1810\\1790\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{9\times 8-10\times 11}&-\frac{10}{9\times 8-10\times 11}\\-\frac{11}{9\times 8-10\times 11}&\frac{9}{9\times 8-10\times 11}\end{matrix}\right)\left(\begin{matrix}1810\\1790\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{19}&\frac{5}{19}\\\frac{11}{38}&-\frac{9}{38}\end{matrix}\right)\left(\begin{matrix}1810\\1790\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{19}\times 1810+\frac{5}{19}\times 1790\\\frac{11}{38}\times 1810-\frac{9}{38}\times 1790\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}90\\100\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=90,y=100$$

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.

$$9x+10y=1810,11x+8y=1790$$

To make 9x and 11x equal, multiply all terms on each side of the first equation by 11 and all terms on each side of the second by 9.

$$11\times 9x+11\times 10y=11\times 1810,9\times 11x+9\times 8y=9\times 1790$$

Simplify.

$$99x+110y=19910,99x+72y=16110$$

Subtract 99x+72y=16110 from 99x+110y=19910 by subtracting like terms on each side of the equal sign.

$$99x-99x+110y-72y=19910-16110$$

Add 99x to -99x. Terms 99x and -99x cancel out, leaving an equation with only one variable that can be solved.

$$110y-72y=19910-16110$$

Add 110y to -72y.

$$38y=19910-16110$$

Add 19910 to -16110.

$$38y=3800$$

Divide both sides by 38.

$$y=100$$

Substitute 100 for y in 11x+8y=1790. Because the resulting equation contains only one variable, you can solve for x directly.

$$11x+8\times 100=1790$$

Multiply 8 times 100.

$$11x+800=1790$$

Subtract 800 from both sides of the equation.

$$11x=990$$

Divide both sides by 11.

$$x=90$$

The system is now solved.

$$x=90,y=100$$