Solve {l}{55x+70y=9000}{60x+90y=10500}

$$\left\{ \begin{array} { l } { 55 x + 70 y = 9000 } \\ { 60 x + 90 y = 10500 } \end{array} \right.$$

Solve for x, y

$$x=100$$

$$y=50$$

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.

$$55x+70y=9000,60x+90y=10500$$

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

$$55x+70y=9000$$

Subtract 70y from both sides of the equation.

$$55x=-70y+9000$$

Divide both sides by 55.

$$x=\frac{1}{55}\left(-70y+9000\right)$$

Multiply \frac{1}{55} times -70y+9000.

$$x=-\frac{14}{11}y+\frac{1800}{11}$$

Substitute \frac{-14y+1800}{11} for x in the other equation, 60x+90y=10500.

$$60\left(-\frac{14}{11}y+\frac{1800}{11}\right)+90y=10500$$

Multiply 60 times \frac{-14y+1800}{11}.

$$-\frac{840}{11}y+\frac{108000}{11}+90y=10500$$

Add -\frac{840y}{11} to 90y.

$$\frac{150}{11}y+\frac{108000}{11}=10500$$

Subtract \frac{108000}{11} from both sides of the equation.

$$\frac{150}{11}y=\frac{7500}{11}$$

Divide both sides of the equation by \frac{150}{11}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=50$$

Substitute 50 for y in x=-\frac{14}{11}y+\frac{1800}{11}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{14}{11}\times 50+\frac{1800}{11}$$

Multiply -\frac{14}{11} times 50.

$$x=\frac{-700+1800}{11}$$

Add \frac{1800}{11} to -\frac{700}{11} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=100$$

The system is now solved.

$$x=100,y=50$$

Steps Using Matrices

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

$$55x+70y=9000,60x+90y=10500$$

Write the equations in matrix form.

$$\left(\begin{matrix}55&70\\60&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9000\\10500\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}55&70\\60&90\end{matrix}\right).

$$inverse(\left(\begin{matrix}55&70\\60&90\end{matrix}\right))\left(\begin{matrix}55&70\\60&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}55&70\\60&90\end{matrix}\right))\left(\begin{matrix}9000\\10500\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}55&70\\60&90\end{matrix}\right))\left(\begin{matrix}9000\\10500\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}55&70\\60&90\end{matrix}\right))\left(\begin{matrix}9000\\10500\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{90}{55\times 90-70\times 60}&-\frac{70}{55\times 90-70\times 60}\\-\frac{60}{55\times 90-70\times 60}&\frac{55}{55\times 90-70\times 60}\end{matrix}\right)\left(\begin{matrix}9000\\10500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{25}&-\frac{7}{75}\\-\frac{2}{25}&\frac{11}{150}\end{matrix}\right)\left(\begin{matrix}9000\\10500\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{25}\times 9000-\frac{7}{75}\times 10500\\-\frac{2}{25}\times 9000+\frac{11}{150}\times 10500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\50\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=100,y=50$$

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.

$$55x+70y=9000,60x+90y=10500$$

To make 55x and 60x equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 55.

$$60\times 55x+60\times 70y=60\times 9000,55\times 60x+55\times 90y=55\times 10500$$

Simplify.

$$3300x+4200y=540000,3300x+4950y=577500$$

Subtract 3300x+4950y=577500 from 3300x+4200y=540000 by subtracting like terms on each side of the equal sign.

$$3300x-3300x+4200y-4950y=540000-577500$$

Add 3300x to -3300x. Terms 3300x and -3300x cancel out, leaving an equation with only one variable that can be solved.

$$4200y-4950y=540000-577500$$

Add 4200y to -4950y.

$$-750y=540000-577500$$

Add 540000 to -577500.

$$-750y=-37500$$

Divide both sides by -750.

$$y=50$$

Substitute 50 for y in 60x+90y=10500. Because the resulting equation contains only one variable, you can solve for x directly.

$$60x+90\times 50=10500$$

Multiply 90 times 50.

$$60x+4500=10500$$

Subtract 4500 from both sides of the equation.

$$60x=6000$$

Divide both sides by 60.

$$x=100$$

The system is now solved.

$$x=100,y=50$$