Solve {l}{8x+20y=64}{35x-20y=65}

$$\left. \begin{array} { l } { 8 x + 20 y = 64 } \\ { 35 x - 20 y = 65 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=2$$

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.

$$8x+20y=64,35x-20y=65$$

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

$$8x+20y=64$$

Subtract 20y from both sides of the equation.

$$8x=-20y+64$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-20y+64\right)$$

Multiply \frac{1}{8} times -20y+64.

$$x=-\frac{5}{2}y+8$$

Substitute -\frac{5y}{2}+8 for x in the other equation, 35x-20y=65.

$$35\left(-\frac{5}{2}y+8\right)-20y=65$$

Multiply 35 times -\frac{5y}{2}+8.

$$-\frac{175}{2}y+280-20y=65$$

Add -\frac{175y}{2} to -20y.

$$-\frac{215}{2}y+280=65$$

Subtract 280 from both sides of the equation.

$$-\frac{215}{2}y=-215$$

Divide both sides of the equation by -\frac{215}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=2$$

Substitute 2 for y in x=-\frac{5}{2}y+8. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{2}\times 2+8$$

Multiply -\frac{5}{2} times 2.

$$x=-5+8$$

Add 8 to -5.

$$x=3$$

The system is now solved.

$$x=3,y=2$$

Steps Using Matrices

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

$$8x+20y=64,35x-20y=65$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&20\\35&-20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}64\\65\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&20\\35&-20\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&20\\35&-20\end{matrix}\right))\left(\begin{matrix}8&20\\35&-20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&20\\35&-20\end{matrix}\right))\left(\begin{matrix}64\\65\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&20\\35&-20\end{matrix}\right))\left(\begin{matrix}64\\65\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&20\\35&-20\end{matrix}\right))\left(\begin{matrix}64\\65\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{20}{8\left(-20\right)-20\times 35}&-\frac{20}{8\left(-20\right)-20\times 35}\\-\frac{35}{8\left(-20\right)-20\times 35}&\frac{8}{8\left(-20\right)-20\times 35}\end{matrix}\right)\left(\begin{matrix}64\\65\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{43}&\frac{1}{43}\\\frac{7}{172}&-\frac{2}{215}\end{matrix}\right)\left(\begin{matrix}64\\65\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{43}\times 64+\frac{1}{43}\times 65\\\frac{7}{172}\times 64-\frac{2}{215}\times 65\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=2$$

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.

$$8x+20y=64,35x-20y=65$$

To make 8x and 35x equal, multiply all terms on each side of the first equation by 35 and all terms on each side of the second by 8.

$$35\times 8x+35\times 20y=35\times 64,8\times 35x+8\left(-20\right)y=8\times 65$$

Simplify.

$$280x+700y=2240,280x-160y=520$$

Subtract 280x-160y=520 from 280x+700y=2240 by subtracting like terms on each side of the equal sign.

$$280x-280x+700y+160y=2240-520$$

Add 280x to -280x. Terms 280x and -280x cancel out, leaving an equation with only one variable that can be solved.

$$700y+160y=2240-520$$

Add 700y to 160y.

$$860y=2240-520$$

Add 2240 to -520.

$$860y=1720$$

Divide both sides by 860.

$$y=2$$

Substitute 2 for y in 35x-20y=65. Because the resulting equation contains only one variable, you can solve for x directly.

$$35x-20\times 2=65$$

Multiply -20 times 2.

$$35x-40=65$$

Add 40 to both sides of the equation.

$$35x=105$$

Divide both sides by 35.

$$x=3$$

The system is now solved.

$$x=3,y=2$$